The tangent

The trajectory is the path a moving object follows through space. Let’s analyse this curve ζ from the geometric point of view. The trajectory curve can be defined by

If we take two consecutive positions,  and , the average speed is the vector carried by the rope  The instantaneous speed on her side is a vector tangent to the trajectory.

The direction of the speed vector defined thus the tangent to the trajectory. We can get rid of the length of the speed vector and keep its direction by dividing it by its intensity. As a result, we obtain a vector of length 1 tangent to the trajectory.

In the Cartesian system, it corresponds to

In this last expression, the time is not explicitly present. If we defined ds as the instantaneous displacement  we can write

The length L of the trajectory is the sum of the lengths of the average displacements and, in the limit, to the integration of the lengths of the instantaneous displacements.

The length of the trajectory s is a parameter intrinsically related to the curve ζ as it doesn’t depend upon the choice of the origin point O, nor of the parameterisation of the curve ζ. It shows, in its infinitesimal version ds, the distance to make on the tangent of the trajectory.

The normal

The vector  being defined, how do we express the geometric aspect related to the acceleration? By construction, the speed vector is tangent to the trajectory of the position vector. The acceleration vector is thus also tangent to the trajectory of the speed vector. How do we represent that in the space?

Let’s derivate the speed vector, giving explicitly its norm V and the direction :

The first term is proportional to  and is thus parallel to the speed vector. The second term is necessarily perpendicular to it. Indeed, the trajectory of  cannot be anywhere else than on a sphere because, by definition, its length does not change. As its derivative is tangent to this trajectory, it is thus tangent to the sphere and thus perpendicular to the radius of the sphere, i.e. to the vector .

We can define the normal vector  as the derivative of the tangent vector.

λ is still to be determined.

• Algebraically

The acceleration  is thus the sum of one tangent part  and one normal part , perpendicular to the first part.

The length of the square of the acceleration is thus given by (the vector product of two perpendicular vectors gives zero)

And then

• Geometrically

Given two consecutive positions of the point P on the trajectory ζ:  and , and the direction of the speed  and  at the corresponding positions. The tangent unit vector , by definition, doesn’t change its length. However, its direction changes: it turns by an angle .

The two tangent vectors delimitate a plane P in which is also their variation . With these three vectors, we can build an isosceles triangle and thus evaluate the length of : 2 sin(/2) and its angle with : π/2-/2.

At the limit, when P₂ tends towards P₁, the length of  becomes  and the angle becomes thus π/2, i.e. becomes perpendicular to . As a result we can write

λ corresponds thus to the angular speed of rotation of the tangent vector. To be more accurate, we see that if we extend the normal in P₁ and in P₂, they meet at the point C, and that this point C is, at the limit of , the centrum of the tangent circle of the curve ζ in P₁ and P₂. The angle  equals the length of the arc ∆s divided by the radius R of the circle. As a result,

The radius R is called the curvature radius of the trajectory ζ at a given point. 1/R is called the curvature of ζ and the limit plane P formed by the tangent vector and the normal is called the osculating plane of the trajectory ζ.

The binormal

As the tangent  and the normal  are defined, we can build a vector perpendicular to those two vectors: the binormal . To do so, we make the vector product:

This vector is perpendicular to the plane P of the trajectory and is oriented so that the three vectors  can be overimposed to the three Cartesian vectors . Knowing the definitions of the speed and of the acceleration, we also can say that

What allow us to determine the equation of the curvature radius R

The dot over a variable means that it is its derivative over the time. Two dots will mean the second derivative, etc. The readability of the equation can be eased this way. In terms of components of the speed and of the acceleration, we have

Frenet formulas

The three vectors obtained this way form the referential of Frenet that follows the point P on its trajectory. The referential is thus intrinsically related to the trajectory. The relation we have seen earlier

becomes, if we use the parameter s of length on the trajectory

What happens to the other vectors of the referential in the movement on the trajectory? As the vectors are orthonormal to each other, we can determine them from scalar products.

As a result, the derivative of the binormal vector is perpendicular to the binormal  and to the tangent  at the same time. It is thus only proportional to the normal . We define the torsion radius To by analogy with the curvature radius R:

Remember that to define the vector  we chose R positive; to define the vector , we used the two vectors  and . As a result, we don’t have the choice for the sign of To, that can be positive or negative.

The derivative of  can be easily calculated:

We can now write down the equations of Frenet:

We can use the fact that the vectors are the scalar product of each other.

In terms of speed , acceleration  and over-acceleration , we have

What leads to

That was the result obtained previously. We also find that

Canonic applications

The uniform circular motion

The point P moves with a constant speed on a circle of radius ρ and of centre O. The angle the point travels by unit of time is thus constant. We define the angular speed ω as

If we choose the x-y plane to place the circle and O as the centre of the Cartesian system, the position of the point P is given by

The speed, acceleration and over-acceleration are

The osculating plane is the one we chose, obviously.

The distance s made on the curve is given by the length of the arc ωρ. Indeed:

The curvature radius R equals ρ by construction, confirmed algebraically:

The torsion 1/To equals zero because the trajectory is in the plane z=0.

We can observe that

• The acceleration is perpendicular to the speed,
• The speed (not the angular speed) is not constant
• The modulus of the speed is constant
• The acceleration is not constant
• The modulus of the acceleration is constant
• The acceleration points towards the centre O (it is centripetal)
• The acceleration is proportional to the radius of the circle (at a fixed ω) but inversely proportional to it at fixed V.

The regular helix

If we turn uniformly in the x-y plane while moving in the z direction with a constant speed, we obtain a movement of coiling around a cylinder. The time required to make one turn of the cylinder is the period τ and the distance made in the direction z over this time is the step H. The equation of the position of the point P is given by

At the exception of the movement on the z axis, this equation is thus the same than the one or the circle, given that ω=2π/τ. We can easily write the equations for the speed, acceleration and over-acceleration:

We note that some of the simple relations we had for the uniform circular motion are no longer valid: . However, the acceleration is still perpendicular to the speed () and the over-acceleration perpendicular to the acceleration (). The acceleration points towards the axis of the cylinder and is proportional to its radius ρ (which is different from the radius of the trajectory). Concerning the curvature radius R and the torsion radius To, we have

The torsion and the curvature are thus constant during the movement.

The Frenet vectors are given by

 The length L of the helix is easy to calculate because the modulus of the speed is constant:

The following figure shows an example where the three Frenet vectors, the position of the centre of curvature C and a few circles of centre C and of radius R in the osculating plane P.

We can visualise the torsion by the rotation of the osculating plane. The normal vector is the one pointing towards the centre of the curvature, the tangent vector is the one tangent to the trajectory and the vector product of these two vectors gives the binormal vector, perpendicular to the circule tangent to the trajectory.

The post Chapter 4: the trajectory appeared first on BORZUYA UNIVERSITY.

Several observations can be made at this point:

• the position  is the addition of the previous position  with the displacement .

• the displacement  equals the sum of the displacements  and .

• it may look obvious, but there is always one position more than there are displacements. As a result, it is impossible to determine the position of the object from its displacements alone. We will need at least one of its positions.

We introduced here the addition operator + on vectors. In the Cartesian system, it is easy to see that the addition and the subtraction operators are simply the mathematical + and – operators applied on the numbers corresponding to the coordinates.

The average speed

Now that the displacements are defined, we can consider that it doesn’t take the same time to make the displacement between P₁ and P₅ directly or with the intermediate steps. The average speed is defined as the displacement by unit of time. At the initial time t₁, the object was at the position P₁ defined by the position vector . At the time t₂, the particle has moved from this position to the position P₂, defined by the position vector . The average speed between these two positions is thus the displacement of the object  divided by the interval of time t₂-t₁.

The speed is said to be average because it depends upon the two chosen positions. One common notation in mathematics is to write differences of two values by the sign Δ. We can thus write

That we can also write, with the explicit dependence over the time:

The instantaneous speed

One concept is now to get rid of the two positions to define an instantaneous speed, associated to a given position and a given time. To do so, we will bring the second point closer from the first and see what happens. We can’t only chose one point because the division by zero is not allowed. However, if this ratio continues to exist when the two points are almost identical, then the instantaneous speed is this value.

It is also the definition of the derivative of the position  with regards to the time, calculated at t₁. The limit towards zero of Δ will be noted d to obtain the notation corresponding to the usual derivative.

To indicate at which time we look for the speed, we use the notation

Each component of the speed is given by the derivative of the corresponding component of the position with regards to the time.

The acceleration

The same way we defined the speed as the variation of the position over time, the acceleration a is the variation of speed over time. We define an average acceleration as

and the instantaneous acceleration as the derivative of the speed vector with regards to the time.

The acceleration can be obtained directly from the position by taking its second derivative. We introduce the notations

Note the difference of position of the square exponent: it is the derivative d/dt that is taken twice, not the object of the derivative. It is more evident if we use the definition of the derivative explicitly:

From acceleration to speed

As the acceleration is a difference of speed, there is thus always one acceleration less than there are speeds. From a speed at a time t₁, on can determine the speed at a time t₂ if we are given the acceleration during the interval of time.

From point to point, we can thus determine the speed:

We can generalise this as

Obviously, the sum of the intervals of times equals the elapsed time:

At the limit of the discrete time jumps, the sum becomes an integration

From the speed to the position

We can apply the same method on the speed to obtain the position:

Let insist on the physical meaning of this:

• the integration∫ is a sum,
• this sum is always made on a product,
• this product always contain an infinitesimal difference (here the time dt) and one finite quantity (here the speed V(t)),
• the sum ∫ is defined between a point of beginning and a point of end.

The uniform rectilinear motion

A uniform rectilinear motion (URM) is obtained when the speed of the object doesn’t depend upon time. The speed is thus constant and is defined at the initial time t₀.

As the speed is constant, the acceleration is thus zero.

The equations for the position in each direction correspond to the equation of a straight line in the three dimensions space.

The uniformly accelerated rectilinear motion

Another simple problem is when the acceleration is constant. Similarly to the URM, here we easily find the speed:

And the position:

The post Chapter 3b: the motion, speed and acceleration appeared first on BORZUYA UNIVERSITY.

In this section, we consider that every object can be considered as a dot without volume.

The position

The motionis the history of the position of an object, the succession of the positions of this object over the time. To define the position of the object, we need a reference system in which we can give the position: coordinates. Several systems of reference exist and there are several ways to calculate the coordinates of objects. All of them are corrects but some are more convenient than the others. During a trip, you won’t give your position with regard to the sun, the same is true in physics.

Several coordinates are generally necessary to determine the exact position of one object. If you say that you are 10km from Paris, you give an information on your position but we are lacking at least one coordinate to determine your position. Usually you need one coordinate by dimension of the system.

One dimension

We choose one origin to the coordinate system, the zero point. It is convenient to choose the initial position of the object A as the origin but it is not mandatory. Next we choose a direction that will be the positive positions. In the opposite direction we have the negative positions. Still for convenience, the positive positions are placed in the direction we guess the object will move towards. Imagine that the object moves towards another object B placed at a distance d. The position vector indicates the distance between an object and the origin, and points towards the object with an arrow. The symbol for vectors is topped by an arrow pointing to the right. The position vector for B is

Where  is the unit vector in the direction x (the single direction in this problem).

Two dimensions

The second coordinate is usually orthogonal, perpendicular to the first coordinate to avoid a maximum of angle problems and to benefit the simplicity of the calculation for right triangles. Each coordinate has a direction.

The position vector is now defined by two components from which we can calculate its length if desired.

The addition sign between the two unit vectors is seen as “followed by” and not like the addition of two usual numbers. Another way to write coordinates is to put them in brackets. If we do this, then we don’t write the unit vectors.

This system of reference, the Cartesian system, is not the single one that can be used at two dimensions to determine the position of an object. We can also position the object from its distance r to the origin point and an angle θ from one axis. This reference system is called the polar system.

It is possible to determine the relation between the coordinates X;Y in the Cartesian system and r;θ in the polar system using the relation defining the cosine, the sinus and the tangent:

The unity vector  can thus be calculated.

or

We can also define the unit vector :

Both polar unity vectors depend thus upon θ that should thus be chosen conscientiously.

Three dimensions

A third coordinate is added in the reference systems. In the Cartesian system, we add a coordinate that is orthogonal to the two previous ones.

In the polar system, we need a second angle to determine the position of an object. We take the first from the axis x in the xy plane and the second angle is taken from the z axis in the zr plane of the object.

Again, the Cartesian system can be associated to the polar system.

The unit vectors are defined as

The post Chapter 3a: the motion- the position appeared first on BORZUYA UNIVERSITY.

On the other hand, we can measure a distance in meters or in miles, the weight in grams or in pounds. All these units have a well-defined value and can serve as reference units. While some countries as the UK or the USA use some different units, international conventions defined an international system of units (SI base units). Their biggest advantage is the simple relation between the units of different quantities. In SI, one millilitre of water occupies one cubic centimetre, weighs one gram, and requires one calorie of energy to heat up by one degree centigrade, which is one percent of the difference between its freezing point and its boiling point. In the American system, you will need to make huge calculations to calculate how much energy it takes to boil a room-temperature gallon of water because you can’t directly relate any of those quantities.

The International Bureau of Weights and Measures (French: Bureau international des poids et mesures) is an intergovernmental organization, established to maintain the International System of Units (SI) under the terms of the Metre Convention (Convention du Mètre, May 20^th 1875). The organisation is usually referred to by its French initialism, BIPM. Its role is to

• establish fundamental standards and scales for the measure of main physical quantities and to conserve the international prototypes;
• compare international standards with national standards;
• ensure the coordination of the corresponding techniques of measurement;
• measure and coordinate the measures of the fundamental, physical constants involved in the above activities.

SI Units

The definitions of the reference units are mainly made to give them a well-known and fixed value.

These definitions fix the speed of light c at 299792458 m/s et the permeability of the void μ₀ at 4π 10^-7 H/m exactly. They also sometimes require some precisions. For instance, we point out that the cesium atom is at rest, that the carbon atoms are not connected, are at rest and in their fundamental state.

Deriving units

By commodity, some units are the combination of the SI units to express frequently used units.

Finally, there are some quantities without specific unit names

We can point out a few specific units:

To put an end to this section, we will list the prefix of multiples of the SI units.

Dimensional analysis

The existence of the IS system means that all the other physical quantities have units that are homogeneous functions of these base units. A function is homogeneous if, making a scale change on all of its variables: x₁ → λ₁x₁, x₂ → λ₂x₂, x₃ → λ₃x₃, … the function itself changes of scale: f(λ₁x₁, λ₂x₂, λ₃x₃,…) = λ₁^α1 λ₂^α2 λ₃^α3…f(x₁, x₂, x₃,…). The units of any quantity, let’s call it Q_r is thus always expressed by a relation like

So if we know the units of one physical quantity, and admitting that there is a relation between this quantity and other variables, and knowing the units of those variables, we can guess the relation between the quantities.

For instance, we observe the swinging of one object attached to one string: a pendulum. The pendulum oscillates because it falls and it is restrained by the string. We want to determine the relation between the times the pendulum takes to make one oscillation, i.e. the period, to the parameters we guess as important: the mass M of the object, the length L of the string, and the gravity acceleration g that affects each object on the planet. The units of the variables are kg for the mass, m for the length, and m/s² for the acceleration. The period is a time and thus its unit is s. The relation between the variables and the period should be something like this:

As there is no kg at the left of the equality, and it is present at the right side of the equation with the exponent a, then we conclude that a=0. Looking at the seconds, their exponent is 1 at the left and -2b at the right, b is thus b=-1/2. Finally, there is no m at the left while it is present at the right side of the equation, thus 0=b+c. As we determined the value of b, we have that c=1/2 and that the global relation is

From our dimensional analysis, we determined that the mass of the object has no influence on the period of the oscillation of the pendulum. Note that we did not write the equal sign: the dimensional analysis doesn’t give the true law; it gives clues on the variables but there can still be numerical factors that can be determined experimentally. On the other hand, the dimensional analysis allows to identify wrong laws not respecting the units of the quantities.

Let’s analyse a second example: the period T of revolution of planets around the Sun. First we identify the important parameters involved in the problem: the mass M of the Sun, the distance R between the planet and the Sun, and a constant G giving the gravitational force. The units of the parameters are respectively kg, m and kg^-1m³s^-2. The period is given in seconds s. The law should be of the form T ~ M^aG^bR^c. For the units, we have the relation

The next step is to identify the exponent of each unit at the left and the right side of the equation:

The law is thus written

This relation is the expression of the Kepler’s law that describes the trajectory of planets of the solar system.

Scales and orders of magnitude

An experimental approach is to estimate the order of magnitude of variables that appear in physical processes. Either we measure the characteristics of one well known property of the matter, or we determine a transition zone between two models of description.

For instance, we can regroup the matter as solids, liquids and gases. One major difference between these three phases is their density, i.e. the mass of the matter for a given volume, given in kg/m³. We can thus regroup liquids as matter with a density with the order of magnitude around 10³kg/m³ (at T=293K, water: 1003kg/m³, olive oil: 910kg/m³, sulfuric acid: 1834kg/m³,…)while gases have a density of order 1 (at T=273K, air: 1.2kg/m³, CO₂: 1.98kg/m³, methane: 0.72kg/m³,…). Between solids and liquids, there is a factor 10 in density (at T=293K: iron: 7893kg/m³, copper: 8954kg/m³, gold: 19320kg/m³,…). We will want to determine the temperature at which a solid becomes liquid, i.e. its melting temperature.

As the interactions between particles of a solid differ from the interactions between particles in a gas, laws are not the same at the microscopic scale than in the astronomic scale, not because the interactions mysteriously disappear, but because we can neglect some interactions. For instance, imagine an interaction between particles that depends directly on the distance between two particles and one interaction that depends on the third power of the distance. If the distance is small, both interactions will have an effect, but as soon as the distance gets large, we can neglect the first interaction.

The post Chapter 2: quantities and units appeared first on BORZUYA UNIVERSITY.

“One of the noblest desire of the man is to know the laws ruling the Universe, and those who contributed to enlighten some of the mysteries were always admired by their peers; they appear as privileged, wearing on them the divine light, a through centuries the generations gaze upon their indelible work and rank them first amongst the glories of the humanity.”

From Achille Cazin (Hachette, 1881).

Physics can be defined as a science that studies the general properties of matter, space, and time, and establishes laws that describe natural phenomena. We will point out the word law first, and ask ourselves if we can clearly, and without ambiguity, define all the objects we are discussing, i.e. space, time and matter. Their definitions are (Oxford dictionaries)

• Space: A continuous area or expanse which is free, available, or unoccupied.
• Matter: Physical substance in general, as distinct from mind and spirit; (in physics) that which occupies space and possesses rest mass, especially as distinct from energy.
• Time: The indefinite continued progress of existence and events in the past, present, and future regarded as a whole.

It is not what we can call clearly defined terms and it doesn’t indicate any of the relations that may exist between them. If we go back in time, Isaac Newton gave its definitions (1687):

• The absolute space, which is without any relation with anything from the surroundings, is always unchanging and immobile. The relative space is any measure or mobile dimension of this space, which is defined with regards to its position with regards to objects that we consider as the immobile space…
• The quantity of matter is the measure that we obtain from its density and its volume…
• The absolute time, true and mathematic, is without relation to anything from the surroundings, and by its nature flows uniformly. The relative time, is any measure, accurate or not, of the duration of an event, that we use in place of the true time. I.e. the hour, day, month,…

It is difficult to avoid cross references and circular definitions. We will abandon the idea to define everything and accept the fact that some notions can make sense without being explicit. Physics is seeking for the laws that rule the reciprocal actions between one object and its surrounding. Enouncing those laws is not an easy job, and we have seen many changes in the history, as the understanding of the scientists evolved. For instance, L. Wouters gives in a school book of 1916 the law for the dilatation of bodies due to heat as

“The first effect that heat produces on bodies is to increase their volume, to dilate them”

This law is based upon its hypothesis on the nature of heat:

“Based on the modern hypothesis on the nature of heat, it results from the vibratory movement of the smallest molecules of the ponderable matter, and is transmitted via a fluid called aether. Aether is a subtle, perfectly elastic, substance that fills the intermolecular spaces as well as the so-called interplanetary voids. Heat is, at the end, a particular state of movement.”

Today, it is obvious that this law is false, mostly because of the notion of aether it is based on. The law itself is true with some exceptions. So, shall we give no definition nor law? Obviously we will, but laws should results from the simple relations obtained from experimentations: if I modify one physical quantity of my experiment, then another physical quantity changes, at this effect is repeated consistently if I repeat the experiment.

To end this introduction, I will extend the citation of Achille Cazin (Hachette, 1881) I opened the introduction with.

“No matter the study on which we work, there are some general rules that one must follow to avoid falling in annoying confusions. …

In physics, we observe all the circumstances around a natural phenomenon; we measure all the available quantities; we seek relations between theses quantities, and these relations are called the law of the phenomenon. When the phenomenon is too complex, and it looks like it is impossible to state one unique law, we modify the phenomenon, we make an experiment. Some circumstances seeming ancillary, are made negligible, and we observe the dominating quantities. From this experiment we obtain an approximate law, and by extension we seek the influence of the neglected circumstances, in which way they alter the law. Doing that, we find the limit law towards which tends the observed law when the circumstances become more and more negligible. Such a law is then used as a fundamental principle, seen as the temporary expression of a physical truth, temporary because one more accurate observation or a new phenomenon can modify the conclusions admitted as truth until now.”

The post Chapter 1: Elementary physics – Introduction appeared first on BORZUYA UNIVERSITY.

The hydroxides and sulphides of calcium, barium, magnesium, potassium and sodium are soluble and belong to the barium-magnesium group.

Description of the elements of the silver group

Iron (Fe³⁺,Fe²⁺)

Iron in its compounds is ordinarily found in the 2+ (ferrous) or the 3+ (ferric) states. The latter is more common, since most ferrous compounds oxidize in the air, particularly if water is present. Iron(II) compounds are usually found as hydrates are light green. Iron(III) salts are also ordinarily obtained as hydrates are often yellow or orange. Both states form many complexes as Fe(CN)₆^4- or Fe(CN)₆^3-. Cyanide complexes are perhaps the most stable iron complexes and have characteristic deep red colours.

Metallic iron is a good reducing agent, dissolving readily in 6M HCl with evolution of hydrogen. Conversion of Fe(II) to Fe(II) or the reverse is easily accomplished by common oxidizing agents (air of H₂O₂ in acid) or by reducing agents such as H₂S, Sn²⁺ or I^–.

Cobalt, Co²⁺

Like iron, cobalt can exist in two common oxidation states in aqueous solutions, Co²⁺ and Co³⁺. The 3+ state is stable only in presence of strong coordinating liquids such as NH₃; the general chemistry of cobalt schemes of qualitative analysis is usually that of Co²⁺.

Cobalt(II) salts in water solution are characteristically pink, the colour of the hydrated cobalt ion Co(H₂O)₆²⁺, but the colour is too delicate to be used to characterise the ion. Cobalt(II) forms several complex ions. Cobalt chloride solution turns from pink to blue when heated to boiling because of a change in coordination number: the pink form arises from octahedrally coordinate Co(II), whereas the blue form is tetrahedrally coordinated Co(II). Cobalt sulphide does not dissolve readily in 6M HCl even when heated. Cobalt sulphide does not dissolve readily in 6M HCl even when heated. Co(II) reacts with thiocyanate solutions to form a blue complex, Co(SCN)₄^2-, which is much more stable in ethanol than in water. Addition of KNO₂ to solutions of Co²⁺ produces a characteristic yellow precipitate of K₃Co(NO₂)₆ in which the metal has been oxidized:

Under strongly oxidizing conditions, Co(II) can be converted to Co(III), which has a stability that is enhanced in a complex species like Co(NH₃)₆³⁺ or an insoluble substance such as the yellow cobalt nitrite produced in the reaction above.

Nickel, Ni²⁺

Nickel salts, like those of most of the other members of group III, are typically coloured: hydrates are green, Ni(NH₃)₆²⁺ is blue and many other nickel complexes have characteristic colours, and nickel forms a very characteristic rose red precipitate with dimethylglyoxime (N₂C₄H₈O₂, structure shown below), an organic chelating agent.

Chromium, Cr³⁺

The common oxidation states of chromium are 3+ and 6+. Cr³⁺ forms several complexes, all of which are coloured. Cr(H₂O)₆³⁺ is reddish violet in solution. The chromium-containing precipitate obtained in the group III precipitation scheme is the hydroxide Cr(OH)₃ rather than the sulphide. Chromium(III) hydroxide is amphoteric: it dissolves in excess base to form the green chromite Cr(OH)₄^– and in acid to form the hydrate chromium(III) ion. Chromium(III) can be oxidized to Cr(VI) with several oxidizing agents, such as ClO₃^– in 16M HNO₃, H₂O₂ in 6M NaOH, and ClO^– in 6M NaOH. Chromium is usually identified in the form of a Cr(VI) species in neutral or basic solutions. CrO₄^2- is bright yellow and if acid is added, the orange Cr₂O₇^2- ion is formed.

If H₂O₂ is added to a solution containing the dichromate ion, a transitory blue colour due to CrO₅ is observed. It is the basis for the confirmatory test for chromium. The following structure has been proposed for CrO₅.

The oxidation number in CrO₅ can be calculated to be 10. However, it does not mean that 10 electrons are involved in the bonding.

Aluminium, Al³⁺

In aqueous chemistry, the only important oxidation state of aluminium is the 3+ state. All its simple salts are colourless (Al is not a member of the transition series of elements). Most aluminium compounds are soluble in water (exceptions: its oxide and hydroxide). The hydrated ion Al(H₂O)₆³⁺ is colourless too. Aqueous solutions of aluminium salts are extensively hydrolysed and are acidic.

In extreme cases, hydrolysis can be sufficiently extensive to yield cloudy solutions containing hydrated hydroxides. The suspension can be reversed by making the solution acidic, reversing the chemical equation.

Manganese, Mn²⁺

The aqueous chemistry of manganese incorporates several common oxidation states: Mn(II), Mn(IV) and Mn(VII). Aqueous solutions of Mn²⁺ are coloured pink. MnO₂ is insoluble in water and KMnO₄ is deeply purple. Mn(II) can be oxidized to Mn(IV) in basic solution but with stronger oxidizing agents in acid solution, oxidation to Mn(VII) can occur.

We don’t make confirmatory tests based on the pink colour of Mn(II) but well on the characteristic colour of the permanganate ion.

Zinc, Zn²⁺

Compounds containing the Zn²⁺ ion are usually colourless and many are soluble in water. The ion complexes to for the colourless Zn(H₂O)₄²⁺ which can be hydrolysed following the equation:

Other complexes can be formed with NH₃ or OH^–. Zinc hydroxide is amphoteric, dissolving in both acid and base.

Precipitation and analysis of the aluminium-nickel group

The conditions for the precipitation of the aluminium-nickel group of ions must be carefully maintained to keep ions from the following groups from precipitating prematurely. The most problematic ion is Mg²⁺ as Mg(OH)₂. We avoid its precipitation by manipulating the equilibrium associated with NH₄Cl. NH₃ is a weak base that ionizes to give enough OH^– to form a precipitate with Mg²⁺ if it is present in a solution. Ammonium chloride is a salt and hence a strong electrolyte. Consequently, NH₄Cl provides NH₄⁺ ions to keep the equilibrium

to the left. There won’t be enough OH^– ions to form a precipitate of Mg(OH)₂. Ammonium chloride also prevents the precipitation of Mn(OH)₂ in exactly the same way, but MnS is quite insoluble so that manganese precipitates as MnS when (NH₄)₂S is added, while MgS is soluble and does not precipitate.

The hydroxides of aluminium, chromium and iron(III) are so insoluble that they precipitate even in the presence of the low concentration of OH^– even in presence of NH₄Cl. The hydroxides of cobalt, nickel and zinc are likewise so insoluble that they also will precipitate but they redissolve in excess NH₄OH. The behaviour of iron(II) is complicated in aqueous solution because it is readily oxidized by air to Fe(III).

A solution containing Fe(OH)₂ will initially show a green precipitate that gradually darkens as more Fe(OH)₂ is oxidized to Fe(OH)₃, brownish red.

The colour of the precipitate can thus give precious information about the presence or absence of certain ions.

If NH₄Cl and aqueous NH₃ give no precipitate, aluminium, chromium and iron(II and III) are definitely absent. In interpreting colours, we must recognize that a dark colour will obscure or cover up a lighter colour. Fe(OH)₃ can thus cover up Cr(OH)₃ or Al(OH₃).

If the addition of (NH₄)₂S gives no further precipitation, cobalt, nickel, manganese and zinc ions are absent. A black precipitate does not give a good clue on which ions are present but a peach or white precipitate indicate respectively the presence of manganese and of zinc.

The hydroxide of iron is converted to the black sulphide.

The interest of the separate additions of NH₄Cl and (NH₄)₂S is the observation of the colours of the precipitates.

The behaviours of zinc, nickel and cobalt ions when treated with aqueous NH₃ is the same as that already noted for copper and cadmium ions. The hydroxides of the three metals are first formed. They are dissolved by excess NH₄OH to form Zn(NH₃)₄²⁺, Ni(NH₃)₆²⁺ and Co(NH₃)₆³⁺, respectively colourless, pale violet blue and deep amber. Note that cobalt(II) has been oxidized to cobalt(III) in the process: in alkaline solutions, cobalt(II) is a fairly strong reducing agent and is slowly oxidized to cobalt(III) by atmospheric oxygen.

Ammonium sulphide reacts with the Zn(NH₃)₆²⁺, Ni(NH₃)₆²⁺ and Co(NH₃)₆³⁺ complexes to form the corresponding sulphide ZnS, NiS and CoS.

Also, MnS and FeS are formed by direct combination of the two ions.

Ferrous sulphide may also be formed by reduction of Fe(III) as described by (*).

If (NH₄)₂S is added to a cold neutral solution containing aluminium, chromium and iron(III) ions, Al(OH)₃, Cr(OH)₃ and Fe₂S₃ will be precipitated.

At higher temperatures, Fe₂S₃ is hydrolysed, giving a brownish red precipitate.

and FeS is formed (reaction (*)).

Procedure 15

The decantate from Procedure 5 is placed in a casserole and is evaporated down to a volume of 8 to 10 drops. Transfer to a test tube, centrifuge and decant into another test tube, discarding the precipitate. To the decantate, add 4 drops of 2M NH₄Cl, mix thoroughly, and then add 15M aqueous NH₃ drop by drop and with constant stirring, until the solution is just alkaline. Then add one more drop and 20 drops of hot water. Mix thoroughly, centrifuge but do not decant. Note the colour of the precipitate and of the supernatant liquid. Add 8 or 9 drops of ammonium sulphide and mix again.

Heat the tube carefully for one or two minutes in the boiling water bath, taking care to avoid any overflow. Centrifuge and test for complete precipitation with a drop of ammonium sulphide. When we are sure that the precipitation is complete, wash down the sides of the tube with a few drops of hot water, centrifuge and note the colours of the precipitate and of the supernatant liquid, then decant, saving the decantate for Procedure 2’. Wash the precipitate 3 times with 20-drops portions of a hot solution prepared by mixing equal volumes of water and 1M NH₄C₂H₃O₂ and analyse according to Procedure 16.

Notes

1.  The decantate from the copper-arsenic group is boiled down before the aluminium-nickel group is precipitated in order to drive off all H₂S and to precipitate any sulphide of the copper-arsenic group that may have gone in the decantate. If the H₂S were not removed, the sulphides and hydroxides of the aluminium-nickel group would all precipitate together when aqueous NH₃ was added. As a result, the desirable valuable observation of colours of precipitates to which we referred earlier could not be made.

2. If phosphate, borates, fluorides, oxalates, silicates or tartrates are present in the solution and that the solution also contains calcium, barium, or magnesium ions as well as cations of the aluminium-nickel group, a special series of procedures must be used instead of the ones presented in this section. However, the combination of ions just mentioned is rare.

3.  NH₄Cl is a strong electrolyte that helps coagulate any hydroxides or sulphides, thereby preventing them from becoming colloidal. Similarly, the addition of NH₄C₂H₃O₂ to the washing solution prevents peptization of the precipitate being washed.

4.  NH₃ is added drop by drop and the pH is verified because a large excess of NH₄OH tends to cause dispersion of the precipitated sulphides and hydroxides. If it is difficult for the precipitates to settle down, the mixture should be boiled for one minute before centrifuging.

Separation of the aluminium subgroup from the nickel group

The hydroxides of aluminium, chromium and zinc are amphoteric and are, therefore soluble in NaOH. In contrast, the hydroxides of iron, manganese, cobalt, and nickel are not amphoteric and therefore not soluble in NaOH.

Of the species precipitated as the aluminium-nickel group of ions, Al(OH)₃, Fe(OH)₃, Cr(OH)₃, MnS, FeS and ZnS are readily soluble in HCl while NiS and CoS are not. However they are soluble in aqua regia (HCl+HNO₃).

Aqua regia oxidises Fe(II) to Fe(III).

When NaOH is added to a solution containing the members of the aluminium-nickel group, the hydroxides of all seven metals are first precipitated. The hydroxides of aluminium, chromium, and zinc are amphoteric and dissolve in an excess of NaOH to form the complex ions Al(OH)₄^–, Cr(OH)₄^–, and Zn(OH)₄^2-, respectively. The hydroxides of iron, manganese, cobalt, and nickel are not amphoteric and do not dissolve in excess NaOH.

A careful observation of colours of solutions and precipitates formed in Procedure 16 may give information regarding the cations present.

The precipitate formed when NaOH is added to Co²⁺ may vary in colour from blue to pink to tan to light brown. Co(OH)₂ exists in two forms, one blue (believed to be more dispersed) and one pink. It changes to the coarser pink form on standing and heat speeds up the process. Air oxidizes Co(OH)₂ to black Co(OH)₃. A pink solution of CoCl₂ turns blue when heated but regains its pink colour when cooled. It also turns blue when treated with concentrated HCl and the pink colour is restored by dilution with water. These colour changes are due to changes in the composition and structure of the complex ions present in the solution. Pink is due to Co(H₂O)₆²⁺ and blue to Co(H₂O)Cl₃^– and CoCl₄^2- ions. The colours of all the solution given above are probably due to similar complex ions.Pure, freshly precipitated Mn(OH)₂ is tan but it turns brown in contact with air when MnO₂ oxidises into Mn₂O₃.

The fact that nickel and cobalt do not precipitate in the copper-arsenic group indicates that NiS and CoS are soluble in dilute HCl. In the analysis of the aluminium-nickel group, however, CoS and NiS do not dissolve in concentrated HCl. This contradictory behaviour is explained as follows: freshly precipitated NiS and CoS exist in a form readily dissolved by dilute HCl. On standing, this soluble modification changes into a second form that is insoluble in both dilute and concentrated HCl.

Separation of the aluminium subgroup from the nickel subgroup

Procedure 16

Treat the precipitate from Procedure 15 with 10 drops of 12M HCl, mix thoroughly, transfer to a casserole, and boil gently for 30s. If the precipitate is not completely dissolved, add 3 drops of 16M HNO₃, mix thoroughly, and boil until a clear solution is obtained. Add 10 drops of cold water, transfer to a test tube, centrifuge to remove any precipitate of sulphur, and decant into a casserole. Note the colour of the solution. Make the solution strongly alkaline with 8M NaOH and mix thoroughly. If the quantity of precipitate is so large that the product is mushy or non-fluid, add 10 to 20 drops of water. Note the colour of the solution and of the precipitate. Then add 2 drops of 3% H₂O₂, stir for one minute and then boil for two minutes, replenishing the water lost. Transfer to a test tube before the precipitate has had a chance to settle and centrifuge. Decant, saving the decantate for Procedure 21. Note the colour of the precipitate and of the decantate. Wash the precipitate three times with hot water and analyse according to Procedure 17.

Notes

1.  Hydrogen peroxide is added to oxidize chromate(III) (Cr(OH)₄^–, green) to chromate(VI) (CrO₄^2-, yellow).

If chromium is present, the colour of the solution will thus change. A blackening of the precipitate may indicate the formation of MnO₂ or Co(OH)_3.

2.  Any excess of H₂O₂ present in the solution must be decomposed by boiling; otherwise it will interfere with Procedure 21.

Detection of iron, cobalt, nickel and manganese

Up to this point in the analysis, the general practice has been to separate a cation from all other cations before identifying it. The exceptions were Cu²⁺ and Cd²⁺, which were identified in each other’s presence, and Sb³⁺, which was identified in the presence of Sn⁴⁺.

In the identification of Fe³⁺, Co²⁺, Ni²⁺ and Mn²⁺, separate samples of the same solution will be used in making each confirmatory test; the four ions will not be separated from each other prior to the test. This can be done because in this particular quartet, none of any three interferes seriously with the confirmatory test for the fourth.

Fe(OH)₃ and Ni(OH)₂ dissolve readily in 2M H₂SO₄ to give Fe³⁺ and Ni²⁺, but MnO₂ and Co(OH)₃ are very slowly dissolved by 2M H₂SO₄. Hydrogen peroxide speeds up the solution process by reducing MnO₂ and Co(OH)₃ to MnO and Co(OH)₂. These latter compounds, in which Mn and Co are in their lower oxidation states; are more basic in character and as a result are readily dissolved by H₂SO₄. The net equations for the reduction by H₂O₂ and the subsequent solution by H₂SO₄ are

Note that in basic solutions, H₂O₂ oxidizes Mn(OH)₂ and Co(OH)₂ to MnO₂ and Co(OH)₃ (Procedure 16), what is the opposite that we seek here. In Procedure 17, H₂O₂ functions as a reducing agent and reduces MnO₂ and Co(OH)₃ to Mn²⁺ and Co²⁺, respectively. The role H₂O₂ takes (reducing or oxidizing agent) depends on the relative reducing and oxidizing power of the substance with which it reacts and the relative rates of the reactions involved. In basic solutions, the potential of the half reaction

is only slightly lower than that of the half reaction

but very much higher than that of the half-reaction

Accordingly, the tendency for H₂O₂ to oxidize Mn(OH)₂ to MnO₂ will be very much greater than its tendency to reduce MnO₂ to Mn(OH)₂. This is equivalent to saying that, in basic solution, the reducing power of Mn(II) toward H₂O₂ is much greater than the oxidizing power of Mn(IV) toward H₂O₂. In acid solutions, the situation is essentially reversed.

Although the potential of the first half reaction differs from that of the second by about the same amount it does from that of the third, the rate of the reaction of H₂O₂ involving the former is considerably greater than that of the latter. As a result,  the tendency for H₂O₂ to reduce MnO₂ to Mn²⁺ is greater than its tendency to oxidize Mn²⁺ to MnO₂. The same rationale applies to the behaviour of Co(II) and Co(III).

In an acid solution, H₂O₂ will thus not oxidize Co²⁺ to Co³⁺, but the tendency for Co³⁺ to be reduced to Co²⁺ will be great. In basic solution, the potential of the half reaction

is -0.17v. Accordingly, the tendency for H₂O₂ to oxidize Co(II) to Co(III) will be very much greater than its tendency to reduce Co(III) to Co(II). During the reactions, molecules of O₂ gas are formed.

Procedure 17

To the precipitate from Procedure 16, add 20 drops of 2M H₂SO₄, mix thoroughly, and transfer to a casserole. Boil gently for one minute, add a drop of 3% H₂O₂, and continue boiling for one minute after the precipitate is completely dissolved. Add 10 drops of water, allow to cool, note the colour of the solution, and divide into four approximately equal portions to use to test for iron, cobalt, nickel and manganese.

Test for iron: to one portion, add one or two drops of 0.2M KSCN. A blood red solution, due to Fe(SCN)₆³⁺ proves the presence of iron.

Notes

1.  To test for iron in a solid, proceed as follows: dissolve a small sample of the solid in 10-15 drops of dilute HCl. Divide the solution into three parts. Test one part for iron(III) ions according to Procedure 17. To the second part, add a few drops of potassium hexacyanoferrate(II), K₄Fe(CN)₆. A dark blue precipitate (Prussian blue) proves the presence of Fe³⁺ ions.

To the third part, add a few drops of potassium hexacyanoferrate(III), K₃Fe(CN)₆.  A dark blue precipitate (Turnbull’s blue) proves the presence of Fe²⁺ ions.

Note that the oxidation states are different.

Although the Fe(SCN)₆^3- ion is widely accepted as the predominant species responsible for the red colour in the test for Fe³⁺, there is little doubt that complexes with a smaller number of SCN^–, such as FeSCN²⁺, Fe(SCN)₂⁺, …, are present too. When the concentration of SCN^– is high, as in the case when 0.2M KSCN is added in making the test, the predominant species is likely to be Fe(SCN)₆^3-.

Detection of cobalt

Procedure 18

To the second portion of the solution prepared in Procedure 17, add enough solid NaF to form a saturated solution; mix well by stirring. Then add 10-20 drops of a saturated solution of NH₄SCN in ethyl alcohol. The formation of a blue solution due to Co(SCN)₄^2- proves the presence of Co²⁺.

Notes

1.  The identification of Fe³⁺ and Co²⁺ in Procedures 17 and 18 is unique in that each is identified by the use of the same reagent and each forms the same type of complex ion, Fe(SCN)₆^3- and Co(SCN)₄^2-. The test for iron is carried out in dilute aqueous solution, in which the red complex ion Fe(SCN)₆^3- is very stable; the blue complex ion Co(SCN)₄^2- is unstable in aqueous solution. As a result, cobalt ions do not interfere with the test for iron. The test for cobalt is carried out in alcoholic solution and in the presence of fluoride ions. The blue complex ion Co(SCN)₄^2- is stable in alcoholic solution; the Fe³⁺ ions are tied up in the form of the colourless stable ion FeF₆^3- (or, possibly, FeF²⁺, FeF₂⁺, FeF₄^–, or FeF₃^2-) and cannot, therefore, interfere with the test for cobalt.

2.  The reaction of Co²⁺ with SCN^– is incomplete, due to the instability of the complex ion.

An excess of SCN^– is necessary to give a good test.

Detection of Nickel

Procedure 19

Make the third portion of the solution prepared in Procedure 17 basic with 5M aqueous NH₃. If a precipitate of Fe(OH)₃ or Mn(OH)₂ forms, centrifuge and decant; to the clear decantate, add 2-4 drops of dimethyl-glyoxime, mix thoroughly, and allow to stand for one minute. A strawberry red precipitate (NiC₈H₁₄N₄O₄) proves the presence of nickel.

Notes

1.  The following reaction takes place in the confirmatory test for nickel:

Since the red compound is readily soluble in acids, the reaction must be carried out in alkaline solution. Iron(II) ions react with (CH₃)₂C₂(NOH)₂ to form a soluble red compound; however, no precipitate is formed.

Detection of Manganese

Procedure 20

Dilute the fourth portion with an equal volume of water, add 2 drops of 3M HNO₃, mix thoroughly, then add a few grains of solid sodium bismuthate, mix thoroughly, and allow to stand for 1min. Centrifuge. A pink to reddish purple solution, due to the permanganate ion (MnO₄^–) proves the presence of manganese.

Notes

1.  Bismuth is predominantly metallic in character; its common oxidation number is 3 and its compounds ionize to give Bi³⁺ In the compound NaBiO₃, bismuth has an oxidation number of 5 and has the property of a non-metal in that it is present in the negative ion BiO₃^–. This behaviour of bismuth illustrates the rule that as the oxidation number of an element increases, it becomes more non-metallic in character.

The behaviour of NaBiO₃ illustrates the rule that those compounds of polyvalent elements in which the element exists in one of its higher oxidation states generally can act as oxidizing agents. Other compounds (oxidizing agents) that illustrate this rule are KMNO₄ and K₂Cr₂O₇.

2.  The oxidation of Mn²⁺ to MnO₄^– by sodium bismuthate (NaBiO₃) takes place as follows:

Reducing agents of any kind, such as chlorides and sulphides, will interfere with the confirmatory test for manganese because they reduce the violet MnO₄^– to the practically colourless Mn²⁺.

Separation and detection of aluminium

Procedure 21

Treat the decantate from Procedure 16 with 16M HNO₃ until slightly acid; then add 15M aqueous NH₃ until distinctly alkaline. Continue stirring the ammoniac solution for 1 min. Centrifuge and decant, saving the decantate for Procedure 22. Because the Al(OH)₃ that precipitates is gelatinous, highly translucent, very finely divided, and the colour of opaque, bluish white glass, its presence, suspended in the solution, is not easy to detect. On centrifuging, however, it will appear in the bottom of the test tube as a whitish, jellylike, opaque precipitate. Wash the precipitate three times with hot water, then dissolve, remove it by centrifuging and decantation. Add 2 drops of aluminon solution, mix thoroughly, make just barely alkaline with 5M aqueous NH₃, again mix thoroughly, and then centrifuge. A cherry red precipitate proves the presence of aluminium.

Notes

1.  It is imperative that all chromate(III) ions (Cr(OH)₄^–) be completely oxidized to chromate(VI) ions (CrO₄^2-) in Procedure 16. Any unoxidized Cr(OH)₄^– will react as follows:

This reaction of chromate(III) ions is the same as that which the aluminate ion (Al(OH)₄^–) will undergo in the course of the separation.

It is thus evident that if Cr(OH)₄^– is not oxidized to CrO₄^2-, Cr(OH)₃ will precipitate when the test for aluminium is made. It is obvious, therefore, that oxidation of Cr(OH)₄^– to CrO₄^2- by H₂O₂ is necessary if a separation of chromium from aluminium is to be accomplished. Although Cr(OH)₃ is green, in small amounts the colour is not marked, and the Cr(OH)₃ precipitate may be mistaken for Al(OH)₃.

Cr(OH)₃ does not form a red lake with aluminon. Excessive amounts of Cr(OH)₃ (green) will, however, mask the red aluminon. Therefore, if much Ce(OH)₃ precipitates with the Al(OH)₃, it should be reoxidized to CrO_42- with H₂O₂.

2.  When HNO₃ is added in the first step in Procedure 21, a precipitate sometimes forms when the solution is just about neutral and then redissolves when more HNO₃ is added. This precipitate may be either Al(OH)₃ or Zn(OH)₂, both of which are white; in case chromium is not completely oxidized to CrO₄^2-, Cr(OH)₃ (green) mau precipitate. Two reactions account for the formation and disappearance of this precipitate.

a) When HNO₃ is added to neutrality,

b) When more HNO₃ is added,

Excess H₂O₂, if present in the decantate from Procedure 16, will interfere with the separation and identification of aluminium and chromium. When HNO₃ is added in the first step in Procedure 21, the following reactions will take place.

The chromate is thus reduced to Cr³⁺, which will then interfere with the test for aluminium is already discussed in Note 1.

3.  When NH₃ + 3H₂O is added in Procedure 21, a white precipitate sometimes forms; it redissolves when more aqueous NH₃ is added. This precipitate is probably Zn(OH)₂. The reactions that account for its formation and redissolution are

4.  Since HCl may reduce chromate to Cr³⁺, the solution is acidified with HNO₃ rather than with HCl.

5.  If lead, tin, and antimony are not completely precipitated in the copper-arsenic group, they appear as insoluble hydroxides or basic salts in the final test for aluminium. The hydroxides, like Al(OH)₃, do not dissolve in excess NH₄ They do not, however form the characteristic red lake with aluminon.

6.  Aluminon is an organic dye. It is the ammonium salt of aurintricarboxylic acid and its chemical formula is

7.  A lake is formed when a coloured compound (a dye) combines with or is adsorbed from a solution by an insoluble gelatinous precipitate. The dye aluminon is preferentially adsorbed by al(OH)₃: this dye is not adsorbed by Cr(OH)₃ or Zn(OH)₂ or by the hydroxides of lead, tin, and antimony.

Separation and detection of chromium and zinc

Procedure 22

Follow (A) is the decantate from Procedure 21 is colourless, follow (B) if it is yellow.

(A)  Decantate; therefore chromium is absent. Place a drop of the decantate on a piece of diphenylthiocarbazone paper and allow to stand for 1 min. A purplish red coloration proves the presence of zinc ions (in the absence of the Zn²⁺ ion, NH₄OH will give a yellowish brown colour). A confirmatory test for Zn²⁺ is given in Note 5.

(B)  Decantate is yellow; chromium is probably present. Add 6-7 drops of 0.2M BaCl₂ to the yellow decantate, mix thoroughly, centrifuge until the supernatant liquid is clear, and decant, saving the decantate for (C). Wash the precipitate, BaCrO₄ mixed with some BaSO₄, twice with hot water, ad 3 drops of 3M HNO₃, heat gently but do not boil vigorously, and stir for about 1 min. Add 10 drops of cold water, mix thoroughly, cool under the cold water tap, and then ad 10 drops of ether and 1 drop of 3% H₂O₂. Mix well by vigorous stirring and allow to settle. A blue coloration of the ether layer due to chromium peroxide (CrO₅) proves the presence of chromium.

(C) Place a drop of the decantate from (B) on a piece of diphenylthiocarbazone paper and allow to stand for about 1 min. A purplish red coloration proves the presence of zinc ions (in absence of Zn²⁺ ions NH₄OH will give a yellowish brown colour). A confirmatory test for Zn²⁺ is given in Note 5.

Notes

1.  The final test for chromium depends on the fact that in dilute acid solution Cr₂O₇^2- interacts with H₂O₂ to form a deep indigo blue compound, chromium peroxide(CrO₅). Yellow chromate is first changed to orange dichromate (Cr₂O₇^2-).

This change of chromate to dichromate, which always take place when the solution is acidified, has already been mentioned in Note 2, Procedure 21.

The dichromate is then converted to CrO₅ by H₂O₂.

The compound CrO₅ is unstable and decomposes on standing to form Cr³⁺. If the concentration of HNO₃ is low, the CrO₅ decomposes very slowly; if, however, the concentration of HNO₃ is high, CrO₅ decomposes so rapidly that the blue colour may not be noticed. Furthermore, CrO₅ is very rapidly reduced to Cr³⁺ by excess H₂O₂. Hence high concentrations of HNO₃ and excess H₂O₂ must be avoided.

CrO₅ is very soluble in ether; HNO₃ is not. Treating with ether partially separates the CrO₅ from HNO₃ and concentrates it in the ether layer. Since separation from HNO₃ will increase the stability of CrO₅, extraction with ether preserves as well as concentrates CrO₅.

2.  The precipitate obtained on addition of BaCl₂ may contain varying amounts of BaSO₄ because of oxidation of H₂S and S to H₂SO₄ in earlier procedures. BaSO₄ is white, whereas BaCrO₄ is light yellow. Very finely divided BaCrO₄ is such a pale yellow, however, that the precipitate may appear white even though it consists largely of BaCrO₄. The confirmatory test for chromium should therefore be completed as directed, even if the precipitate appears white.

BaCrO₄, being the salt of the relatively weak acid H₂CrO₄, is soluble in HNO₃, whereas BaSO₄, being the salt of a strong acid, is insoluble.

3.  Diphenylthiocarbazone paper is also called dithizone paper.

4.  The purplish red colour is believed to be due to the complex ion formed by interaction Zn²⁺ ions with diphenylthiocarbazone molecules, which have the following structure.

5.  Make the decantate from (B) acid with 6M HCl. Then add 3-4 drops of 0.2M K₄Fe(CN)₆ and mix thoroughly. The resulting mixture should be acidic. A greyish white to bluish green precipitate, Zn₃K₂[Fe(CN)₆]₂, proves the presence of zinc.

The post Chapter 10c: The Aluminium-Nickel group appeared first on BORZUYA UNIVERSITY.

The sulphides are formed with the weak acid H₂S. Example with Hg²⁺:

The metal sulphide begins to precipitate when the product of the concentration of the cation and the sulphide ion equals the solubility product for the sulphide and the precipitation continues until it is not the case anymore. It is an equilibrium. The pH of the solution plays a role: the two first steps of the reaction are the deprotonation of H₂S. If we increase the pH of the solution by the addition of HCl, the equilibrium

is shifted to the left and the S^2- concentration decreases. On the contrary, if we dilute the solution but keep it saturated in H₂S, the concentration of the S^2- ion increases. We want thus a small H⁺ concentration to have a large concentration of S^2-. The H⁺ concentration can be decreased by dilution or by addition of the salt of a weak acid (NH₄C₂H₃O₂ for instance, NH₄⁺ replaces H⁺ in the solution with a negligible effect on the pH of the solution).

Two factors modify the application of the rule that we want a high S^2- concentration and thus an H⁺ concentration as low as possible. Arsenic, usually present as arsenate, is converted into As₂S₅. To obtain it, the reaction requires a high concentration of H⁺. At the other extreme, if the acidity is too low, the sulphide ion concentration will be so high that the solubility products for the sulphides of iron, zinc, manganese, cobalt and nickel (not in the copper-arsenic group) will be exceeded and will also be precipitated.

Let’s now take a look at the ions of the copper-arsenic group.Due to the successive dilutions we make in the experiments, the Co²⁺ concentration should not be larger than 0.02M, so the S^2- concentration should not exceed 3 10^-20M. The overall ionisation constant for H₂S is 1.3 10^-20. Since the solubility of H₂S in water at room temperature is about 0.1M, the value of the ion product [H⁺]²[S^2-]=1.3 10^-21, meaning that if we want to keep the S^2- concentration lower than 3 10^-20M, [H⁺] must be no lower than about 0.2M. So, even if the analyses are not quantitative, we have to be careful to the quantities we use in this procedure.

Description of the elements of the silver group

Mercury(II)

It is the highest common oxidation state of mercury in aqueous solution. Aqueous solution containing mercury are poisonous. The soluble forms of Hg²⁺ are the halides, acetate and cyanide. The nitrate dissolves in dilute acids. Stable complexes are formed between Hg²⁺ and the halide ions or sulphide ions but do not bind well with NH₃ and OH^–. Few mercury(II) compounds are coloured.

Lead(II)

Lead(II) was already discussed in the previous section.

Bismuth(III)

It is the most common oxidation state of bismuth in aqueous solutions. The free ion is strongly hydrolysed, forming insoluble basic salts. The following equation represents the hydrolysis of BiCl₃:

Solutions containing Bi³⁺ must be highly acidic to keep the cation in solution. In water, a milky white suspension may result from the formation of the insoluble white basic salt of bismuth and antimony. Bismuth can be oxidised to the 5+ state (ex: NaBiO₃) by very strong oxidizing agents and is sufficiently powerful to oxidize Mn²⁺ to MnO^4-.

The majority of bismuth compounds are colourless (exception: brown Bi₂S₃) Complexes can be formed in concentrated solutions of halides: Bi₂S₃ dissolves in 12M HCl.

Copper(II)

There are two oxidation states for the copper: 2+ is the most common and 1+ the second possibility. Cupric chloride, nitrate and sulphate are soluble in aqueous, and aqueous solutions of Cu(II) salts are blue, the colour of the always present complex ion Cu(H₂O)₄²⁺. Most Cu(II) solids are also blue (hydrates) and contain this ion. Other usual ligands are NH₃, Cl^– and CN^–.

All Cu(I) salts are insoluble in water and are colourless solids (exceptions: Cu₂O (red) and Cu₂S (black)). They are often precipitated from the reduction of Cu(II) salts. For example,

Cu(I) forms stable complexes with halide and cyanide ions and with ammonia, complexes oxidized in solution by atmospheric oxygen.

Cadmium(II)

The only common oxidation state of cadmium in aqueous systems is 2+. Halides, nitrate, sulphate and acetate of cadmium are soluble in water. The salts are colourless (exceptions: CdS (yellow), CdO (yellow brown). Cadmium forms complex ions with common ligands (halides, NH₃ and CN^–).

Arsenic

It exists in either the 3+ or 5+ oxidation states. It is usually not present as its ion but as an arsenate (AsO₄^3-) or arsenide (H₂AsO₃^–). If AsCl₃ or AsCl₅ are dissolved in water, they immediately hydrolyse to give the weak acids according to the equations

Sulphides are then precipitated.

These reactions are slow but are speeded up when [H⁺] is increased. The amount of As₂S₃ formed by this series of reaction is small and can be oxidized to As(V) by (NH₄)₂S₂ (procedure 6). We will thus not obtain any As(III) at the end of the process.

Antimony(III)

Antimony exists in the Sb³⁺ and Sb⁵⁺ states in its common compounds. The salts hydrolyse in water to form basic salts that are often insoluble. For instance,

However, antimony(III) tends to form soluble complex ions with halide or hydroxide ion. For example, SbCl₃ is not soluble in water but will dissolve in hydrochloric acid solution, forming SbCl₆^3-, or in bases forming Sb(OH)₆^3-. The majority of antimony compounds are colourless (exceptions: SbS₃ (red) and SbS₅ (yellow)). In water, a milky white suspension may result from the formation of the insoluble white basic salt of bismuth and antimony.

If treated by HCl, the mildness disappears because the equilibrium is shifted to the left. In concentrated HCl solution, antimony exists as the complex, stable tetrachloro-antimonate(III) ion, SbCl₄^–.

The hydrolysis stakes place when a strongly acid solution of antimony chloride is added to water.

Tin (II) and (IV)

Pure compounds containing Sn(II) and Sn(IV) are common. In aqueous solutions, both are stable but Sn(II) is slowly converted into Sn(IV) by air oxidation. Both tin oxidation states readily undergo hydrolysis to yield stannite’s (Sn(II)) or stannate’s (Sn(IV)). Thus common tin salts are insoluble in water because the hydrolysis products are insoluble basic salts such as SnOCl₂.

Tin compounds are usually colourless (exceptions: SnS (brown) and SnS₂ (light yellow)). Both oxidation state forms complex ions, with the usual halide and hydroxide ions. As a result, although tin (IV) salts undergo hydrolysis to form insoluble salts, it forms the soluble, complex ions SnCl₆^2- and Sn(OH)₆^2- in hydrochloric acid and strong bases. The Sn(IV) hydroxide complexes are sufficiently stable to cause SnS₂ to dissolve in 6M NaOH.

Procedure 5

Place the decantate from Procedure 1 in a casserole or in a lens and add two drops of 3% H₂O₂ and two drops of 2M HCl. Carefully boil down to a volume of 1 or 2 drops and then allow to cool. Add 6 drops of 6M HCl. Carefully evaporate the contents of the casserole down to a pasty mass (do not bake the residue). Cool and then we can either use H₂S gas as the precipitating reagent (A) of a thiocetamide solution (B). To the residue of the casserole, add exactly 4 drops of 2M HCl, Swirl until all the residue is dissolved. If necessary, stir the mixture and warm slightly. Transfer the solution to a test tube.

(A) Heat carefully until it shows signs of effervescence and then treat with H₂S under the hood for 20 to 30s.

Dilute with 10 drops of hot water and continue to treat with H₂S for another 20-30s.

Add 1 drop of 1M NH₄C₂H₃O₂ and treat with H₂S for another 20-30s.

Finally, add 25 drops of cold water and treat with H₂S for another 20-30s.

Each time, make sure that H₂S is bubbled all the way to the bottom of the solution. Centrifuge and test for complete precipitation by first noting the appearance of the solution and then passing H₂S into the top of the supernatant liquid (do not disturb the precipitate). If precipitation is not complete, add 5 more drops of water and treat with H₂S for an additional 20-30s. Centrifuge and repeat until the precipitation is complete. When it is the case, wash down the walls of the test tube with 2 or 3 drops of water, centrifuge and decant into a casserole. Boil it for one minute and save it for the analyses of the next groups (Procedure 15). Wash the precipitate twice with 15 drops of hot water and pass to Procedure 6. If the wash water peptizes the precipitate (makes a colloidal suspension from it, that does not settle when centrifuged), then add 10 drops of 1M NH₄C₂H₃O₂, mix well and heat nearly to boiling before centrifuging.

(B) Add 4 drops of 1M thioacetamide solution, mix thoroughly and heat in a boiling water bath for four minutes. Add then

– 8 drops of hot water,

– 8 drops of 1M thiocetamid and

– 1 drop of 1M NH₄C₂H₃O₂.

Mix well and heat in the boiling water bath for 4 minutes. Centrifuge and decant into a test tube. Save both the precipitate and the decantate. Test the decantate for complete precipitation by adding two more drops of 1M thioacetamide, mixing well and allowing to stand for 1 minute. If a precipitate forms (the previous precipitation was not complete), add two drops of hot water and two drops of 1M thioacetamide. Mix well and heat in the boiling water bath for 2 minutes. When a test shows that precipitation is complete, transfer the decantate to a casserole. Boil the decantate for 1 minute and save it for the test for next groups (Procedure 15).

Precipitates from the precipitation and from the tests for complete precipitation are combined by using a few drops of water to flush the latter into the test tube containing the first precipitate. Wash the precipitate 3 times: first with 10 drops of hot water and then twice with 20 drops of hot solution prepared using equal volumes of hot water and of 1M NH₄C₂H₃O₂. Analyse the content of the tube according to Procedure 6. As for (A), if the wash solution leads to a colloidal suspension, add 10 drops of 1M NH₄C₂H₃O₂ to the suspension, mix well and heat nearly to boiling, centrifuge, saving the precipitate for Procedure 6.

Notes

1. In Procedure 5, the initial H⁺ concentration is 2M. In the course of the precipitation, about a tenfold dilution occurs, which should reduce the concentration to 0.2M. However, H⁺ ions are liberated during the precipitation. For instance,

As a result, the final H+ ion concentration is slightly higher than 0.2M. It ranges from 2M to 0.2M, which allows for complete precipitation of the copper-arsenic group without precipitation of the cations of the aluminium-nickel group.

The sulphides of mercury, lead and copper are black. SnS and Bi₂S₃ are dark brown but look black when wet. CdS, SnS₂ and As₂S₅ are yellow. Sb₂S₃ is orange.

Lead may first form an orange precipitate of PbS.PbCl₂ that changes to black PbS on continued H₂S treatment. Mercury first precipitates as white HgS.HgCl₂ and changes of colour through yellow, orange and brown to black HgS. The intermediate colours results from a mixture of the two precipitates in varying proportions. CuCl₂ is green in concentrated solution and blue in dilute solution. The initial colour of the precipitates is thus not a good indication of the ion. The absence of any green of blue colour in an unknown solution shows the absence of copper.

2. If only members of the copper-arsenic group are present, we can use a different procedure which saves considerable time. Place 4 drops of the solution in a test tube. Add 6 drops of H₂O₂ and 10 drops of ammonium sulphide solution. Then add more ammonium sulphide until the solution turns yellow (about 30 drops). Stir the content of the tube.

Heat for 3-4 minutes in the boiling water bath while stirring. Avoid to heat to the point where excessive frothing of the contents occurs. Centrifuge and decant into a test tube. The decantate may contain AsS_43-, SbS_33- and SnS_32- for Procedure 11. Repeat the treatment of the precipitate with a second 10-drop portion of ammonium sulphide solution, heating and stirring for 2 minutes. Centrifuge and add the decantate to the first. Save it for Procedure 11. Wash the precipitate twice with 20 drops of a hot solution prepared by mixing equal volumes of water a 1M NH₄C₂H₃O₂ and analyse the precipitate, which may consist of the sulphides of mercury(II), lead, bismuth, copper and cadmium (Procedure 6).

3. Evaporations in casseroles are made as follows: hold the casserole in your hand and pass it back and forth over the top of the flame of a Bunsen. At the end of every two or three back-and-forth passes, tilt the casserole slightly so that the solution will run to its lower edge. Be careful not to overheat to the point where the residue begins to bake. If brown areas develop on the bottom of the casserole, indicating baking, then swish the remaining solution around until the brown area is removed. When only 2 or 3 drops of liquid remain, remove from above the flame and let the heat of the casserole complete the evaporation. Baking may sublime off the chlorides of arsenic, mercury and tin.

4. Four drops of 6M HCl are added to the residue in the casserole before dryness in order that nitrate ions may be reduced in accordance with the equation

If present, nitrate ions will oxidize the H₂S and precipitate S:

Other oxidizing agents may be present in a solution to be analysed and will react with H₂S in acid solutions.

Fe³⁺ will not be affected by evaporation with 6M HCl but Cr₂O₇^2- is reduced:

5. To treat a solution with H₂S, attach a clean glass bubbling tube to the rubber outlet tube at the source of H₂ This bubbling tube is made by drawing down a glass tube of suitable diameter to a fairly fine constricted end. The overall length of the bubbling tube should be about 5in. Insert the end of the bubbling tube from which H₂S is escaping into the surface of the solution in the test tube and then gradually bring it down to the bottom of the solution. The constricted tube will deliver a stream of very fine bubbles; large bubbles would tend to throw the solution out of the small test tube. If the tip of the bubbling tube is brought suddenly all the way to the bottom of the solution, the sudden rush of gas may throw the solution out of the test tube. A very rapid rate of H₂S bubbling should be avoided.

An ordinary Kipp generator, placed in a hood, is a very satisfactory source of H₂S. A trap, in the form of a bottle or flask should be placed between the generator valve and the point where the bubbling tube is attached.

One of the most satisfactory sources of H₂S is the commercial mixture of sulphur, hydrocarbon mix, and asbestos, of which ”Aitch-Tu-Ess” is an example. This mixture can be purchased in bulk or in small capsules (in bulk is recommended). It is used for generation of H₂S as follows. Set up a generator under the hood, as illustrated below.

Fill the 6in tube A ½ to ¾ full of H₂S-generating mixture. Insert the rubber stopper B, to which is attached the delivery tube C, the rubber connecting tube D, and the clean bubbling tube E, which is inserted into the solution to be treated. The heating should be gentle, just strong enough to give a fairly rapid evolution of H₂S bubbles. The burner with which A is being heated can be held in one hand, while the test tube containing the solution is being held in the other hand. Do not heat so strongly that sulphur is distilled over. The evolution of H₂S will cease when heating is stopped. The tube E should be removed from the solution being treated as soon as heating is stopped to avoid solution being drawn from the test tube up into E by the contracting gas.

The tube E should be removed and cleaned after each series of precipitations. Components A, B, C and D can be kept together when not in use. When the mixture no longer yields H₂S on heating, tube A can be replaced by new mixture.

Hydrogen sulphide is a poison. Therefore all H₂S treatment should be carried out under a good hood. Hydrogen sulphide is insidious since the sense of smell may become fatigued and fail to give warnings of high concentration. Low concentrations may produce irritation of the mucous membranes. Headaches, dizziness, nausea, and lassitude may appear after exposure.

6. Care should be taken so that the tube is not heated to the point at which excessive frothing results in loss of solution by spillage. If the tube is grasped with a test-tube holder, it can be withdrawn from the boiling water if and when spillage is imminent.

7. The organic compound thioacetamid, CH₃CSNH₂, hydrolyses in water (particularly at higher temperatures) to yield H₂S:

It therefore serves as a convenient source of H₂S. The hydrolysis is so slight at room temperature that a 1M solution of the compound, when preserved in a stoppered bottle, undergoes very little deterioration. At temperatures of about 80°C, the hydrolysis is sufficient so that a 1M solution yields a solution saturated with H₂S.

Separation of the copper subgroup from the arsenic subgroup

We can separate the two groups in a solution of ammonium sulphide: sulphides of arsenic, antimony and tin(IV) are soluble and sulphides of mercury, lead, bismuth, copper and cadmium are insoluble.

As a general rule, the water-insoluble salt of a weak acid will dissolve in a strong acid. By this rule the metal sulphides should dissolve in HCl. The fact that they do not dissolve means that their normal solubility in water is so small that not even a strong acid can give a concentration of H⁺ sufficiently high to cause them to go into solution. For instance, the solubility of CuS is very low (4 10^-35). It means that the concentration of sulphide ions in the equilibrium

is extremely low. The equilibrium reaction

which is set up when a strong acid is added to the sulphide, is not capable of giving a sulphide ion concentration lower than that in equilibrium with CuS and Cu²⁺. Consequently, CuS does not dissolve. On prolonged boiling with HCl, CuS will dissolve, very slowly. In this case the S^2- ions are actually removed from solution as H₂S. It should be noted at this point that the solubility principle does not tells at which speed an insoluble substance dissolves.

Ammonium sulphide ((NH₄)₂S) can dissolve As₂S₃, As₂S₅, Sb₂S₃ and SnS₂, but not SnS. Therefore, if tin is in solution we need it as SnS₂, thus present as Sn(IV) during treatment with H₂S. To guarantee it, the solution is boiled with H₂O₂ before precipitation with H₂S (see Procedure 5).

The separation of arsenic, antimony and tin from mercury(II), lead, bismuth, copper and cadmium depends on the fact that As₂S₃, As₂S₅, SnS₂ and SbS₃ are soluble in a solution of (NH₄)₂S, whereas HgS, PbS, BiS₃ CuS and CdS are insoluble. To accelerate the process, the content of the tube in Procedure 6 are kept hot.

The soluble compounds are strong electrolytes and ionize to give NH₄⁺ ions and SnS₃^2-, AsS₃^3-, AsS₄^3- and SbS₃^3- ions. The formation of the latter ions can be represented by the net equations

These ions may be classed as complex ions. Mercury(II), lead, bismuth, copper and cadmium do not form complex thiol ions. An examination of the formulas for the thiol salts suggests why these sulphides can be dissolved by (NH₄)₂S: (NH₄)₂SnS₃ is the salt of thiostannic acid H₂SnS₃. It is the analogue of the ternary salt (NH₄)₂SnO₃ in the oxygen system.  Any ternary salt in the oxygen system may be considered to have been formed as a result of the reaction of a basic oxide with an acidic oxide. The oxides of metals are basic oxides, the oxides of nonmetals are acidic oxides.

The sulphides of the elements bear the same relationship to HS that the oxides bear to H₂O.

Therefore, CuS, PbS, HgS, CdS, SnS, Bi₂S₃ and (NH₄)₂S are basic sulphides in the sulphur system. SnS₂, As₂S₃, As₂S₅ and Sb₂S₃ are acidic sulphides in the sulphur system. Basic sulphides react with acidic sulphides to form a salt in the sulphur system. Therefor we expect the acidic sulphides to react with (NH₄)₂S to form salts and HgS, PbS, Bi₂S₃, CuS, CdS and SnS should not react.

There remains the question of why SnS is basic while SnS₂ is acidic. It is generally true that as the oxidation number of a polyvalent element increases, it becomes more non-metallic in character, and as an element becomes more non-metallic in character, its compounds become more acidic. Oxidation with H₂O₂ guarantees that the tin is present in its higher oxidation state.

Antimony(V) is very unstable and does not ordinarily exist in solution.

Procedure 6

To the test tube containing the precipitate from Procedure 5, add 10 drops of ammonium sulphide solution. Stir the contents of the tube well; then heat for 3-4 min in the boiling water bath, stirring the contents meanwhile. Avoid any excessive frothing of the content. Centrifuge, decant in a test tube and save the decantate which contains any present species of AsS₄^3-, SbS₃^3- and SnS₃^2- for Procedure 11. Repeat the treatment of the precipitate with a second 10-drop portion of (NH₄)₂S, heating for 2 minutes. Combine the decantates together. Wash the precipitate twice with 20 drops of a hot solution prepared by mixing equal volumes of water and 1M NH₄C₂H₃O₂. The precipitate will be analysed in Procedure 7 to isolate mercury from lead, bismuth, copper and cadmium.

Separation of mercury from lead, bismuth, copper and cadmium

We use warm dilute HNO₃ to separate HgS, insoluble in its presence, from the sulphides of copper, cadmium and lead that are soluble: the dissolve in accordance with the equations

The second equation removes S^2- from solution, shifting the first equation to the right.

Procedure 7

Add 15 drops of 3M HNO₃ to the test tube containing the precipitate from Procedure 6, mix thoroughly, transfer to a casserole and boil gently for about one minute. Replenish the HNO3 when necessary. Transfer to a test tube, centrifuge, and decant in a test tube (save it for Procedure 8). Wash the precipitate (HgS and S) twice with 15-drop portions of water made acidic with one drop of 3M HNO₃. Add the first washing to the decantate in the test tube but discard the second washing.

Treat the precipitate with 6 drops of 12M HCl and 2 drops of 16M HNO₃, mix thoroughly and heat for 1 minute in the boiling water bath. Add 10 drops of hot water, transfer to a casserole, boil gently for 30s and then transfer back to a test tube. Cool by holding the test tube under the water tap, then centrifuge and decant into another test tube. Add 2-5 drop of 0.2M SnCl₂ solution to the cool decantate in the test tube. A black (Hg) or grey (Hg2Cl2 + Hg) precipitates proves the presence of mercury(II).

Notes

1. If the HNO₃ is too concentrated, it will dissolve some of the HgS. In addition, it may oxidise PbS to PbSO₄, which will remain as a residue mixed with the HgS.

2. HgS is black. The presence of a white or yellow residue after digestion with HNO₃ does not, however, eliminate the possibility of mercury being present. The residue, whatever its colour, should be tested for mercury.

3. HgS is insoluble in both concentrated HNO₃ and concentrated HCl, when each is used separately. However, a mixture of the two concentrated acids dissolves HgS readily and quickly.

HgCl₄^2- is pale yellow and is largely dissociated into colourless Hg²⁺ and Cl^– ions when the solution is diluted with water. The Hg²⁺ ions can then be identified by means of the reaction written in note 4.

4. The extent to which Hg²⁺ is reduced by Sn²⁺ depends on the relative amounts of the two kinds of ions present. If Hg²⁺ is present in large excess, the reduction is mostly to Hg₂Cl₂:

Hg₂Cl₂ is white. If more Sn²⁺ ions are added, Hg₂Cl₂ that is first formed according to the reaction above is further reduced to Hg, as follows:

Hg is black. If Sn²⁺ is in excess, the following reaction takes place:

The combination of Hg₂Cl₂ (white) and Hg (black) forms a dark grey to black precipitate. The equations above show that tin(II) compounds are strong reducing agents.

The solution is boiled to remove chlorine formed by oxidation of Cl^– by NO₃^–. The chlorine would oxidise Sn²⁺ to Sn⁴⁺, destroying the reductant (Sn²⁺) and preventing it from reducing the Hg²⁺.

Separation of lead from bismuth, copper and cadmium

Lead sulphate is insoluble in water. The sulphates of bismuth, copper, and cadmium are soluble. This fact is the basis for the separation of lead ions from the other cited ions.

Procedure 8

Add 4 drops of 18M H₂SO₄ to a casserole containing the decantate from Procedure 5 and evaporate under a hood until the volume is about 1 drop and dense white fumes of SO₃ are formed. These fumes are dense and will obscure the view of the casserole inside walls. Paler fumes (from HNO₃) must not be mistaken for the SO₃ fumes.

Cool, add 15 drops of cold water, and stir the contents until all material in the casserole is dissolved or suspended; then transfer it to a test tube before the suspended material has a chance to settle.  Swirl the casserole with 4 drops of cold water and transfer the washing to the same tube. Cool under the water tap. A white precipitate in the form of a fine suspension shows the presence of lead (PbSO₄). Centrifuge until the supernatant liquid is clear and decant into a test tube (save for Procedure 9). Wash the precipitate twice with 10-drop portions of cold water. To the washed precipitate in the test tube, add 4 drops of 1M NH₄C₂H₃O₂ and stir for 20s. Then add 2 drops of 0.2M K₂CrO₄. A yellow precipitate confirms the presence of lead (PbCrO₄).

Notes

1. PbSO₄ is appreciably soluble in concentrated HNO₃ due to formation of the hydrogen sulphate ion in the manner discussed in note 2. For this reason, HNO₃ must be removed before PbSO₄ will precipitate. When a solution containing H₂SO₄, HNO₃ and water is boiled, the water and HNO₃ are first driven off because they boil at comparatively low temperatures (100-120°C). Further heating results in boiling the H₂SO₄ (boiling point: 338°C). At its boiling temperature the H₂SO₄ decomposes to a slight extent.

SO₃ fumes strongly in moist air. Therefore formation of dense white fumes of SO₃ at the end of the evaporation gives assurance that all HNO₃ has been removed.

2. When the solution is cooled down, sulphates of bismuth, copper and cadmium may crystallize out. However they are soluble in dilute H₂SO₄ and will dissolve if water is added. PbSO₄, on the other hand is soluble in concentrated H₂SO₄ due to formation of HSO₄^–.

On dilution with water, the equilibria are shifted to the left and PbSO₄ precipitates, precipitate finely divided while sulphates of bismuth, copper and cadmium form relatively large crystals.

3. PbSO₄ dissolves the ammonium acetate because the complex ion Pb(C₂H₃O₂)₄^2-.

This complex ion is very stable and the solution contains thus a very low concentration of lead ions.

PbCrO₄ is less soluble than PbSO₄. Therefore, when K₂CrO₄ is added to the solution formed by adding NH₄C₂H₃O₂ to PbSO₄, a precipitate of PbCrO₄ is formed even though the concentration of lead ions in the solution is very low.

As lead is largely removed in the silver group, only very small quantities will ordinarily appear in the copper-arsenic group. For this reason, the test for lead may not be as strong as the test for other cations in this group.

Separation of bismuth from copper and cadmium

Addition of NH₄OH to a solution containing bismuth, copper and cadmium ions first precipitates the hydroxides of all three metals. The hydroxides of copper and cadmium, however, dissolve in an excess of NH₄OH. The hydroxide of bismuth does not.

Procedure 9

To the decantate from Procedure 8, add 15M NH₃, dropwise and with constant mixing, until it becomes distinctly alkaline. Stir for 2 minutes. Centrifuge and decant (save it for Procedure 10). Wash the precipitate twice with 15 drops of hot water. To the washed precipitate, add 3 drops of 8M NaOH and 2 drops of 0.2M SnCl₂ and stir. A jet-black precipitate proves the presence of bismuth (Bi).

Notes

1. When NH₄OH is added to a solution containing copper ions, Cu(OH)₂ (blue) is first precipitated.

An excess of NH₄OH, however, dissolves the Cu(OH)₂, to give a deep blue solution in which copper is present as the complex ion Cu(NH₄)₄²⁺.

The detailed reaction is

Cadmium behaves in the same manner as copper; Cd(NH₃)₄²⁺ is formed. Cd(OH)₂ (white) dissolves quite slowly in excess NH₄OH; therefore the solution should be stirred for 2 minutes to ensure complete dissolution of Cd(OH)₂. Similar complexes will be met in the case of the ions of nickel, cobalt and zinc. Ag(NH₃)₂⁺ was formed in the analysis of the silver group.

2. The formation of the black precipitate of bismuth results from the action of stannate(II) ions (Sn(OH)₄^2-) on Bi(OH)₃ (white):

The stannate(II) ions were formed when the SnCl2, added as the final reagent, reacted with the excess of NaOH.

The reaction of stannate(II) ions with Bi(OH)₃ to give stannate (IV) and bismuth serves to illustrate the reducing character of tin(II) compounds. Stannate ions Sn(OH)₄^2- are formed from the action of NaOH on SnCl₂ in two steps. In a first time, if NaOH is added to a solution of SnCl₂, a white precipitate forms:

If NaOH is put in excess, the white precipitate redissolves to form a clear, colourless solution containing the stannate ion.

If the white precipitate of Sn(OH)₂ that first forms when NaOH is added is treated with an acid (such as HCl), it dissolves to form a clear solution that can be shown to contain Sn²⁺ ions.

Sn(OH)₂ is thus amphoteric as is will dissolve in either a strong acid or a strong base. Amongst the metals we consider in our analyses, the hydroxides Sn(OH)₂, Sn(OH)₄, Pb(OH)₂, Sb(OH)₃, Al(OH)₃, Cr(OH)₃ and Zn(OH)₂ are amphoteric, and the other are not amphoteric.

It should be pointed out that there is not complete agreement amongst chemists about the composition of the stannate ion: Sn(OH)₄^2-, SnO₂^2-, Sn(OH)₃^– and HSnO₂^– are suggested. Since the formula of the ion may change as the concentration of the solution changes, it may well be that all four are present at one time or another, or even may all be present in equilibrium at one time.

Stannate ions reduce the hydroxides of antimony, lead, copper and cadmium to the corresponding metal, but the process is slow and the metallic deposit is not jet-black. Bi(OH)₃ is also reduced but it forms instantly a jet-black deposit of metallic bismuth.

Finally, the stannate(II) ions react in contact with air and with itself to form stannate(IV) ions.

For these reasons, stannate(II) is not kept as a reagent but is formed at the time it is to be used.

Detection of copper and cadmium

Procedure 10

Detection of copper:

If the decantate from Procedure 9 is colourless, copper may or may not be absent: solutions containing copper in amounts less than 1 part in 25000 may appear colourless. If the decantate is deep blue, copper is present (Cu(NH₃)₄²⁺. It is wise to verify the presence of copper in either case. To do so, place 5 drops of the decantate (not all the decantate, cadmium may be present) in a test tube and add 5M HC₂H₃O₂ until the solution is decolorized. Then add 2 drops of 0.2M Fe(CN)₆. A red precipitate confirms the presence of copper (Cu₂Fe(CN₆).

Detection of cadmium:

If copper is absent, treat the colourless decantate from Procedure 9 with 2 or 3 drops of ammonium sulphide solution, mix thoroughly, and allow to stand for about one minute. The formation of a yellow precipitate proves the presence of cadmium (CdS).

If copper was confirmed before, add 0.2M KCN (caution: KCN is poisonous) dropwise to a 10-drops portion of the blue decantate until the colour disappears. Then treat the solution with 2 or 3 drops of ammonium sulphide solution, mix thoroughly and allow to stand for about one minute. A yellow precipitate proves the presence of cadmium (CdS). As soon as the test is completed, dump the contents of the tube to which KCN was added into the sink and flush it away with a lot of water.

Notes

1. The blue colour of Cu(NH₃)₄²⁺ is visible as little as 1 part of copper is present in 25000 parts of water. The red precipitate of Cu₂Fe(CN)₆ will detect 1 part of copper in 1 million parts of water.

2. Cu₂Fe(CN)₆ and CdFe(CN)₆ are soluble in strong acids (HCl and H₂SO₄) but precipitates readily in presence of a weak acid (such as acetic acid).

3. KCN is poisonous and is thus not kept available as usual reagents, because of potential accidents. It must never be mixed with an acid or an acid solution. Even a very weak acid will react with it and liberate poisonous HCN gas

4. When excess KCN is added to a solution containing Cu(NH₃)₄²⁺ and/or Cd(NH3)₄²⁺, the complex ions Cu(CN)₂^– and/or Cd(CN)₄^– are formed. Cu(CN)₂^– is very stable and only very slightly dissociated into Cu⁺ and CN^–. The concentration of Cu⁺ is so low that we don’t obtain any precipitate of Cu₂S when sulphide ions are added. Cd(CN)₄^2- is less stable and dissociates appreciably into Cd²⁺ and CN^–. The concentration is high enough for a precipitate of CdS to form when sulphide ions are added.

Reprecipitation of the sulphides of arsenic, antimony and tin

Procedure 11

To the test tube that contains the decantate from Procedure 6 (potentially containing AsS₄^3-, SbS₃^3- and SnS₃^2-), add 6M HCl under a hood with constant stirring until the solution shows an acidic reaction when tested with litmus. As long as each drop of 6M HCl keeps on bringing down more precipitate, the solution is still alkaline; when no more precipitate forms, the solution is probably acidic. A large excess of HCl must be avoided (see notes). Centrifuge and decant, discarding the decantate. Wash the precipitate three times, each time with 10 drops of a hot solution prepared by mixing equal volumes of water and 1M NH₄C₂H₃O₂, and analyse according to Procedure 12.

Notes

1. A dark-coloured decantate from Procedure 6 means that some CuS, and perhaps HgS, has been put into a state of colloidal suspension by the (NH₄)₂ If this decantate is allowed to stand for about 24h, the CuS and HgS will ordinarily settle out. The yellow supernatant liquid can then be decanted, leaving the CuS and HgS behind. If the decantate is to be analysed immediately, add 5 drops of 1M NH₄C₂H₃O₂ to coagulate the CuS and HgS, mix thoroughly, centrifuge, and decant, discarding the precipitate.

2. The addition of dilute HCl to the soluble thiol salts of the arsenic subgroup reprecipitates the sulphides of the three elements in accordance with the following equation

The compound H₃AsS₄ is unstable and decomposes.

(NH₄)₂SnS₃ and (NH₄)₃SbS₃ react in a similar manner. H₂SnS₃ and H₃SbS₃ decompose to give SnS₂ and Sb₂S₃, respectively.

H₂S plays thus the same role as H₂O and As₂S₅, Sb₂S₃ and SnS₂ are the equivalents of acid anhydrides in the sulphur acid system.

3. In reprecipitating the arsenic group, a large excess of HCl must be avoided, since SnS₂ and Sb₂S₃ are appreciably soluble in high concentrations of this acid. Also, the liquid in contact of their precipitates should be decanted off at once since the SnS₂ and SbS₃ will dissolve on long standing.

4. The precipitate that is formed in Procedure 11 may be only sulphur. Ammonium sulphide may contain some ammonium polysulfide (NH₄)₂S₂. When HCl is added to a solution containing it, sulphur is liberated according to the reaction

Since (NH₄)₂S₂ is a moderately strong oxidizing agent, it will oxidize As(III) to As(V) in Procedure 6.

Separation of arsenic from antimony and tin

Arsenic sulphide is insoluble in concentrated HCl. As written in the note 3 above, sulphides of tin and antimony dissolve in the HCl to form soluble chlorides.

The Sn⁴⁺ and Sb³⁺ ions react with the Cl^– ions present to form the complex chlorous ions SnCl₆^2- and SbCl₄^–.

Procedure 12

To the precipitate from Procedure 11, add 15 drops of 12M HCl, mix thoroughly and heat the test tube in the boiling water bath for 3-4 minutes, stirring frequently. Add 7 drops of hot water, mix well and continue the heating for 20-30s. Centrifuge and decant into a test tube. Save the decantate (for Procedure 14) and wash the precipitate once with 10 drops of 6M HCl and then 3 times with 10-drops portions of a hot solution prepared by mixing equal volumes of water and 1M NH₄C₂H₃O₂. Analyse the precipitate according to Procedure 13.

Note

• The precipitate (As₂S₅) is first washed with HCl rather than with water to remove all traces of Sb³⁺ because they will react with water to form insoluble SbOCl, which will remain with the As₂S₅. The final washing with hot water is to remove all traces of HCl. Chloride would interfere with the confirmatory test for arsenic (Procedure 13) by forming AgCl.

Detection of Arsenic

Procedure 13

Add 10 drops of 16M HNO₃ to the test tube containing the precipitate from Procedure 12 and heat in the boiling water bath for one minute, or until the original precipitate is disintegrated and a deposit of sulphur is formed. Add 4-5 drops of water, centrifuge, and decant into a casserole, discarding any precipitate that remains in the test tube. Evaporate the solution in the casserole very carefully to complete dryness, but do not bake. Allow to cool. Add 4 drops of 0.2M AgNO₃ and swish around in the casserole for 10 seconds. A reddish brown precipitate proves the presence of arsenic (Ag₃AsO₄). If no reddish brown precipitate appears, add 0.2M NaC₂H₃O₂, a drop at a time and with thorough mixing, until precipitation is complete or a maximum of 30 drops has been added. A reddish brown or chocolate brown precipitate proves the presence of arsenic.

Notes

1. As₂S₅ and As₂S₃ are dissolved by HNO₃ this way:

If the H₃AsO₄ is heated to 160-200°C, As₂S₅ is formed.

Ag₃AsO₄ is reformed when the aqueous solutions of AgNO₃, and NaC₂H₃O₂ are added.

2. Ag₃AsO₄ is formed in the confirmatory test for arsenic according to the equation

Ag₃AsO₄ is the salt of a weak acid and is, therefore, soluble in HNO₃. However, it is insoluble in a weakly acidic or neutral solution. Evaporation to dryness should remove all trace of HNO₃. As a precaution against the possibility of some HNO₃ being left, NaC₂H₃O₂ is added to the reaction mixture. As the salt of a weak acid, NaC₂H₃O₂ will serve as a buffer for any HNO₃ that might have remained. The C₂H₃O₂^– ions from the strong electrolyte NaC₂H₃O₂ combine with H⁺ ions from HNO₃ to form the weak acid HC₂H₃O². This reaction maintains the H⁺ ion concentration at such a low value that Ag₃AsO₄ will precipitate.

3. The formation of a white precipitate (AgCl) means that the As₂S₅ was not washed free of chloride in Procedure 12. AgC₂H₃O₂ (white) is sparely soluble but may also precipitate.

Detection of antimony and tin

Procedure 14

A. Detection of antimony

Transfer the decantate from Procedure 12 to a casserole, boil for 1 min to remove all H₂S, then add 4-5 drops of cold water and mix thoroughly. Place 4 drops of this solution in a test tube and add 5M NH₄OH drop by drop with constant stirring until the solution is barely alkaline (showed by a litmus test).

If antimony or tin is present, a white precipitate or milky suspension, SbOOH and/or Sn(OH)₄ will form in the alkaline solution. If there is no precipitate (or suspension), then tin and antimony are absent (no need to continue Procedure 14 then).

Add 5MHC₂H₃O₂ drop by drop with constant stirring until the solution is acidic to litmus. The white precipitate should remain in the weakly acidic solution and does not interfere with the test for antimony. Add one crystal of Na₂S₂O₃•5H₂O and heat by placing the test tube in the boiling water bath for 2-3 min. An orange red precipitate proves the presence of antimony (Sb₂OS₂).

B. Detection of tin

To the remainder of the solution in the casserole, add a 1-in. piece of 26-gauge aluminium wire. Warm gently until the wire has dissolved; then boil gently for about 2 minutes, or until the black precipitate either has all dissolved or appears not to be dissolving anymore, replenishing the solution with 6M HCl if necessary. If the black precipitate dissolved completely, then antimony is absent. If after 2 minutes there is still a black residue, Sb is present. Transfer the contents of the casserole immediately to a test tube, cool, centrifuge and decant.

Immediately add 2-3 drops of 0.1M HgCl₂ solution to the decantate, mix thoroughly and allow to stand one minute.  The presence of tin is showed by the presence of a white (Hg₂Cl₂) or grey (Hg₂Cl₂ + Hg) precipitate.

Notes

1. The solution used in making the confirmatory test for antimony must be free from H₂S; otherwise SnS₂, as well as Sb₂S₃, will precipitate when the middle is made alkaline.

2. Addition of NH₄OH precipitates SbOOH and Sn(OH)₄ according to the following equations:

The detailed reaction are between the Sb³⁺ and Sn⁴⁺ ions in equilibrium with SbCl₄^– and SnCl₆^2- and the OH^– ions in equilibrium with NH₄OH.

3. The thiosulfate ion S₂O₃^2- disproportionates sparingly to yield H₂S.

Although the concentration of S^2- ions provided by this H₂S is not high enough to precipitate tin as SnS₂, it will precipitate antimony as Sb₂OS₂.

4. A flame test for tin may be carried out as follows: use cold water to fill a clean (outside and inside) test tube. Dip the bottom of the tube into the solution to be tested; then hold the bottom of the tube in a hot, non-luminous Bunsen flame. A blue coloration of the flame, which appears to cling to the wall of the test tube, proves the presence of tin. A trial flame test for tin should first be run on a sample of tin salt solution from the reagent shelf so that the characteristic blue coloration can be recognized.

5. The aluminium is added to reduce Sn⁴⁺ to Sn.

Excess aluminium is dissolved by the HCl. The tin then dissolves in HCl to form Sn²⁺ ions.

The tin does not dissolve until all the aluminium has dissolved or the excess is removed. Furthermore, since tin is not very active, it dissolves less readily than the aluminium. For that reason the mixture must be boiled.

Since the tin(II) formed by the reaction of tin with HCl may eventually be oxidised by the oxygen of the air to tin(IV), the test should be completed as rapidly as possible.

6. The chemistry of the final test for tin, in which Sn²⁺ reduces Hg²⁺ to Hg+Hg₂Cl₂, is the same as that involved in the confirmatory test for mercury (Procedure 7, note 4).

The metallic aluminium added to the solution containing Sn⁴⁺ and Sb³⁺ replaces antimony (black), as well as tin (grey). Antimony, being less active than hydrogen, does not dissolve in HCl however.

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It should be obvious to any chemist, but laboratory manipulations involve the security of each person in the lab and the careful manipulation of equipment and compounds: wear the appropriate protective clothing (laboratory jacket, gloves and glasses) and be aware of your environment.

Mineral solutions

The method involves a separation of the compounds by the formation, or the absence of formation, of a precipitate in given conditions (temperature, reagents). The concentration of the sample should typically be about 0.1M with a volume about 1ml. Careless work may produce precipitates of the wrong colour or precipitates when there should be none, and eventually lead to the loss of too much solute in the sample to obtain results. The identification tests are ineffective if you do not have at least 1mg of solute present (~10% of the initial sample mass). The goal is not a quantitative identification so the volumes we will use don’t have to be very precise and involve small quantities. As a result, we don’t use graduated cylinders to measure the reagents but droppers. They are tested to determine how many drops of volume are needed to give 1 ml of liquid. It should typically be between 20 and 23.

When we obtain precipitates, we will separate the precipitate from the solution (centrifugation is preferred to filtration) and keep both for further studies. If we obtain a precipitate, we want to be sure that all the species supposed to precipitate actually does because it can interfere with further experiments. We let enough time for the precipitate to form (it can sometimes be slow), separate it from the solution and then add a few droplets of reagent to make sure everything is in the precipitate. It is possible that two different ions form precipitates with a single reagent, but they have not the same constant of formation, so one species will precipitate first and then the second if there is an excess of reagent. In our procedures it is usually not important because we will separate the two species later and identify them.

The separation is made by centrifugation that causes the heavy precipitate to be thrown to the bottom of the tube while the lighter liquid remains on top. The test tube must always be balanced in the centrifuge by a tube containing a similar volume of water. 15 to 30 seconds of centrifugation should be enough for most of the compounds but it is not always the case. The liquid should be a clear supernatant liquid. If it is not the case we continue the centrifugation. The last drop of decantate should be removed by gentle tapping from the lip of the tube if the precipitate is packed enough. If the precipitate is light and fluffy, it is better to allow the supernatant liquid to remain behind with the precipitate. In the worst case of a light precipitate, badly packed, we take the top 75-80% of the liquid with a medicine dropper. Absolutely avoid particles of precipitate if further tests are to be made on the solution. The solid is then washed (often with water) to avoid contamination and centrifuged again. It is best to wash the precipitate twice. Centrifugations can also be repeated on the decantate if furthers investigations have to be made on it.

To transfer a solid, for instance into a tube for centrifugation, we stir the solid well to form a slurry and then quickly pour it into the other container. It should move 90% of the sample and as the analyses are not quantitative, we can forget the last 10%.

Several operations will require a heating. Test tubes are never directly put in contact to the flame of a Bunsen: the content would become overheated and splatter out of the tube. It is safer to heat a volume of water and to place the test tube in this Bain-Marie. When we want to evaporate a liquid (to concentrate the species or remove a volatile reagent), we perform it under the hood (hotte). Never heat to dryness or the sample may decompose or become inert.

To control the pH of the solution, we simply touch a litmus paper with the tip of a stirring rod. The acidic or basic reagent is added drop by drop until you think the pH is about right. Mix the solution with the stirring rod and check the pH. Several properties of the solution may change because of the pH variation: colour, “transparency” or apparition or dissolution of a precipitate.

For our experiments, we keep some materials available:

• a beaker of hot distilled water,
• medicine droppers (use more than one to avoid contamination),
• small bottles (ideally with their own medicine dropper) of each reagent
• a tube holder: the initial sample will be separated in several tubes that we will analyse and separate. We will thus need several tubes and wash and rinse them after they have been used (we put them upside-down to let the droplets fall).
• a towel: we want to have a clean work environment to avoid contamination, clean tubes, stirring rods, etc.

Experiments

Now that all of this is said, we can detail the experiments. Let’s consider a solution containing an unknown number of unknown cations. We separate the ions into several groups in function of the formation or the absence of formation of a precipitate during the reaction with one specific reagent. Several cations may form a precipitate with the same reagent, forming a group that we separate next. Cations that don’t form precipitates with this reagent belong to other groups of cations that form precipitates with other reagents in further reactions. The separation by group must be done in the correct order to avoid any disturbance from the ions that should not be in solution anymore. The global scheme of reactions is showed in the following figure.

Each procedure gives a precipitate (double line) and a decantate (simple line). Doted lines indicate reactions of separations of the cations of the group.

The silver group

The first group includes Ag⁺, Pb⁺ and Hg₂²⁺. Let’s first take a look at the properties of the ions of this group.

Description of the elements of the silver group

Ag⁺

The most common oxidation state of silver is 1+. Silver compounds are, for most of them, insoluble in water (common exceptions are AgNO₃ and AgF that are very soluble, and Ag₂SO₄ slightly soluble). To dissolve them, we add cold 6M HNO₃ (exceptions: AgSCN, Ag₂S and silver halides). The compounds are usually white, some exceptions are

Silver ions can form stable complexes in which the coordination number of the silver is 2 (AgL₂). The best known is Ag(NH₃)²⁺ that can be produced from AgCl or AgSCN treated with 6M NH₃. It is a reaction useful to dissolve those solids. AgBr and AgI are less soluble than AgCl (respectively soluble in 15M NH₃ and insoluble even in aqueous NH₃).

Pb⁺

Pb has two oxidation states, Pb²⁺ (most common) and Pb⁴⁺. Most compounds of Pb²⁺ are insoluble in water. Lead nitrate and acetate are the only well-known soluble lead salts. PbCl₂ is less soluble than AgCl and becomes moderately soluble if the solution is heated. Lead forms a stable hydroxide complex ion Pb(OH)₄²⁺ and a weak chloride complex PbCl₄^2-. Common lead(IV) compounds are PbO₂ (insoluble in most reagents, but well in 6M HNO₃ with H₂O₂).

Hg₂²⁺

Mercury has two stable oxidation states Hg₂²⁺ and Hg²⁺, both common. Mercury(I) is a member of the Silver group and Mercury(II) belongs to the copper-arsenic group. Most mercury(I) salts are insoluble (exception: Hg₂(NO₃)₂). Mercury(I) is susceptible to a simultaneous oxidation-reduction process (disproportionation)

This reaction occurs readily in the presence of species that form insoluble precipitates or stable complex ions with mercury(II). When disproportionation of Hg₂²⁺ occurs, metallic mercury is formed as very small droplets, giving a grey colour to the mixture.

Analysis of the silver group

Chlorides of ions of this group are insoluble in cold water and cold, dilute HCl. The other cation chlorides are soluble in this kind of solution so we can separate the silver group ions from ions from the metallic ions of the other groups with the addition of 6M HCl (or some other soluble chloride) to a cold sample. The procedures will be described in detail later in this section. They are given in the correct order to separate all the ions of the silver group. The same will be done for all the analysed groups.

AgCl, Hg₂Cl₂ and PbCl₂ precipitate and all the other metals remain in solution as soluble chlorides. It is possible that lead chlorides aren’t completely precipitated in the process. It will be fully precipitated in the next group. We separate the precipitate from the solution but we discard none because both parts may contain ions of interest. We will focus first on the precipitate as it contains the silver group’s elements, the analysis of the decantate will be made in the sections of the other groups.

We want now to isolate the ions from the Silver group one from each other. AgCl, Hg₂Cl₂ and PbCl₂ precipitates are all white so we cannot discard any possibility based on the colour of the precipitate. We have to proceed to additional investigations. The first step is to add hot water to the solution. In any precipitation process, the insoluble compound will keep precipitating until the concentrations of its ions remaining in solution reach values at which their product just equals the solubility product. When this point is reached, the precipitate will be in equilibrium with its ions, which means that the rate at which the precipitate is forming equals the rate at which it is dissolving and dissociating into ions. As any equilibrium, it can be shifted by changes of concentrations (excess of Cl^– to remove a maximum of the ions from the solution) or a change of temperature. Note that a too large excess of Cl^– may generate a soluble complex (PbCl₄^2-, AgCl₂^–) from the precipitate.

Procedure 1

Place 5 drops of the sample solution in a 3-in. test tube and add 5 drops of water. Add 2 drops of 6M HCl and mix thoroughly by stirring with a glass stirring rod. Wash down any precipitate that adheres to the inside of the tube above the solution with a few drops of cold water and then centrifuge. Test for complete precipitation by adding another drop of 6M HCl to the supernatant solution. When precipitation is complete, centrifuge and decant. Save both the precipitate and the decantate. Wash the precipitate with 5 drops of cold water, adding the washings to the decantate. Except if specified (as here), the washing is always discarded. The precipitate will be analysed in procedure 2 while the decantate will be analysed in Procedure 5.

Notes

• If no precipitate is formed, then the absence of silver and mercury(I) is proved. Lead may be present in small quantity.
• Wash precipitates as follows: Add the water (or other washing liquid) to the precipitate in the test tube, mix thoroughly by stirring the contents of the tube with a glass stirring rod, centrifuge and decant. A bad washing is one of the main source of error in qualitative analysis.

Procedure 2: separation of lead from the other silver’s group ions

As PbCl₂ is more soluble than AgCl and Hg₂Cl₂, it can be separated from the other precipitates if we add hot water. 100ml of water at 0°C dissolves 0.673g of PbCl₂. At 100°C, it dissolves 3.34g of PbCl₂.

Add 15-20 drops of hot water to the test tube containing the precipitate, stir well until all the precipitate is in suspension, and then heat the tube in a beaker of boiling water for 1 minute. Stir frequently.

Centrifuge and decant into another test tube. The decantate may contain Pb²⁺, possibility that we will confirm in Procedure 3. Wash the precipitate twice with 10-drop portions of hot water. It will be analysed in Procedure 4.

Procedure 3: detection test for Pb²⁺

Cool the decantate from Procedure 2 and divide into two parts. To one part, add one drop of 0.2M K₂CrO₄. A yellow precipitate proves the presence of lead (PbCrO₄). To the second part, add one drop of 2M H₂SO₄. A white precipitate, which may form slowly, is a second proof of the presence of lead (PbSO₄).

Notes

1. The solubility product of PbSO₄ is 1.3 10^-8 and that of PbCrO₄ is 2 10^-16 at 20°C.
2. Obviously, the first test is much more sensitive than the second. We show two tests here to demonstrate the difference of sensitivity that two different tests can have. Just as the two confirmatory tests for lead differ in sensitivity, so may confirmatory tests for other ions differ greatly in sensitivity. We will only use the best tests known for each ion.

Procedure 4: detection and separation of Ag⁺ from Hg₂²⁺

Silver chloride, AgCl, is soluble in aqueous NH₃. Also Hg₂Cl₂ reacts with aqueous NH₃ to form Hg and HgNH₂Cl, both of which are insoluble. These facts are the basis for the separation of silver from mercury(I) ions.

Silver chloride dissolves in aqueous ammonia according to

When a solution containing the complex ion Ag(NH₃)₂⁺ and Cl^– is acidified with nitric acid, AgCl is precipitated in accordance with

Formation of the complex is an example of a type of reaction that will be encountered in connection with the detection of other ions. Similar complexes are Cu(NH₃)₄²⁺, Co(NH₃)₆³⁺,Ni(NH₃)₄²⁺, and Cd(NH₃)₄²⁺.

Aqueous ammonia, NH₄OH, reacts with Hg₂Cl₂ to produce a mixture of black, finely divided mercury (Hg) and white mercury(II) amido chloride (HgNH₂Cl).

The compound HgNH₂Cl may be considered to be a derivative of HgCl₂ in which the amino group NH₂ has replaced one atom of chlorine. The formation of Hg and HgNH₂Cl lay be considered to be taking place in two steps: first Hg₂Cl₂ disproportionates

Then, HgCl₂ reacts with aqueous NH₃ to form HgNH₂Cl.

Procedure

Add 4 drops of 15M aqueous NH₃ to the precipitate from Procedure 2, mix thoroughly, centrifuge, and decant into another test tube, saving the decantate for testing for silver. A grey residue proves the presence of mercury(I). To the decantate, add 16M HNO₃ drop by drop and mix constantly with a stirring rod until slightly acid. A white precipitate proves the presence of silver.

Notes

1. Test for acidity or alkalinity of a solution with litmus paper, on a clean towel or drape it over the edge of a beaker. Touch the paper with the extremity of the rod used for stirring the solution.
2. If the PbCl₂ is not completely removed from the precipitate of AgCl and Hg₂Cl₂, it is converted by the NH₄OH into the finely divided insoluble white basic salt Pb(OH)Cl, which may give a turbid decantate. The basic salt will dissolve in HNO₃ and will not interfere with the confirmatory test for silver.

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The formed salt is a racemic melange of diaseteoisomers that can now be separated.

The separation of the enantiomers can also be done via the kinetic splitting that is based on the difference of reaction speed of the enantiomer with a given chiral reactant.

 The speed of reaction has to be very different for the two enantiomers. If we put 0.5 equivalent of the chiral reactant, the enantiomer that reacts faster will from the diastereoisomer and the other enantiomer remains intact (ideal case). The conversion ratio is the proportion of the intact enantiomer over the proportion of the formerd diastereoisomer.

The second way is to directly produce the chiral species. This technique evolved over the years and began with the use of a chiral substrate. Around 1980 a chiral auxiliary was used. In this decade, they also used chiral reactants and after 1990 we started to use chiral catalysts. We will describe those techniques soon. First we will talk about the chiral pool.

The chiral pool

The chiral pool refers to natural compound that are optically active and can be produced at large scale (and thus are cheap). It can be aminoacids, peptides, metabolites, etc. We can use those compounds as starting reactants to obtain a chiral final product. The stereocentres already present on the natural compound will control the stereoselectivity of the reactions and lead to one single enantiomer. We talk about a hemisynthesis (because the starting reactant has not be chemically produced).

Amino-acids

At the exception of the glycine, the 20 amino-acids are chiral and available with a good ee (enantiomeric excess). They are bifunctional and levogyre but an enzymatic method can reverse their configuration.

Hydroxy-acids

They wear one or more OH and COOH and can have several stereocentres.

Carbohydrates

They are very cheap but we have to modify them and to protect some groups to use them correctly.

Terpenes

Most of them are used as precursors of agent of resolution or as ligands during asymmetric catalysis but they can also be used as starting reactant of other terpenes. They are cheap and available in large quantities. Their cyclic structure allows a better control of the strereoselectivity. However, they don’t have a lot of functional groups and their optic purity is not always very good.

Alcaloïdes

These secondary metabolites possess a nitrogen aromatic and are used as chiral catalysts or agents of resolution.

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• the GC with a chiral stationnary phase,
• HPLC with a chiral stationnary phase,
• NMR with chiral lanthanides.

Gazeous chromatography

The stationnary phase of the column is chiral and only composed of one enantiomer (for instance the R configuration). When a racemic melange passes through the column, the enantiomers R’ and S’ have a different interaction with the chiral stationnary phase with which they form diastereoisomeric complexes.

The main interactions can be H bonds, sterical interactions or dipole-dipole interactions.  As the interactions have not the same strength, the enantiomers R’ and S’ have a different retention time and exit the column separately, giving two peaks.

The advantages of this method is that one the condition of experimentation are setted up, the analysis is fast and has a high sensibility. It can detect an enantiomer even if its proportion in the solution is about 0.5%. Moreover, the analysis separates the species composing the sample and we can select the portion of the solution that we want to analyse next.

When we use this method, the racemic melange has first to be tested to be sure that the enantiomers can be separated and that the ratio of the areas of the peaks is 50:50.

NMR

This technique is not as sensible as the GC but it is very fast and cheap.

Normally, enantiomers have the same NMR spectrum because the interatomic distances are identical for the two enantiomers. Diastereoisomers have spectra that can be distinguished. To distinguish two enaintiomers we will thus transform them into diastereoisomers.

To do so, we use two types of chiral reactants.

• chiral agent of derivatisation
• chiral lanthanides

A chiral agent of derivatisation is a chiral compound optically pure that interacts with the enantiomers. The most used one is the Mosher’s acid (α-methoxy-α-trifluoromethylphenylacetic acid (MTPA)).

This reactant forms a covalent liaison with the enantiomer to form diastereoisomers with an important difference of chemical displacement. The ratio between the areas gives the enantiomeric excess.

The method with chiral lanthanides is more used than the method with a chiral agent of derivation.

Lanthanides allow a widening of the NMR spectrum  by the displacement of the signals towards the low fields. The salt of  lanthanide is added bit by bit to the sample to observe the separation of the signals. They form a complex with the enantiomer without covalent liaison but still give an important separation of chemical displacement.

The disadvantages of this method is that it takes more time (the salt is added slowly) and that the salt is expensive.

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