Question

1. Suppose you have a solution that is 0.044 M in Mn²⁺and 0.31 M in Cr³⁺.  What range of pH values will allow you to separate these ions?  In other words, for what range of pH values will one of the hydroxides precipitate but not the other?

Answer 1

One of the most studied types of reactions are precipitation reactions, which lead to the formation of a non-soluble precipitate or solid that separates from the solution. Precipitation reactions usually require the presence of ions.

In order to know if an ion present in solution will form a precipitate when reacting with a certain compound that has been added to the solution containing the ions, is necessary to know the "solubility", which refers to the maximum amount of solute that can be dissolved in a certain amount of liquid . The term "saturated solution" is also usually used, which refers to that solution that contains the highest concentration of solute possible in a given volume of liquid.

Many times, it is desired to separate a type of ions from a solution, leaving other ions in it. This separation can be achieved by adding an appropriate reagent to the solution, which will react with the ion of interest to form an insoluble precipitate or solid. This solid can be separated from the solution and in this way we can separate the ion of interest.
However, if both types of ions present in the solution precipitate forming insoluble solids, they can also be separated if both of ions form precipitates of different solubilities, that is, if an ion forms the precipitate first (the least soluble, the one that reaches first the maximum amount of solute that can be dissolved) and then the most soluble ion. In this way it is possible to separate both ions.

The solubility of many compounds also depends on the pH. By adding OH^- ions (by adding a basic substance) to the solution containing the ions of this problem, we obtain the following solubility equilibria:

Mn(OH)_{2 (s)} ------> Mn²⁺_(aq) + 2 OH^-_(aq)

Cr(OH)_{3 (s)} ------> Mn³⁺_(aq) + 3 OH^-_(aq)

These solubility equilibria refers to a saturated solution of magnesium hydroxide and chromium hydroxide which is in contact with solid magnesium hydroxide and solid chromium hydroxide. The equilibrium constant for both compounds formed can be written as follows:

K_ps= [Mn²⁺] [OH^-]²

K_ps=[Cr³⁺] [OH^-]³

K_ps is the "constant of the product of solubility" and is the product of the molar concentrations of the ions in solution, each of them raised to the power of its stoichiometric coefficient (remember that the concentration of the solid is taken as a constant, so it is not written in the expression of the constant of equilibrium).

It can be observed that when OH^- ions are added, the equilibrium will shift to the left, increasing Mn(OH)₂ and Cr(OH)₃, so that the solubility decreases. By precipitating both ions at different times, they can be separated.

Mn(OH)₂

By stoichiometry it can be observed that 1 mol of Mn(OH)₂ produces 1 mol Mn²⁺ and 2 mol of OH^- in solution. If the problem tells us that the solution is 0.044 M in Mn²⁺ then, the moles of OH^- obtained in solution are:

[OH^-]=2 x 0.044 M=0.088 M; then we can calculate the pOH and then the pH:

pOH= -log (0.088 M)=1.055; pH=14-1.055=12.945

From pH 12.945 will begin to precipitate the Mn⁺².

Cr(OH)3

By stoichiometry it can be observed that 1 mol of Cr(OH)³ produces 1 mol Cr³⁺ and 3 mol of OH^- in solution. If the problem tells us that the solution is 0.31 M in Cr³⁺ then, the moles of OH^- obtained in solution are:

[OH^-]=3 x 0.31 M=0.93 M; then we can calculate the pOH and then the pH:

pOH= -log (0.93 M)=0.03; pH=14-0.03=13.97

From pH 13.97 will begin to precipitate the Cr⁺³.

The range of pH values that allows these ions to be separated is: 12.945-13.97 In this range of pH values the magnesium hydroxide will precipitate but not the chromium hydroxide.