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>> Applications of Solubility Product-Part1
Applications of Solubility Product
(i) Purification of common salt
Natural common salt consists of many insoluble and soluble impurities. Saturated solution of common salt is prepared and insoluble impurities are filtered off. Hydrogen chloride gas (HCl) is circulated through the saturated solution. HCl and NaCl dissociate into their respective ions as:
NaCl ↔ Na+ + Cl-
HCl ↔ H+ + Cl-
The concentration of Cl- ions increases considerably in solution due to ionization HCl. Hence, the ionic product [Na+][Cl-] exceeds the solubility product of sodium chloride and, therefore, pure sodium chloride precipitates out from solution.
(ii) Salting out of soap
Soap is a sodium salt of higher acids. From the solution, soap is precipitated by the addition of concentrated solution of sodium chloride. Soap and sodium chloride are present in the form of ions.
CnH2n+1 COONa ↔ CnH2n+1 COO- + Na+
NaCl ↔ Na+ + Cl-
Thus, the concentration of Na+ ions increases considerably on addition of NaCl solution. Hence, the ionic product [CnH2n+1COO-] [Na+] exceeds the solubility product of soap and, therefore, soap precipitates out from the solution.
(iii) Manufacture of sodium bicarbonate (baking soda):
In Solvay's soda process. CO2 gas is passed through ammonical brine to precipitate out NaHCO3.
NH4OH + CO2 → NH4HCO3
NH4HCO3 + NaCl → NaHCO3 + NH4Cl
NaHCO3 is precipitated first because of its lower solubility product as compared to those of NH4Cl, NH3HCO3 and NaCl.
Thus, baking soda (NaHCO3) can be quantifiably estimated.
(iv) Application of solubility product in quantitative analysis
1. Estimation of barium as barium sulphate:
H2SO4 as precipitating agent is added to the aqueous solution of BaCl2.
BaCl2 + H2SO4 → BaSO4 + 2HCl
Precipitation of BaSO4 takes place when its ionic product exceeds solubility product. H2SO4 is added in slight excess to ensure complete precipitation. Large excess of H2SO4 is harmful for complex formation.
2. Estimation of silver as silver chloride:
NaCl solution is added to the silver nitrate solution, slight excess of NaCl is added to ensure complete precipitation.
AgNO3 + NaCl → AgCl + NaBO3
Again, precipitation of AgCl takes place when ionic product of AgCl exceeds its solubility product.
3. In a similar manner. We estimate lead as lead chromate, calcium as calcium oxalate, etc.
(i) Precipitation of the sulphides of group II and IV
Hydrogen sulphide is a weak electrolyte and is used for the precipitation of various sulphides of group II and IV in quantitative analysis.
It ionizes to a small extent in water:
H2S ↔ 2H+ + S2-
Applying law of mass action,
K = ([H+ ]2 [S2- ])/[H2 S]
The concentrations of S2- ions can be decreased by increasing concentration of H+ions and it can be increased by decreasing concentration of H+ ions. In group II, lower concentration of sulphide ions is required as the solubility products of the sulphides of group II are low while higher concentration of sulphide ions is required in group IV as the solubility products of the suphides of group IV are high. The values of solubility products of various sulphides are given below.
The concentration of S2- ions in group II is lowered by maintaining acidic medium. In the presence of HCl, the ionization of H2S is suppressed due to common ion effect. The concentration is so adjusted that only ionic products of the sulphides of group II exceed their solubility products and, therefore, get precipitated. However, CdS has somewhat higher value. For its precipitation, dilution of the solution is done which increases ionization of H2S and thereby increasing concentration of S2- ions.
In group IV, higher concentration of S2- ions is needed. This is done by changing the medium for acidic to alkaline. Ammonium hydroxide is added, the OH- ions furnished by NH4OH remove H+ ions from solution in the form of water molecules as,
H+ + OH- ↔ H2O
More of the ionization of H2S occurs and, thus, concentration of S2- ions increases. It becomes so high that ionic products of the sulphides of group IV exceed their solubility products add they get precipitated.
(i) Precipitation of III group hydroxides
When NH4OH is added in presence of NH4Cl then precipitation of II group hydroxides takes place, i.e., Al(OH)3, Fe(OH)3 and Cr(OH)3 are precipitated. Solubility product of III group hydroxides is less than those of higher group hydroxides.
NH4Cl → NH+4 + Cl-
NH4OH ↔ NH+4 + OH-
NH+4 ion furnished by NH4Cl lowers the ionisatin of NH4OH and hence the concentration of hydroxide ion OH-. At low concentration of hydroxide ion only III group hydroxides precipitate.
It is a technique of separating two or more ions from a solution by adding a reagent that precipitates first one ion and then the second.
Let us suppose 0.1 M Ba2+ and 0.1 M Sr2+ in aqueous solution. K2CrO4 is added as precipitating agent. Ksp BaCrO4 is 1.2 × 10-10 and Ksp SrCrO4 is 3.5 × 10-5.
[CrO2-4]concentration required to precipitate BaCrO4
= Ksp/[Ba2+] = 1.2 * 10-10/0.1 = 1.2 * 10-9
BaCrO4 will precipitate first because it requires low concentration of CrO2-4 ions. On addition of chromate ions, BaCrO4 starts precipitating when chromate ion concentration reaches 1.2 × 10-9 M. When CrO2-4 ion concentration reaches upto 3.5 × 10-4 M, then SrCrO4 also starts precipitating.
Remaining concentration of Ba2+ when SrCrO4 starts precipitation.
=(Ksp BaCrO4)/[CrO42-] = (1.2×10-10)/(3.5×10-4)= 3.4×10-7 M
% remaining concentration = (3.4 * 10-7)/0.1 * 100
Let us consider dissociation of the ion FeBr+.
FeBr+ ↔ Fe2+ + Br-
Dissociation constant for above equilibria may be given as
Kd = [Fe2+ ][Br- ]/[FeBr+ ]
Reciprocal of dissociation constant is called stability constant.
Ks = [FeBr+ ]/([Fe2+] [Br- ])
Let us consider the formation of complex K2Cd(CN)4, Complex ion is Cd(CN42-) where oxidation state of central metal Cd2+ is (2+). Complexing process proceeds in four steps as
Here Ks = K1K2K3K4.
Significance of stability constant:
Greater will be the value of stability constant more stable will be the complex.
(a) If on addition of a common ion in a salt solution (sparingly soluble), formation of complex ion takes place, then ionization increases, i.e., equilibrium shifts towards right hand direction to maintain the value of Ksp constant. It means, addition of common ion in the case of complex formation increases the solubility of the sparingly soluble salt which is against the concept of common ion effect.
(b) When we add an electrolyte to another electrolyte solution having no common ion, then ionization of the later increases.
(c) For a given electrolyte solubility product is always constant at a particular temperature.
Solubility of Metal Hydroxides in Acid Medium
H+ ion furnished by the medium effects the solubility of metal hydroxide, say M(OH)2, because of neutralization of OH- ion by H+ ion.
M(OH)2 ↔ M2+ + 2OH-
Ksp of M(OH)2 = [M2+][OH-]2
[M2+] = Ksp/[OH-]2
[H+] [OH-] = Kw = 10-14
[OH-]2 = 10-28/[H+]2
From Eqs. (i) and (ii), we have
[M2+] = Ksp[H+]2/10-28