Question

Solubility product of a sulphide MS is 3 × 10^-25 and that of another sulphide NS is 4 × 10^-40. In ammonia solution
A. Only NS gets precipitated
B. Only MS gets precipitated
C. No sulphide precipitates
D. Both sulphides precipitate

Answer 1

Hint: In this question, the concepts of common ion effect, ionic product and solubility product are used. An increase in the concentration of one of the ions dissociated in the solution by the addition of another species containing the same ion will result in an increase in the degree of association of ions in a solution where there are several species associating with each other via a

Complete step by step solution:
By keeping the medium acidic while NH₄Cl (ammoniacal chloride) is present, the concentration of S^2- ions is reduced. Due to the common ion effect, H₂S does not ionise. The ionic product is therefore less than the solubility product.
Thus, both sulphides precipitate.
Thus, Option (D) is correct.

Additional Information: The equilibrium constant for a solid material dissolving in an aqueous solution is the solubility product constant, Ksp. It stands for the degree of solute dissolution in solution. A substance's Ksp value increases with how soluble it is. For the reaction
a A(s)$\rightleftharpoons $cC(aq)+dD(aq)
${{K}_{sp}}={{[C]}^{c}}{{[D]}^{d}}$

Ionic product is the sum of the ion concentrations in a salt solution, each raised to the power indicated
by its stoichiometric coefficient. In a saturated solution, the solubility product is the result of the molar concentrations of the ions, whereas the ionic product is true of any solution.

Note: When
Ionic Product = Solubility Product is when the solution is saturated but there is no precipitation.
Ionic Product < Solubility Product is when the solution is diluted and there is no precipitation.
Ionic Product > Solubility Product is when the solution is supersaturated and contains more ions than it can keep in a solution.