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Use the References to access important values il needed for this question. The molar solubility of zinc sulfide in a 0.155 M ammonium sulfide solution is Submit Answer Retry Entire Group 9 more group attempts remaining

Answer 1

Ans :- 1.187 x 10^-24 M

Explanation :-

Let Molar solubility of ZnS in (NH₄)₂S = S mol/L

Partial dissociation of sparingly soluble salt ZnS in aqueous solution is :

...........................ZnS (s) <-------------------> Zn²⁺ (aq) .............+...............S^2- (aq)

At equilibrium ..............................................S mol/L................................S mol/L

Also, Complete dissociation of strong electrolyte i.e. (NH₄)₂S is :

....................(NH₄)₂S (aq) -----------------------> 2 NH₄⁺ (aq) .........+...........S^2- (aq)

......................0.155 mol/L.....................................2x0.155 mol/L...............0.155 mol/L

So,

Total concentration of S^2- = [S^2-]_Total = (S + 0.155 ) mol/L

Here,

S^2- act as common ion, which will increase the concentration of S^2- in products side and therefore according to Le-Chatelier's principle equilibrium will shifts in backward direction that's why solubility of ZnS starts decreases.

Solubility product (K_sp) is equal to the product of the molar concentration of products in which stoichiometric coefficients of products are raised to the power at equilibrium stage of the reaction.

At 25 ⁰C, K_sp value of sparingly soluble salt ZnS = 1.84 x 10^-25

Expression of K_sp of ZnS is :

K_sp = [Zn²⁺].[S^2-]_Total

1.84 x 10^-25 = S (S + 0.155 )

because S<<<0.155, therefore neglect S as compare to 0.155.

So,

1.84 x 10^-25 = S (0.155 )

S = 1.84 x 10^-25 / 0.155

S = 1.187 x 10^-24 mol/L

Hence, Molar solubility of ZnS in 0.155 M (NH₄)₂S solution = 1.187 x 10^-24 M