Question

A saturated solution of silver benzoate has a pH of 8.63. ${{K}_{a}}$ for benzoic acid is $6.5\text{ x 1}{{\text{0}}^{-5}}$. ${{K}_{sp}}$ of silver benzoate will be
(A) $1.4\text{ x 1}{{\text{0}}^{-2}}\text{ }mo{{l}^{2}}litr{{e}^{-2}}$
(B) $1.9\text{ x 1}{{\text{0}}^{-2}}\text{ }mo{{l}^{2}}litr{{e}^{-2}}$
(C) $2.4\text{ x 1}{{\text{0}}^{-2}}\text{ }mo{{l}^{2}}litr{{e}^{-2}}$
(D) $none of these$

Answer 1

Hint:. Find pOH for silver benzoate. Silver benzoate is a salt of a weak acid and strong base. Then find hydrolysis constant for the reaction. Equate the value of hydrolysis constant to the formula for hydrolysis. With this, you can find the concentration and substitute the value of concentration in the formula for ${{K}_{sp}}$ to obtain the answer.

Complete step by step answer:
Solubility is defined as the property of a substance (solute) to get dissolved in a solvent in order to form a solution.
The solubility product constant is the equilibrium constant for the dissolution of a solid solute in a solvent to form a solution. It is denoted by the symbol ${{K}_{sp}}$. We can form an idea about the dissolution of an ionic compound by the value of its solubility product.
We will now find the value of ${{K}_{sp}}$ of silver benzoate with the values given to us.
pH + pOH = 14 (Ionic product of water)
pOH = 14 - 8.63 = 5.37
$[\text{O}{{\text{H}}^{-}}]$ = Antilog( -pOH) = Antilog( -5.37) = $4.27\text{ x 1}{{\text{0}}^{-6}}\text{ mol/d}{{\text{m}}^{3}}$
We know that Silver benzoate is a salt of a weak acid and strong base.

Hence,
${{\text{K}}_{h}}\text{ = }\dfrac{{{\text{K}}_{w}}}{{{\text{K}}_{a}}}\text{ = }\dfrac{{{10}^{-14}}}{6.5\text{ x 1}{{\text{0}}^{-5}}}\text{ = 1}\text{.54 x 1}{{\text{0}}^{-10}}$
${{\text{K}}_{h}}\text{ = }{{\text{h}}^{2}}\text{C}$
Where,
h is the degree of hydrolysis
C is the concentration of solution
So, h = $\sqrt{\dfrac{{{\text{K}}_{h}}}{\text{C}}}$
Also,
$[\text{O}{{\text{H}}^{-}}]$ = nhC = $\text{n}\sqrt{{{\text{K}}_{\text{h}}}\text{C}}$
C = $\dfrac{\text{( }\!\![\!\!\text{ O}{{\text{H}}^{\text{-}}}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}\text{)}}{{{\text{K}}_{\text{h}}}}$ = $\dfrac{{{(4.27\text{ x 1}{{\text{0}}^{-6}})}^{2}}}{(1.54\text{ x 1}{{\text{0}}^{-10}})}$ = 0.1178
For Silver benzoate,
${{\text{K}}_{\text{sp}}}\text{ = }{{\text{C}}^{\text{2}}}$= ${{(0.1178)}^{2}}$ = $\text{1}\text{.38 x 1}{{\text{0}}^{\text{-2}}}\text{ mo}{{\text{l}}^{\text{2}}}\text{litr}{{\text{e}}^{\text{-2}}}$

Therefore, the correct answer is option (A).

Note: The common ion effect describes the effect of adding a common ion on the ​equilibrium of the new solution. The common ion effect generally decreases the ​solubility of a solute. The equilibrium shifts to the left to relieve off the excess product formed. When a strong electrolyte is added to a solution of a weak electrolyte, the solubility of the weak electrolyte decreases leading to the formation of precipitate at the bottom of the testing apparatus.