Question

If 0.5 M ammonium benzoate is hydrolysed to 0.25 percent, hence its hydrolysis constant is?
A. $2.5\times {{10}^{-5}}$
B. $1.5\times {{10}^{-4}}$
C. $3.125\times {{10}^{-6}}$
D. $6.25\times {{10}^{-6}}$

Answer 1

Hint: A hydrolysis constant is an equilibrium constant for a hydrolysis reaction. We denote it with the symbol ${{K}_{h}}$. An acid dissociation constant ( ${{K}_{a}}$) is a quantitative measure of the strength of an acid in solution. A base dissociation constant ( ${{K}_{b}}$) is a quantitative measure of the strength in solution.

Formula Used: Formula of hydrolysis constant is given by:
H = $\sqrt{\dfrac{{{K}_{h}}}{C}}$; where ${{K}_{h}}$ is the hydrolysis constant
And h is the degree of hydrolysis
And c = concentration

Complete Step by Step Answer:
We know h = $\sqrt{\dfrac{{{K}_{h}}}{C}}$ … (1)
where ${{K}_{h}}$ is the hydrolysis constant
And h is the degree of hydrolysis = $0.25%$= 0.25
And c = concentration = 0.5 M
From equation (1), we get
$\dfrac{0.25}{100}=\sqrt{\dfrac{{{K}_{h}}}{0.5}}$

Solving further, we get
Or ${{\left( \dfrac{0.25}{100} \right)}^{2}}=\dfrac{{{K}_{h}}}{0.5}$
Therefore, ${{K}_{h}}=\dfrac{0.25\times 0.25\times 0.5}{100\times 100}$
Then ${{K}_{h}}=3.125\times {{10}^{-6}}$
Thus, Option (C) is correct.

Note: Solubility product is a type of equilibrium constant whose value depends on the temperature. It is denoted by ${{K}_{sp}}$. It depends upon the temperature. It usually increases with the increase in temperature because of the increased solubility. It is an equilibrium constant that gives the equilibrium between the solid solute and its ions that are dissolved in the solution.