Question

Consider a sparingly soluble salt $AB$ having solubility as $smol{L}^{-1}$. What will be the value of $K_{sp}$ for $AB$ salt.

• A. $smol{L}^{-1}$
• B. $\sqrt[3]{3s}(mol{L}^{-1}{)}^{\frac{1}{3}}$
• C. $\sqrt{s}(mol{L}^{-1}{)}^{0.5}$
• D. $s^2(mol{L}^{-1}{)}^2$

Answer 1

The correct option is D $s^2(mol{L}^{-1}{)}^2$

Consider a saturated solution of sparingly soluble salt AB,
Equilibrium is established between undissolved AB and its ions in the solution
$AB\left(s\right)\rightleftharpoons A^{+}\left(aq\right)+B^{-}\left(aq\right)$
Applying laws of mass action :
$K_c=\frac{\left[A^{+}\right]\left[B^{-}\right]}{\left[AB\right]}$
$K_c\times \left[AB\right]=\left[A^{+}\right]\left[B^{-}\right]$
Amount of salt in contact with the saturated solution does not change with time
Therefore $\left[AB\right]$ remains constant. So,
$K_{sp}=\left[A^{+}\right]\left[B^{-}\right]$ where $K_{sp}=K_c\times \left[AB\right]$
$K_{sp}=\text{solubility product constant}$

Calculation of $K_{sp}$ for $AB$ type sparingly sobule salt:

Let $smol{L}^{-1}$ be the solubility then,
$AB\left(s\right)\rightleftharpoons A^{+}\left(aq\right)+B^{-}\left(aq\right)$
At $t=0C00$
At $t=t_{eq}C-sss$

$K_{sp}=\left[A^{+}\right]\left[B^{-}\right]K_{sp}=s\times s$
$K_{sp}=s^2(mol{L}^{-1}{)}^2$