Question

If the molar solubility $\left(inmol{L}^{-1}\right)$ of a sparingly soluble salt $A{B}_3$​ is $s$. The corresponding solubility product is $K_{sp}$. The solubility $\left(s\right)$ can be written in terms of $K_{sp}$ by the relation:

• A. $s=(128K_{sp}{)}^{\frac{1}{4}}$
• B. $s={\left(\frac{K_{sp}}{27}\right)}^{\frac{1}{4}}$
• C. $s={\left(\frac{K_{sp}}{256}\right)}^{\frac{1}{5}}$
• D. $s={\left(\frac{K_{sp}}{3}\right)}^{\frac{1}{4}}$

Answer 1

The correct option is B $s={\left(\frac{K_{sp}}{27}\right)}^{\frac{1}{4}}$

Let "s" $mol{L}^{-1}$ be the solubility of the $A{B}_3$
$A{B}_3\left(s\right)\rightleftharpoons A^{3+}\left(aq\right)+3B^{-}\left(aq\right)$
$t=t_{eq}c-ss3s$
so,
$K_{sp}=\left(s{)}^1\right(3s{)}^3{K}_{sp}=27s^4$
Thus the expression for solubility becomes,
$s^4=\left(\frac{K_{sp}}{27}\right)$
$\Rightarrow s={\left(\frac{K_{sp}}{27}\right)}^{\frac{1}{4}}$