The solubility product of a sparingly soluble salt $A_2 B$ is $3.2 \times 10^{-11}$. Its solubility in…

The solubility product of a sparingly soluble salt $A_2 B$ is $3.2 \times 10^{-11}$. Its solubility in $\mathrm{mol} \mathrm{L}^{-1}$ is
  1. $4 \times 10^{-4}$
  2. $2 \times 10^{-4}$
  3. $6 \times 10^{-4}$
  4. $3 \times 10^{-4}$

Solution

For the salt of type $A_2 B$, solubility product $ \left(K_{\mathrm{sp}}\right)=x^x \cdot y^y \cdot S^{x+y} $ where, $x$ and $y$ are the number of moles of $A$ and $B$ respectively. $ S=\text { solubility. } $ Thus, $\quad K_{\mathrm{sp}}=2^2 \times 1^1 \cdot S^{2+1} \Rightarrow K_{\mathrm{sp}}=4 S^3$ Given, $K_{\mathrm{sp}}$ for $A_2 B=3.2 \times 10^{-11}$ $ \begin{array}{ll} \therefore & 3.2 \times 10^{-11}=4 S^3 \\ \text { or, } & S^3=\frac{3.2}{4} \times 10^{-11}=8 \times 10^{-12} \\ \text { Thus, } & S=2 \times 10^{-4} \end{array} $ Hence, option (b) is the correct answer

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

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