Question

In a saturated solution of the sparingly solution strong electrolyte ${Agl}_3$ (Molecular mass = 283), the equilibrium which sets in is: ${AglO}_3\left(s\right)\rightleftharpoons {Ag}^{+}\left(aq\right)+{lO}_3^{-}\left(aq\right)$
If the solubility product constant $K_{sp}$ of ${AglO}_3$ at a given temperature is $1.0\times {10}^{-8}$ what is the mass of ${AglO}_3$ contained in 100 mL of its saturated solution?

• A. $1.0\times {10}^{-7}g$
• B. $1.0\times {10}^{-4}g$
• C. $28.3\times {10}^{-2}g$
• D. $2.83\times {10}^{-3}g$

Answer 1

Answer: (D)

$2.83\times {10}^{-3}g$

$K_{sp}=1.0\times {10}^{-8}$

Here, $K_{sp}=S^2$

$\Rightarrow \sqrt{1\times {10}^{-8}}=S$

$\Rightarrow S={10}^{-4}mol/L$

Now, Molarity= $\frac{m}{283}\times \frac{1000}{100}$

$\Rightarrow {10}^{-4}=\frac{m}{283}\times 10$

$\Rightarrow m=\frac{{10}^{-4}\times 283}{10}=2.83\times {10}^{-3}g$