Solubility product of AgCl is 10^{-10}, M^2 . minimum volume (in m^3 ) of water required to dissolve

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6-2.Equilibrium-II (Ionic Equilibrium)

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The solubility product of $AgCl$ is $10^{-10}\, M^2$. The minimum volume (in $m^3$) of water required to dissolve $14.35\, mg$ of $AgCl$ is approximately

A

$10$

B

$0.1$

C

$100$

D

$0.01$

Solution

$A g C l \rightleftharpoons A g^{+}+C l^{-}$

$\quad S \quad S$

$\Rightarrow K_{s p}=S^{2}$

$\Rightarrow 10^{-10}=S^{2} \quad ; \quad S=10^{-5} \,M$

$\Rightarrow$ Molecular weight of $A g C l=143.5 \,g$

$\Rightarrow 10^{-5}=\frac{14.35 \times 10^{-3}}{143.5} \times \frac{1000}{V(m l)}$

$10^{-1}=\frac{1000}{V(m l)}$

$V_{m l}=10000 ml =10\,litres$

Standard 11

Chemistry

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