Equilibrium - Result Question 88
####92. The solubility product of $AgI$ at $25^{\circ} C$ is $1.0 \times 10^{-16} mol^{2} L^{-2}$. The solubiliy of $AgI$ in $10^{-4} N$ solution of $KI$ at $25^{\circ} C$ is approximately (in $mol L^{-1}$ )
(a) $1.0 \times 10^{-8}$
(b) $1.0 \times 10^{-16}$
(c) $1.0 \times 10^{-12}$
(d) $1.0 \times 10^{-10}$
[2003]
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Solution:
- (c) $K _{s p}$ for $AgI=1 \times 10^{-16}$
In solution of $KI, I^{-}$would be due to the both $AgI$ and $KI, 10^{-4}$ solution $KI$ would provide $=10^{-4} I^{-}$
AgI would provide, say $=x I^{-}(x$ is solubility of AgI)
Total $I^{-}=(10^{-4}+x), K _{\text{sp }}$ of $AgI=(10^{-4}+x) x$
$\Rightarrow K _{s p}=10^{-4} x+x^{2}$
as $x$ is very small
$\therefore \quad x^{2}$ can be ignored
$\therefore \quad 10^{-4} x=10^{-16}$
or $\underset{\text{ (solubility) }}{x}=\frac{10^{-16}}{10^{-4}}=10^{-12}(molL^{-1})$
