Question

The Solubility $s$ of $AgCl\left(s\right)$ in pure water and in $0.1M$ $NaCl$ at ${25}^{\circ }C$ respectively are:

$K_{sp}\left(AgCl\right)$ = $2.8\times {10}^{-10}$

• A. $1.673\times {10}^{-5}mol/litre$ in water, $1.4\times {10}^{-9}mol/litre$ in $0.1MNaCl$
• B. $2.8\times {10}^{-9}mol/litre$ in water, $1.673\times {10}^{-5}mol/litre$ in $0.1MNaCl$
• C. $1.673\times {10}^{-5}mol/litre$ in water, $2.8\times {10}^{-9}mol/litre$ $0.1MNaCl$
• D. $1.732\times {10}^{-9}mol/litre$ in water, $2.8\times {10}^{-5}mol/litre$ $0.1MNaCl$

Answer 1

The correct option is C

$1.673\times {10}^{-5}mol/litre$ in water, $2.8\times {10}^{-9}mol/litre$ $0.1MNaCl$

$AgCl\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$

Let the solubility of $AgCl\left(s\right)$ in pure water be $s$

$K_{sp}\left(AgCl\right(s\left)\right)$ = $s^2$ = $2.8\times {10}^{-10}$

Solving, $s$ = $1.673\times {10}^{-5}mol/litre$ in water.

Now, $NaCl$ is a strong electrolyte. It will readily dissociate completely and hence [$C{l}^{-}$] = $0.1M$

This is very large compared to what might be the [$C{l}^{-}$] contributed by the solid $AgCl$. Neglecting the [$C{l}^{-}$] contributed by $AgCl\left(s\right)$, we have

$K_{sp}$ = $\left[A{g}^{+}\right]\left[C{l}^{-}\right]$ = $s\left(0.1\right)$

Solving for $s$, we get $s_{new}$ = $2.8\times {10}^{-10}mol/litre\ast 0.1MNaCl$