Question

The solubility product of silver chloride is $1.8\times {10}^{-18}$ at $298K$. The solubility of $AgCl$ in $0.01$ M $HCl$ solution in mol/dm${}^3$ is:

• A. $8.6\times {10}^{-8}$
• B. $2.4\times {10}^{-9}$
• C. $1.8\times {10}^{-8}$
• D. $9\times {10}^{-9}$

Answer 1

The correct option is C $1.8\times {10}^{-8}$

Let solubility of AgCl in 0.01 M HCl = x mol/L

$AgCl\to A{g}^{+}+C{l}^{-}$

$\therefore \left[A{g}^{+}\right]=xmol/L$

$C{l}^{-}=\left[HCl\right]=0.01M={10}^{-2}M$

$k_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

or $1.8\times {10}^{-10}=\left(x\right)\times \left({10}^{-2}\right)$

$\therefore x=1.8\times {10}^{-8}$
Ans-option C