Question

The molar solubility of $AgCl$ in $1.8$ M $AgN{O}_3$ solution is:
$(K_{sp}$ of $AgCl=1.8\times {10}^{-9})$

• A. ${10}^{-5}$
• B. ${10}^{-9}$
• C. $1.8\times {10}^{-5}$
• D. $1.8\times {10}^{-10}$

Answer 1

The correct option is B ${10}^{-9}$

The expression for the solubility product of AgCl is $K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

In presence of 1.8 M silver nitrate solution,

$\left[A{g}^{+}\right]=1.8+s$ and $\left[C{l}^{-}\right]=s$

But $K_{sp}=1.8\times {10}^{-9}$

Substitute values in the expression for the solubility product.

$1.8\times {10}^{-9}=(1.8+s)s=s^2+1.8s$

Hence, $s={10}^{-9}$