Question

A sample of AgCl was treated with 5.00 ml of 1.5 M $N{a}_{2}C{O}_3$ solution to give $A{g}_{2}C{O}_3$ .The ramaining solution contained 0.0026 g of $C{l}^{-}$ per litre. Calculate the solubility product of AgCl. $[K_{sp,A{g}_{2}C{O}_3}=8.2\times {10}^{-12}]$

• A. $1.1\times {10}^{-2}$
• B. $1.71\times {10}^{-11}$
• C. $1.71\times {10}^{-10}$
• D. $1.32\times {10}^{-9}$

Answer 1

The correct option is C $1.71\times {10}^{-10}$

The expression for the solubility product of $A{g}_{2}C{O}_3$ is

$K_{sp,A{g}_{2}C{O}_3}=\left[A{g}^{+}{]}^2\right[C{O}_3^{2-}]$

Substitute values in the above expression.

$8.2\times {10}^{-12}=[A{g}^{+}{]}^2\times 1.5$

Hence, $\left[A{g}^{+}\right]=2.3\times {10}^{-6}$

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

Substitute values in the above expression.

$K_{sp,AgCl}=2.3\times {10}^{-6}\times \frac{0.0026}{35.5}=1.71\times {10}^{-10}$.