Question

At ${25}^{\circ }C$, the $K_{sp}$ value of $AgCl$ is $1.8\times {10}^{-10}$. If ${10}^{-5}$ moles of $A{g}^{+}$ are added to solution then $K_{sp}$ will be :-

• A. $1.8\times {10}^{-15}$
• B. $1.8\times {10}^{-10}$
• C. $1.8\times {10}^{-5}$
• D. $1.8\times {10}^{10}$

Answer 1

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

$K_{sp}$ of $AgCl$ = $1.8\times {10}^{-10}$

$K_{sp}$ - solubility product

When $AgCl$ is dissolved in water.

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

When ${10}^{-5}$ mole of $A{g}^{+}$ ions are added it shifted backward and maintains equilibrium condition.

Therefore, $K_{sp}$ will remain the same.

Hence, option (B) is the correct answer.