Question

Given that, the solubility product $K_{sp}$ of $AgCl$ is $1.8 \times 10^{-10}$, the concentration of $Cl^-$ ions that must be exceeded before $AgCl$ will precipitate from a solution containing $4 \times 10^{-3}\, M \,Ag^+$ ions is

• A. $4.5 \times 10^{-8}\, M$
• B. $4 \times 10^{-8}\, M$
• C. $1.8 \times 10^{-8}\, M$
• D. $1 \times 10^{-8}\, M$

Answer 1

Answer: (A)

$K_{sp}=\left[Ag^{+}\right]\left[Cl^{-}\right]$
$\left[Cl^{-}\right]=\frac{1.8\times10^{-10}}{4\times10^{-3}}$
$=4.5\times10^{-8}\,M$
If concentration of $Cl^{-}$ exceeds $4.5\times10^{-8}\,M$ then ionic product becomes greater than $K_{sp}$ and precipitation takes place.