Question

If the concentration of $A^{+}$ ions in a solution of a sparingly soluble salt, $AB$ is ${10}^{-3}M.$ Find the concentration of $B^{-}$ ions at which the precipitation of $AB$ starts.
Given : Solubility product of AB is $1.0\times {10}^{-8}$

• A. $<{10}^{-8}M$
• B. $>{10}^{-5}M$
• C. Between ${10}^{-8}Mto{10}^{-7}M$
• D. Between ${10}^{-7}Mto{10}^{-8}M$

Answer 1

The correct option is B $>{10}^{-5}M$

Precipitation of salt will occur when $Ksp<Q_{sp}{Q}_{sp}=\left[A^{+}\right]\left[B^{-}\right]$

$K_{sp}<\left[A^{+}\right]\left[B^{-}\right]$
It is given that $\left[A^{+}\right]={10}^{-3}M.$ and Solubility product of AB is $K_{sp}=1.0\times {10}^{-8}{M}^2$

So, ${10}^{-8}<\left({10}^{-3}\right)\times \left[B^{-}\right]$
$\left[B^{-}\right]>{10}^{-5}M$