Question

The solubility in water of a sparingly soluble salt $A{B}_2$ is $1.0\times {10}^{-5}mol{L}^{-1}.$ Its solubility product number will be

• A. $4\times {10}^{-10}$
• B. $1\times {10}^{-15}$
• C. $1\times {10}^{-10}$
• D. $4\times {10}^{-15}$

Answer 1

The correct option is D $4\times {10}^{-15}$

The dissociation of sparingly soluble salt is given by $A{B}_2\left(s\right)\rightleftharpoons A^{+2}\left(aq\right)+2B^{-}\left(aq\right)-ss$
$\left[A\right]=1.0\times {10}^{-5},\left[B\right]=[1.0\times {10}^{-5}],$
$K_{sp}=\left[B^{-}{]}^2\right[A^{+2}]$
$K_{sp}=[2\times {10}^{-5}{]}^2[1.0\times {10}^{-5}]$$K_{sp}=4\times {10}^{-15}$