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

Answer 1

The correct option is A

$4\times {10}^{-15}$

$A{B}_2\leftrightharpoons A^{+}+2B^{-}$
At equilibrium, $\left[A^{+}\right]=1\times {10}^{-5},\left[B^{-}\right]=2\times {10}^{-5}$

$K_{sp}=[1\times {10}^{-5}][2\times {10}^{-5}{]}^2=4\times {10}^{-15}$