Question

The solubility in water of a sparingly soluble salt $AB_{2}$ 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

Answer: (A)

$AB_{2} \rightleftharpoons \underset{s}{A^{2+}}+\underset{2s}{2B^{-}}$

$ K_{s p} =\left[A^{2+}\right][B]^{2}$

$=(S)(2 S)^{2}=4 S^{3} $

$=4\left(1 \times 10^{-5}\right)^{3} $

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