Question

The molar solubility of $Cd(OH)_2$ is $1.84 \times 10^{-5} \; M$ in water. The expected solubility of $Cd(OH)_2$ in a buffer solution of $pH = 12$ is :

• A. $ 6.23 \times 10^{-11} M$
• B. $ 1.84 \times 10^{-9} M$
• C. $\frac{2.49}{1.84} \times 10^{-9} M$
• D. $ 2.49 \times 10^{-10} M $

Answer 1

Answer: (D)

$K_{sp } = 4 (s)^3$
$ = 4 \times ( 1.84 \times 10^{-5})^3$
$Cd(OH)_2 \rightleftharpoons Cd^{2+} + 2OH^{-}$
$S' S' \; (10^{-2} + S') \simeq 10^{-2}$
$ S' \times (10^{2})^2 = 4 \times (1.84 \times 10^{-5} )^3 $
$S' = 4 \times (1.84)^3 \times 10^{11}$
$(S') = 2.491 \times 10^{-10} M $