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. $\frac{2.49}{1.84}\times {10}^{-9}$
• D. $2.49\times {10}^{-10}M$

Answer 1

The correct option is D $2.49\times {10}^{-10}M$

We know,

$Cd(OH{)}_2\rightleftharpoons C{d}^{+2}+2O{H}^{-}$

Now,

$K_{sp}=4(S{)}^3$

$=4\times (1.84\times {10}^{-5}{)}^3$

$Cd(OH{)}_2\rightleftharpoons C{d}^{+2}+2O{H}^{-}$

$S^{\prime }$ $({10}^{-2}+S^{\prime })\simeq {10}^{-2}$

$S^{\prime }\times ({10}^{-2}{)}^2=4\times (1.84\times {10}^{-5}{)}^3$

$S^{\prime }=4\times (1.84{)}^3\times {10}^{-11}$

$S^{\prime }=2.491\times {10}^{-10}M$