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

Answer 1

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

The equilibrium established will be:
$Cd\left(OH{)}_2\right(s)\rightleftharpoons C{d}^{2+}(aq)+2O{H}^{(}aq)$
$s2s$

At equilibrium, $K_{sp}=s(2s{)}^2=4s^2$

Putting the values,
$\Rightarrow K_{sp}=4\times (1.84\times {10}^{-5}{)}^3$


Solubility in buffer solution having $pH=12$

As, $pH=12\Rightarrow pOH=2\Rightarrow \left[O{H}^{-}\right]={10}^{-2}$

In case of buffer solution,
let solubility be s'

$Cd(OH{)}_2\rightleftharpoons \underset{s^{\prime }}{C{d}^{2+}}+\underset{2s^{\prime }+{10}^{-2}\equiv {10}^{-2}}{2O{H}^{-}}$
Here we took, $2s^{\prime }+{10}^{-2}\equiv {10}^{-2}\text{because}{s}^{\prime }<<{10}^{-2}$

Comparing the $K_{sp}$ from both the cases,
$\therefore K_{sp}=4\times (1.84\times {10}^{-5}{)}^3=s^{\prime }({10}^{-2}{)}^2\Rightarrow s^{\prime }=\frac{24.9\times {10}^{-15}}{{10}^{-4}}=2.49\times {10}^{-10}M$