Question

Solubility of $CaC{l}_2$ is $4\times {10}^{-8}$, then what would be its new solubility in the presence of ${10}^{-2}M$ $Ca(OH{)}_2$?

• A. $6\times {10}^{-11}mol{L}^{-1}$
• B. $8\times {10}^{-8}mol{L}^{-1}$
• C. $8\times {10}^{-11}mol{L}^{-1}$
• D. $9\times {10}^{-8}mol{L}^{-1}$

Answer 1

The correct option is C $8\times {10}^{-11}mol{L}^{-1}$

$\underset{1}{CaC{l}_2}\rightleftharpoons \underset{s}{C{a}^{+2}}+\underset{2s}{2C{l}^{-}}(let,initialsolubility=s)$

$\therefore K_{sp}=4s^3=4(4\times {10}^{-8}{)}^3=256\times {10}^{-24}\dots \dots (1)$
Again,
$\begin{array}{ccccc}Ca(OH{)}_2& \to & C{a}^{2+}& +& 2O{H}^{-}\\ C& & 0& & 0\\ 0& & C& & 2C\end{array}$
$\therefore$ $\begin{array}{cccc}CaC{l}_2& \rightleftharpoons & C{a}^{2+}+& 2C{l}^{-}\\ 1& & s^{{}^{\prime }}+C& 2s^{{}^{\prime }}(newsolubility=s^{{}^{\prime }})\end{array}$

$K_{sp}=\left[C{a}^{2+}\right][C{l}^{-}{]}^2$

$\Rightarrow K_{sp}=[s^{{}^{\prime }}+C][2s^{{}^{\prime }}{]}^2$

As we know that,
$s^{{}^{\prime }}<<C$

$\Rightarrow K_{sp}=(4s^{{}^{\prime }}{)}^{2}C\dots \dots (2)$

From eqn (1) and (2)

$(s^{{}^{\prime }}{)}^2=\frac{256\times {10}^{-24}}{4\times {10}^{-2}}=64\times {10}^{-22}$

$\Rightarrow s^{{}^{\prime }}=8\times {10}^{-11}mol{L}^{-1}$