Question

At ${18}^{o}C$, the solubility of $CdS$ in water is $6.33\times {10}^{-15}$ M. What is the concentration of $C{d}^{2+}$ ion in a solution of pH = 1 saturated with $H_2$S gas, in which concentration of $H_{2}S=0.1M$ ? The product of the first and second ionization constants of $H_{2}S$ is $1.1\times {10}^{-22}$ at this temperature.

• A. $6.343\times {10}^{-8}M$
• B. $4.368\times {10}^{-8}M$
• C. $4.34368\times {10}^{-9}M$
• D. $3.636\times {10}^{-8}M$

Answer 1

The correct option is D $3.636\times {10}^{-8}M$

In absence of hydrogen sulphide, $\left[CdS\right]=\left[C{d}^{2+}\right]=\left[S^{2-}\right]=6.33\times {10}^{-15}$ M

$K_{sp}=\left[C{d}^{2+}\right]\left[S^{2-}\right]=(6.33\times {10}^{-15}{)}^2=4\times {10}^{-29}$

For hydrogen sulphide, $K_{a1}\times K_{a2}=\frac{\left[H^{+}{]}^2\right[S^{2-}]}{\left[H_{2}S\right]}$

$1.1\times {10}^{-22}=\frac{\left(0.1{)}^2\right[S^{2-}]}{0.1}$

$\left[S^{2-}\right]=1\times {10}^{-21}$

$\left[C{d}^{2+}\right]=\frac{K_{sp}}{\left[S^{2-}\right]}=\frac{4\times {10}^{-29}}{1.1\times {10}^{-21}}=3.636\times {10}^{-8}$ M.

Option D is correct.