Question

At $25C$, the solubility product values of $AgCl$ and $AgCNS$ are $1.7\times {10}^{-10}$ and $1.0\times {10}^{-12}$ respectively. When water is saturated with both solids, calculate the ratio $\left[{Cl}^{-}\right]/\left[CN{S}^{-}\right]$ and also $\left[{Ag}^{+}\right]$ in the solution.

• A. $1.3\times {10}^2,1.7\times {10}^{-5}M$
• B. $1.7\times {10}^2,1.308\times {10}^{-5}M$
• C. $1.3\times {10}^4,1.308\times {10}^{-4}M$
• D. $1.7\times {10}^3,1.67\times {10}^{-6}M$

Answer 1

The correct option is B $1.7\times {10}^2,1.308\times {10}^{-5}M$

$AgCNS\rightleftharpoons {Ag}^{+}+CNS$
$(1-{\delta }_1)(\delta +{\delta }_2){\delta }_1$
$AgCl\rightleftharpoons {Ag}^{+}+Cl$
$(1-{\delta }_2)(\delta +{\delta }_2){\delta }_2$
${\left(K_{sp}\right)}_{AgCNS}=\left({\delta }_1\right)({\delta }_1+{\delta }_2)$ ................(i)
${\left(K_{sp}\right)}_{AgCl}=\left({\delta }_2\right)({\delta }_1+{\delta }_2)$ ......................(ii)
Adding equation (i) and (ii), we get
${({\delta }_1+{\delta }_2)}^2={\left(K_{sp}\right)}_{AgCNS}+{\left(K_{sp}\right)}_{AgCl}$
$({\delta }_1+{\delta }_2)=\sqrt{{\left(K_{sp}\right)}_{AgCNS}+{\left(K_{sp}\right)}_{AgCl}}=\sqrt{1.0\times {10}^{-12}+1.7\times {10}^{-10}}$
$\left(AgT\right)=({\delta }_1+{\delta }_2)=1.3\times {10}^{-5}$

Divide equation (ii) from (i), we get
$\frac{\left[{Cl}^{-}\right]}{\left[{CNS}^{-}\right]}=\frac{{\delta }_2}{{\delta }_1}=\frac{1.7\times {10}^{-10}}{1.0\times {10}^{-12}}\Rightarrow 1.7\times {10}^2$