Question

Simultaneous solubility of $AgCNS\left(a\right)$ and $AgBr\left(b\right)$ in a solution of water will be

$K_{{sp}_{\left(AgBr\right)}}=5\times {10}^{-13}$ and $K_{{sp}_{\left(AgCNS\right)}}={10}^{-12}$

• A. $a=4.08\times {10}^{-7}mol$ ${litre}^{-1}$; $b=8.16\times {10}^{-7}$ $mol$ ${litre}^{-1}$
• B. $a=4.08\times {10}^{-7}mol$ ${litre}^{-1}$; $b=4.08\times {10}^{-7}$ $mol$ ${litre}^{-1}$
• C. $a=8.16\times {10}^{-7}mol$ ${litre}^{-1}$; $b=4.08\times {10}^{-7}$ $mol$ ${litre}^{-1}$
• D. None of these

Answer 1

Answer: (C)

$a=8.16\times {10}^{-7}mol$ ${litre}^{-1}$; $b=4.08\times {10}^{-7}$ $mol$ ${litre}^{-1}$

Suppose solubility of $AgCNS$ and $AgBr$ in a solution are $a$ and $b$ respectively.

$AgCNS\left(s\right)\rightleftharpoons \underset{a+b}{{Ag}^{+}\left(aq\right)}+\underset{a}{{CNS}^{-}\left(aq\right)}$

$AgBr\left(s\right)\rightleftharpoons \underset{a+b}{{Ag}^{+}}+\underset{b}{{Br}^{-}}$

$\therefore$ $\left[{Ag}^{2+}\right]=(a+b);\left[CN{S}^{-}\right]=a$ and $\left[{Br}^{-}\right]=b$

For $AgCNS:K_{{sp}_{AgCNS}}=\left[{Ag}^{+}\right]\left[{CNS}^{-}\right]$

or $1\times {10}^{-12}=(a+b)\left(a\right)......\left(i\right)$

For $AgBr:K_{{sp}_{AgBr}}=\left[{Ag}^{+}\right]\left[{Br}^{-}\right]$

or $5\times {10}^{-13}=(a+b)\left(b\right)........\left(ii\right)$

By Eqs $\left(i\right)$ and $\left(ii\right)$, we get

$\frac{a}{b}=\frac{{10}^{-12}}{5\times {10}^{-13}}=2$ ($a=2b$)

$\therefore$ By Eq $\left(i\right)$

$(2b+b)\left(2b\right)=1\times {10}^{-12}$

$\therefore$ $6b^2=1\times {10}^{-12}$

or $b=4.08\times {10}^{-7}$ $mol$ ${litre}^{-1}$

Similarly by Eq $\left(ii\right)$

$[a+(a/2\left)\right]a=1\times {10}^{-12}$

$a=8.16\times {10}^{-7}$ $mol$ ${litre}^{-1}$