Question

The number of mole of $AgI$ which may be dissolved in $1.0$ litre of $1.0M$ ${CN}^{-}$ solution whose
$K_{sp}$ for $AgI$ and $K_c$ for $Ag{\left(CN\right)}_2^{-}$ are $1.2\times {10}^{-17}$ and $7.1\times {10}^{19}$ $M^{-2}$ respectively.

• A. $0.40$ $mol$
• B. $0.45$ $mol$
• C. $0.50$ $mol$
• D. None of these

Answer 1

The correct option is C $0.50$ $mol$

$AgI\left(s\right)\rightleftharpoons {Ag}^{+}\left(aq\right)+I^{-}\left(aq\right);$

$K_{sp}=\left[{Ag}^{+}\right]\left[I^{-}\right]=1.2\times {10}^{-17}.....\left(i\right)$

${Ag}^{+}\left(aq\right)+2{CN}^{-}\left(aq\right)\rightleftharpoons \left[Ag{\left(CN\right)}_2^{-}\right]\left(aq\right)$

$K_c=\frac{\left[Ag{\left(CN\right)}_2^{-}\right]}{\left[{Ag}^{+}\right]{\left[{CN}^{-}\right]}^2}=7.1\times {10}^{19}......\left(ii\right)$

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

$\frac{\left[Ag{\left(CN\right)}_2^{-}\right]\left[I^{-}\right]}{{\left[{CN}^{-}\right]}^2}=K_{sp}\times K_c$

$=1.2\times {10}^{-17}\times 7.1\times {10}^{19}=8.52\times {10}^2...\left(iii\right)$

Now,
$\underset{MolebeforereactionMoleafterreaction}{AgI\left(s\right)}+\underset{1(1-2x)}{2{CN}^{-}}\rightleftharpoons \underset{0x}{\left[Ag{\left(CN\right)}_2^{-}\right]}+\underset{0x}{I^{-}}$

Let $x$ mole of $AgI$ be dissolved in ${CN}^{-}$ solution

$\therefore$ $8.52\times {10}^2=\frac{x\cdot x}{{(1-2x)}^2}$ [From Eq. $\left(iii\right)$]

or $\frac{x}{1-2x}=29.2$

or $x\approx 0.50mol$