Question

Determine the amount of $AgI$ which may be dissolved in 1.0 litre of $1.0MC{N}^{-}$ solution. $K_{sp}$ for $AgI$ and equilibrium constant for $\left[Ag\right(CN{)}_2{]}^{-}$ formation are $0.45\times {10}^{-17}{M}^2$ and $5\times {10}^{19}{M}^{-2}$ respectively.

• A. 0.18 mol
• B. 0.48 mol
• C. 4.8 mol
• D. 1.8 mol

Answer 1

The correct option is B 0.48 mol

$AgI\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+I^{-}\left(aq\right)..........\left(1\right)K_{sp}=\left[A{g}^{+}\right(aq\left)\right]\left[I^{-}\right(aq\left)\right]A{g}^{+}\left(aq\right)+2C{N}^{-}\left(aq\right)\rightleftharpoons \left[Ag\right(CN{)}_{2\left(aq\right)}{]}^{-}.......\left(2\right)K_f=\frac{{\left[Ag\right(CN{)}_2]}^{-}}{\left[A{g}^{+}\right][C{N}^{-}{]}^2}$
Now, let x moles of AgI dissolved in $C{N}^{-}$ solution
Adding the equation (1) and (2), we get
$AgI\left(s\right)+2C{N}^{-}\left(aq\right)\rightleftharpoons \left[Ag\right(CN{)}_{2\left(aq\right)}{]}^{-}+I^{-}1mole00(1-2x)molexmolexmole$
As $\left[C{N}^{-}\right]=M.V=1\times 1=1mole\therefore K_c=\frac{x.x}{(1-2x{)}^2}Also,K_c=K_f.K_{sp}\therefore \frac{x^2}{(1-2x{)}^2}=0.45\times {10}^{-17}{M}^2\times 5\times {10}^{19}{M}^{-2}$
$\Rightarrow \frac{x^2}{(1-2x{)}^2}=225$
$\Rightarrow x=0.48mol$