Question

Determine the mol. of $AgI$ which may be dissloved in $1.0L$ of $1.0M$ $C{N}^{⊝}$. solution. $K_{sp}$ for $AgI$ and $K_c$ for $Ag(CN{)}_2^{⊝}$ are $1.2\times {10}^{-17}{M}^2$ and $7.1\times {10}^{19}{M}^{-2}$, respectively.

• A. $0.5$
• B. $0.25$
• C. $0.16$
• D. None of these

Answer 1

The correct option is A $0.5$

Given $Agl\left(s\right)\rightleftharpoons A{g}^{\oplus }\left(aq\right)+I^{⊝}\left(aq\right)$;

$K_{sp}=\left[A{g}^{\oplus }\right]\left[I^{⊝}\right]=1.2\times {10}^{-17}$

$A{g}^{\oplus }\left(aq\right)+2C{N}^{⊝}\left(aq\right)\rightleftharpoons \left[Ag\right(CN{)}_2{]}^{⊝}\left(aq\right)$;

$K_f=\frac{\left[Ag\right(CN{)}_2^{⊝}]}{\left[A{g}^{\oplus }\right][C{N}^{⊝}{]}^2}=7.1\times {10}^{19}$

Let $x$ mole of $AgI$ be dissolved in $C{N}^{⊝}$ solution then,

Now, $AgI\left(s\right)+2C{N}^{⊝}\rightleftharpoons \left[Ag\right(CN{)}_2^{⊝}]+I^{⊝}$
Mole before reaction 1 0 0
Mole after reaction $(1-2x)$ $x$ $x$

By equation (1) and (2), $K_{eq}=K_{sp}\times K_f$

$K_{eq}=\frac{\left[Ag\right(CN{)}_2^{⊝}\left]\right[I^{⊝}]}{[C{N}^{⊝}{]}^2}=1.2\times {10}^{-17}\times 7.1\times {10}^{19}$

$K_{eq}=8.52\times {10}^2=\frac{xx}{(1-2x{)}^2}=\frac{x^2}{(1-2x{)}^2}$

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

Thus, $x=29.2-58.4x$ or $x=0.49mol$