Question

The Copper(I) ion forms a complex ion with $C{N}^{-}$ according to the following equation:
$C{u}^{+}\left(aq\right)+3C{N}^{-}\left(aq\right)\rightleftharpoons \left[Cu\right(CN{)}_3{]}^{2-}\left(aq\right);K=1.0\times {10}^{11}$
The concentration of $C{N}^{-}$ at equilibrium when 1.0 M $NaCN$ (1 L) is added to a sufficient amount of $CuBr\left(s\right)$ was found to be $x\times {10}^{-3}M$. The value of $x$ is

$K_{sp}=1\times {10}^{-5}$

Answer 1

$CuBr$ reacts with $NaCN$ and forms following equilibrium.
$CuBr\left(s\right)\rightleftharpoons C{u}^{+}+B{r}^{-};K_{sp}=1.0\times {10}^{-5}..\left(i\right)$
$C{u}^{+}\left(aq\right)+3C{N}^{-}\left(aq\right)\rightleftharpoons \left[Cu\right(CN{)}_3{]}^{2-}\left(aq\right);K_f=1.0\times {10}^{11}..\left(ii\right)$
Adding eq (i) and (ii), we get
$CuBr\left(s\right)+3C{N}^{-}\left(aq\right)\rightleftharpoons \left[Cu\right(CN{)}_3{]}^{2-}+B{r}^{-}\left(aq\right);K=?$
$K=K_{sp}\times K_f={10}^6$
Since the reaction goes to completion, so $K>{10}^3$.

$CuBr\left(s\right)+3C{N}^{-}\left(aq\right)\rightleftharpoons \left[Cu\right(CN{)}_3{]}^{2-}+B{r}^{-}\left(aq\right)inti100ateq(1-3x)xx$
Here reaction goes to completion, then $1-3x=0\Rightarrow x=0.33M$

$CuBr\left(s\right)+3C{N}^{-}\left(aq\right)\rightleftharpoons \left[Cu\right(CN{)}_3{]}^{2-}+B{r}^{-}\left(aq\right)inti00.330.33ateq3y0.33-y0.33-y$
$K=\frac{(0.33-y{)}^2}{(3y{)}^3}=1\times {10}^6\Rightarrow \frac{(0.33{)}^2}{27y^3}=1\times {10}^6$
(y is too small to eliminate from numerator)
$y=1.6\times {10}^{-3}M$
From the above, $3y=4.8\times {10}^{-3}M$