Ionic equilibrium between multiple ions This is a question from IIT-JEE 1993:


  For the reaction
  
  $$\ce{[Ag(CN)2]^- -&gt; Ag^+ + 2CN^-}$$
  
  the equilibrium constant at $\pu{298 K}$ is $\pu{4e-10}$. The silver ion concentration in a solution which was originally $\pu{0.10 M}$ in $\ce{KCN}$ and $\pu{0.03 M}$ in silver nitrate is?


This is what I've done:

$$
\begin{align}
[\ce{K^+}] &amp;= 0.1 \\
[\ce{CN-}] &amp;= 0.1 - 2x \\
[\ce{Ag+}] &amp;= 0.03 - x \\
[\ce{NO3-}] &amp;= 0.03
\end{align}
$$

$$\frac{(0.03-x)(0.1-2x)^2}{x} = 4\cdot 10^{-10}$$

But this gives $x = 0.03$, hence, $[\ce{Ag+}] \approx 0$.

The answer key says $[\ce{Ag+}] \approx 7.5\cdot 10^{-18}$. Where am I going wrong?
In most of these type problems chemists look for ways to simply the mathematical expression by using approximations based on a knowledge of the chemistry. The gist is that because of the significant figures you assume that the reaction essentially goes to completion. So $\ce{Ag(CN)_2^− \approx 0.0300}$ and $\ce{CN− \approx 0.0400}$. Then you solve for [Ag+]. So:

$4\times10^{-10} = \dfrac{\ce{[Ag^+][CN^-]^2}}{\ce{[Ag(CN)_2^-]}}$  

or rearranging for $\ce{[Ag^+]}$

$\ce{Ag^+} = 4\times10^{-10} \dfrac{{\ce{[Ag(CN)_2^-]}}}{\ce{[CN^-]^2}} = 4\times10^{-10} \dfrac{0.03}{0.04^2} = 7.5\times10^{-9} $

Checking $7.5\times10^{-9} \ll 0.03$ so the assumption that the reaction essentially went to completion is valid. in other words:

$0.0300 - 7.5\times10^{-9} \approx 0.300$