Ionic equilibrium between multiple ions

This is a question from IIT-JEE 1993:



> 
> For the reaction
> 
> 
> $$\ce{[Ag(CN)2]^- -> Ag^+ + 2CN^-}$$
> 
> 
> 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:


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


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


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:


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


 5

You are going wrong *in your choice of $x$.*


You should have recognized in advance that $\ce{CN-}$ is in excess, and hence $\ce{Ag+}$ would go to the complex almost completely, so the equilibrium $\ce{[Ag+]}$ is going to be awfully low, while the concentration of $\ce{[Ag(CN)2]-}$ (which you have unfortunately defined as $x$) is going to be *really* very close to 0.03 and even closer than that, like 0.29999 and more nines until your keyboard runs out of nines and stops working. No wonder that when you are doing your calculations with 3 (or 4, or any sensible number of) significant figures, it just reads as 0.03.


Throw everything away, define $x=\ce{[Ag+]}$ and start from the scratch.


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