# 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?

• Wrong? To me, $10^{-18}\approx0$ looks about right. Mar 22, 2019 at 6:03
• 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. So $0 \approx 10^{-18} \ll 0.03$
– MaxW
Mar 22, 2019 at 6:14
• @MaxW This was an MCQ question. There were 3 options in the order of $10^{-15} to 10^{-18}$. And the 4th was None of the Above. So, when I got 0, I went for the 4th one and was wrong Mar 22, 2019 at 6:46
• 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 $\ce{[Ag^+]}$.
– MaxW
Mar 22, 2019 at 7:36
• That does indeed yield $7.5\times10^{-9}$
– MaxW
Mar 22, 2019 at 7:45

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:

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

• The $\ce{[Ag+]}$ should be $7.5 \times 10^{-8}$, I believe. Mar 22, 2019 at 18:01
• @MathewMahindaratne - Thank you for the correction. There was a mistake. I had the equilibrium constant as $4\times 10^{-9}$ and it should have been $4\times 10^{-10}$. With the correct equilibrium constant the value is $7.5\times10^{-9}$.
– MaxW
Mar 22, 2019 at 18:08

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.

• I wouldn't say that the OP really setup $x$ wrong but rather the OP didn't understand the implication of $x \approx 0.03$.
– MaxW
Mar 22, 2019 at 6:17
• Well, it wouldn't matter what you define as $x$, if not for the strict adherence to the significant figures rules. It is the combination of the two that makes it wrong. Mar 22, 2019 at 6:24
• So, in the Kc equation, $\ce{[Ag^+]} = x$ & $\ce{CN^-} = 0.04 -2x$ & $\ce{[Ag(CN)2]^-} = 0.03 - x$ ? Doing that gives $7.5 * 10^{-9}$. What's causing the error in order of answer Mar 22, 2019 at 7:01
• It is $0.04\color{red}{+}2x$ instead of "-" (not that it matters much). Well, now I believe you are doing it right, and if the answer key says otherwise, there must be a mistake in it. Mar 22, 2019 at 7:35