I'm having difficulties understanding this problem.

We have in the problem that $K_\mathrm a$ for $\ce{HCN} = 6.2\cdot10^{-10}$ and $K_\mathrm b$ for $\ce{NH3} = 1.8\cdot10^{-5}$.

  1. Write chemical equations to represent the dissociation of $\ce{NH4CN}$ in water, and the acid-base equilibrium equation.
  2. Will the salt react as an acid or a base?

I know this is part of the solution:

$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$ $$\ce{CN- + H2O <=> HCN + OH-}$$

I understand Le Chatelier's principle and I know that the reaction will shift to the side of less energy.


1 Answer 1


Fun! weak acid versus weak base!

Ammonium cyanide, $\ce{NH4CN}$, is a solid where the atoms are grouped into the same ions that are generated in solution: $\ce{NH4+}$ and $\ce{CN-}$. Although it can be sublimed with very mild warming, it is fairly unstable. In particular it reacts with atmospheric water vapor, evolving ammonia and HCN. Dangerous stuff!

The first part of the answer to 1. is the decomposition of $\ce{NH4CN}$ into ions in aqueous solution, which should be obvious; one subtlety is that you may want to add (s) and (aq) next to each species, especially if your professor makes a point of doing so. So that's the dissociation. The data in the problem don't allow us to compute the equilibrium constant of the dissociation but it's safe to say that it is extremely high, that is, dissociation is complete: there is no $\ce{NH4CN}$ as such in solution.

The acid-base equilibrium equations can be the ones you wrote. Again, you may want to add more information, like this:

$$\ce{NH4+ (aq) + H2O <=> NH3 (aq) + H3O+ (aq)}\quad K=5.5\times10^{-10}$$

Here the equilibrium constant $K$ is the acid dissociation constant of the acid $\ce{NH4+}$, the protonated form (aka conjugate acid) of ammonia. I got its value by dividing $K_\mathrm w$, the dissociation constant of water ($1.0\times10^{-14}$), by the $K_\mathrm b$ of ammonia, given in the problem.

I encourage you to add similar information for the equilibrium involving cyanide ion. Hint: if you keep the reaction as you wrote it, the appropriate $K$ is not the $K_\mathrm a$ of $\ce{HCN}$ but the $K_\mathrm b$ of the deprotonated form (conjugate base), $\ce{CN-}$. (But you can instead consider the alternate reaction $\ce{HCN + H2O <=> CN- + H3O+}$, and then the equilibrium constant is the $K_\mathrm a$ of $\ce{HCN}$.)

How about part 2? The only correct answer to

Will the salt react as an acid or a base?

is "It can do either, depending on what it's reacting with". But maybe what they mean is whether the solution of the salt is acidic or basic. If so, one way to find out is to combine the two equilibrium equations and their $K$s. I don't want to give problem away completely, but think of it this way:

  • What is the pH of a (say) 1 M solution of $\ce{HCN}$, based on the $K_\mathrm a$ of $\ce{HCN}$?

    (I'm assuming the calculation of the pH of a solution of a pure weak acid or a pure weak base has been covered in the course before getting to the question you posted today.)

  • What is the pH of a 1 M solution of $\ce{NH3}$, based on the $K_\mathrm b$?

  • Is $\ce{HCN}$ a stronger acid than $\ce{NH3}$ is a base, or is $\ce{NH3}$ a stronger base than $\ce{HCN}$ is an acid? Based on this, which would you expect to "win out" if you mix the two in equal proportions?

(The point here is that what you get by dissolving $\ce{NH4CN}$ is indistinguishable from what you'd get by mixing equal amounts of equimolar solutions of $\ce{HCN}$ and ammonia. In other words, if you mix 1 L of a 2 M solution of $\ce{HCN}$ with 1 L of a 2 M solution of ammonia, the result is very nearly 2 L of a 1 M solution of $\ce{NH4CN}$. Footnote: "very nearly" because volume is not always preserved exactly, unlike mass.)

  • $\begingroup$ I think this is a very good answer. The only thing I'd add (which may or not be apparent to the OP) is that combining the two acid-base reactions the OP has already written with the neutralization of hydronium and hydroxide (reversing the water autoionization reaction) gives what I think is probably the dominant equilibrium ($K_{eq} \approx 9.0 \cdot 10^{-1}$). $\endgroup$
    – Greg E.
    Commented Aug 3, 2014 at 22:57
  • $\begingroup$ I'm fairly sure the calculation is right $\frac{K_w}{K_b} \cdot \frac{K_w}{K_a} \cdot \frac{1}{K_w} = \frac{1.0 \cdot 10^{-14}}{1.8\cdot10^{-5} \cdot 6.2\cdot10^{-10}}$. This agrees with the calculation of $10^{\Delta pKa}$ using reference values. $\endgroup$
    – Greg E.
    Commented Aug 3, 2014 at 23:43
  • $\begingroup$ Yes, I realized my mistake in the meantime. OP will appreciate the calculation though :-) $\endgroup$ Commented Aug 3, 2014 at 23:45
  • $\begingroup$ I think it is important to state, that $K_w=1\cdot10^{-14}$ only holds for $25~^\circ\mathrm{C};~1~\mathrm{atm}$. $\endgroup$ Commented Aug 4, 2014 at 4:32

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