If I dissolve the weak acid HCN into water, I get a solution of oxonium and CN ions and since there is an equilibrium there is more HCN left.

So I thought CN ions can consequently react with water to form hydroxide-ions, which is a weak base-reaction, by which we also get the original HCN molecules. I am confused because if its true what I am saying, there are two equilibriums.

Could anyone explain this, and tell me in general whether the bases that are formed when you dissolve an acid can react reversibly to an acid?


2 Answers 2


There are 3 equilibriums, described by 3 equilibrium constants:

The first one is the equilibrium of water dissociation:

$$\ce{H2O <<=>H+ + OH-}$$


The second one is for the ( very weak ) hydrogencyanide acid dissociation:

$$\ce{HCN <<=>H+ + CN-}$$


The third one is for the cyanide anion hydrolysis:

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

$$K_\mathrm{b}=\frac{[\ce{HCN}][\ce{OH-}]}{[\ce{CN-}]}=\frac{[\ce{HCN}] \cdot K_\mathrm{w}}{[\ce{CN-}][\ce{H+}]}=\frac{ K_\mathrm{w}}{K_\mathrm{a}}$$

All 3 equilibriums are therefore related by equation

$$K_\mathrm{w}=K_\mathrm{a}\cdot K_\mathrm{b}$$

Remember all 3 equilibriums are dynamic, with chemical reactions ongoing simultaneously in both directions at equal rate.

In the case the rates were not equal, the dominant reaction shifts the system toward the equilibrium for the rates to be equal again.


If you're dissolving into neutral water, then you are overthinking the problem.

If you dissolve a weak acid into neutral water, the result is an acidic solution.

If you dissolve a weak base into neutral water, the result is a basic solution.

It is easy to confuse yourself if you only choose to look at one component of the equilibrium and try to understand the system. You want to look at the total equilibrium. In your example, yes, the cyanide will react with water but it does that to a smaller extent than hydrogen cyanide will dissociate. That's why hydrogen cyanide is an acid.


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