Using two approximations for the dissociation of a weak acid, we can write

$$K_\mathrm{a}= \frac{[\ce{H+(aq)}]^2}{[\ce{HA(aq)}]}$$

One of these two approximations we use is that the concentration of the acid HA has not changed upon dissociation, because the dissociation of a weak acid is so small compared to the starting concentration of the acid that is negligible.

However, I was taught that this approximation is invalid when the acid is stronger, but also when the solution is very dilute.

I don't understand why this approximation wouldn't work if the solution is very dilute. I have some ideas though:

  • The acid concentration decreases significantly when dissolved in a lot of water, so maybe the H+ would become significant then?

  • Maybe the H+ from water's dissociation affect the equilibrium of the weak acid's dissociation so that the dissociation becomes significant?

Are any of these right? Could someone explain why this approximation is invalid for very dilute solutions?


1 Answer 1


The above simplified formula considers 2 simplifying assumptions:

  • Concentration of hydrogen ions from water autodissociation can be neglected, so $[\ce{H+}] \approx [\ce{A-}]$ .
  • Concentration of an acid $c_0$ , in relation to its $\mathrm{p}K_\mathrm{a}$, is high enough for the fraction of ionized acid to be negligible, so $c_0 \approx [\ce{HA}]$ .

Especially the second assumption is broken for too dilute weak acid solutions. For more diluted solutions, higher fraction of an acid is ionized to match the equilibrium equation.

if we consider $$K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}}$$

then if we dilute the acid 4 times, concentration of the conjugate base decreases just 2 times, so the fraction of the ionized acid form increases 2 times as well.

It also means the stronger the acid is, the higher concentration must be . As for the same concentration, stronger acids are more ionized and $c_0 \approx [\ce{HA}]$ causes a bigger error.

The rule of thumb is, for the formula to be acceptable, this inequality should be true: $$ - \log {c_0} < \mathrm{p}K_\mathrm{a} - 2$$

so maximally about 1% of an acid is ionized.

The first assumption is broken mainly for too weak acids. E.g. if $\mathrm{p}K_\mathrm{a} = 10$, practically all $\ce{H+}$ comes from water autodissociation and the equation fails.


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