# Approximated formula of acid dissociation constant

$K_\mathrm{a}$ can be expressed as $K_{\mathrm{a}} = \frac{[\ce{H^{+}}] [\ce{A^-}]} {[\ce{HA}]}$, however, it can be approximated that $K_{\mathrm{a}} = \frac{[\ce{H^{+}}]^{2}} {[\ce{HA}]}$.

I usually use the approximated equation when I'm dealing with questions that don't provide [$\ce{A-}$]. What I don't understand is why/how can we approximate it to this?

To approximate to $\ce{[H+]^2}$, then $\ce{[H+]}$ and $\ce{[A-]}$ must be pretty much equal, and so this must assume that the acid completely dissociates. However, in practice, do acids dissociate to the point that $\ce{[H+]}$ and $\ce{[A-]}$ are equal? If not, then how can we approximate to $\ce{[H+]}^2$, and are there any mathematics that demonstrate this approximation?

• As an approximation, it's valid when 1) the amount of $\ce{H3O+}$ from the dissociation of water is negligible compared to that from the dissociation of the acid, and 2) dissociation of the acid is almost complete ($\alpha \approx 1$). So in general it works for relatively concentrated ($1 < \mathrm{p}C < 5$) solutions of moderately weak ($3 < \mathrm{p}K_\mathrm{a} < 8$) acids.
– user41033
Apr 25 '18 at 13:06

The acid dissociates in equal amounts of $[\ce{H^+}]$ and $[\ce{A^-}]$ hence we can say they are equal quantities.

We can approximate it like this because we assume the $[\ce{H^+}]$ from the dissociation of water to be negligible compared with the $[\ce{H^+}]$ from the dissociation of the acid.

We also assume the $[\ce{HA}]$ at the start is equal to $[\ce{HA}]$ at equilibrium because the dissociation of weak acids is small.

Here is some mathematics that you requested:

$K_{\mathrm{a}}$ can be calculated using equilibrium concentrations: $$K_{\mathrm{a}} = \frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{equim}}=\frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{start}-[\ce{H+(aq)}]_\text{equim}}$$

Taken from my OCR A Level Chemistry Textbook (The specification under which this year's exams will be.)

Here is an extension to fully show how the two equate. Taken from the same textbook:

$$K_{\mathrm{a}} =\frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{start}-[\ce{H+(aq)}]_\text{equim}}\rightarrow\frac{[\ce{H^{+}(aq)}]_\text{equim}^2} {[\ce{HA(aq)}]_\text{start}}$$ Therefore, $$K_{\mathrm{a}} =\frac{[\ce{H^{+}(aq)}]^2} {[\ce{HA(aq)}]}$$