I am working on problems such as :

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The charges on the side groups at each pH are found by the following rule:

  • at $\ce{pH < pK_a}$, $\ce{H+}$ on, protonated
  • at $\ce{pH > pK_a}$, $\ce{H+}$ off, deprotonated
  • at $\ce{pH = pK_a}$ , neutral

But why can we compare $\mathrm{p}H$ and $\mathrm{p}\ce{Ka}$ directly?

I just started my amino acids topic. In all my acid and base questions we always compared $\mathrm{p}H$ to $\mathrm{p}K_a$ using the Henderson Hasselbalch Equation:

$$\mathrm{p}H =\mathrm{p}K_a +\log \left( \frac{\ce[A^-]}{\ce{[HA]}} \right)$$

So why do we now compare $\mathrm{p}H$ to $\mathrm{p}K_a$ directly?


1 Answer 1


The properties you listed are correct, and all of them follow from the relationship described by the Henderson-Hasselbalch equation.

The equation shows a close relationship between $\mathrm{p}K_a$ and $\mathrm{p}H$. Note that $\ce{log ([A^{-}]/[HA])}$ equals zero when $\ce{[A^{-}]/[HA]}$ equals 1, and when $\ce{log ([A^{-}]/[HA])=0}$, $\mathrm{p}K_a=\mathrm{p}H$. This means that at the equivalence point, $\mathrm{p}H$ equals $\mathrm{p}K_a$.


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