# Amino acids: why is charge found by comparing pH and pKa and not pH to pKa + log [A-]/log[H+]?

I am working on problems such as : 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?

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 $$\mathrm{p}K_a$$.