# When pH > pKa and when pH < pKa and the effectiveness of buffer solutions

I understand that when $$\mathrm{pH}= \mathrm{p}K_\mathrm{a}$$, the buffer solution will be at its maximum capacity, and there will be equal concentrations of the acid/conjugate acid and the base/conjugate base.

However, when $$\mathrm{pH} > \mathrm{p}K_\mathrm{a}$$, why is it that $$\ce{[A^-] > [HA]}$$? Shouldn't it be the other what around since the Henderson-Hasselbalch equation is:

$$\mathrm{pH}= \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[A^-]}}{\ce{[HA]}}$$

So $$\frac{\ce{[A^-]}}{\ce{[HA]}}$$ should be less than $$0$$, so $$\ce{[HA]}$$ must be greater.

Why is it that when $$\mathrm{pH} > \mathrm{p}K_\mathrm{a}$$, the concentration of the conjugate base ($$\ce{[A-]}$$) is greater than the concentration of the acid ($$\ce{[HA]}$$)?

To calculate $$\mathrm{pH}$$ of buffer solutions, Henderson-Hasselbalch equation is a very good assert:

$$\mathrm{pH}= \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[A^-]}}{\ce{[HA]}} \tag{1}$$

Also, you can rewrite the equation as:

$$\mathrm{pH} - \mathrm{p}K_\mathrm{a} = \log \frac{\ce{[A^-]}}{\ce{[HA]}} \tag{2}$$

Accordingly, as you correctly suggested, for monobasic acid, when $$\mathrm{pH} = \mathrm{p}K_\mathrm{a}$$, the equation $$(2)$$ becomes:

$$\mathrm{pH} - \mathrm{p}K_\mathrm{a} = \log \frac{\ce{[A^-]}}{\ce{[HA]}} = 0 = \log 1$$ Thus, $$\ce{[A^-]} = \ce{[HA]}$$. Now, look at the equation $$(2)$$ again for the condition of $$\mathrm{pH} > \mathrm{p}K_\mathrm{a}$$: The term $$\left(\mathrm{pH} - \mathrm{p}K_\mathrm{a}\right)$$ should be positive, and hence $$\log \frac{\ce{[A^-]}}{\ce{[HA]}} > 0$$. Mathematically, the only way this could happen is if and only if $$\ce{[A^-]} > \ce{[HA]}$$.

Similarly, if the condition is $$\mathrm{pH} < \mathrm{p}K_\mathrm{a}$$, the term $$\left(\mathrm{pH} - \mathrm{p}K_\mathrm{a}\right)$$ should be negative, and hence $$\log \frac{\ce{[A^-]}}{\ce{[HA]}} < 0$$. Again, in mathematics, the only way this could happen is if and only if $$\ce{[A^-]} < \ce{[HA]}$$.