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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 ($\ce{[HA]}$)?

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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 $\ce{[A^-]} < \ce{[HA]}$.

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