# The reasoning behind predominance diagrams

Consider a one dimensional predominance diagram for a simple acid-base reaction:

$$\ce{AH <=> A^- + H^+}$$

With equilibrium constant:

$$K_a = \frac{[A^-]_{eq}[H^+]_{eq}}{[AH]_{eq}}$$

If I understand correctly, $[H^+] > K_a$, then $[AH]>[A^-]$. At first I thought I understood why this was so: the $[H^+]$ cancels out on both sides of the inequality, and then you bring $[AH]$ to the other side. Only now I realized that you can't actually cancel $[H^+]$ because on one side of the inequality it's actually $[H^+]_{eq}$, not the current amount of protons but the amount at equilibrium. So now I'm lost. What justifies this rule for determining predominance?

## 2 Answers

I think you might be making a fundamental mistake when you differentiate between the "current amount of protons" and the "amount at equilibrium." All dissociation constants are based on equilibrium concentrations, and predominance diagrams refer to the concentrations present only after equilibrium has already been established.

Consider the concrete example of the dissociation of ammonium: $\ce{NH4+ <=> NH3 + H+}$. The equilibrium constant expression for this chemical process is:

$$K_\text{a} = \frac{\ce{[NH3][H+]}}{\ce{[NH4+]}}$$

The $\mathrm{p}K_\text{a}$ of this process has been experimentally determined to be approximately $9.25$. Therefore, the following equation should hold:

$$-\log(K_\text{a}) = -\log\frac{\ce{[NH3][H+]}}{\ce{[NH4+]}} = \mathrm{p}K_\text{a} = 9.25$$

If a given solution has a measured $\mathrm{pH}$ precisely equal to the $\mathrm{p}K_\text{a}$ of the above process, then what we have is the following (keeping in mind that $\ce{[H+]} = 10^{-\mathrm{pH}}$):

$$\log\frac{[\ce{NH3}]\cdot 10^{-9.25}}{[\ce{NH4+}]} = -9.25$$

Taking the anti-log of both sides of the equation:

$$\frac{[\ce{NH3}]\cdot 10^{-9.25}}{[\ce{NH4+}]} = 10^{-9.25}$$

Then dividing both sides of the equation by $10^{-9.25}$, we find:

$$\frac{[\ce{NH3}]}{[\ce{NH4+}]} = 1$$

That is to say, the concentrations of ammonium and ammonia in the solution are perfectly equal when the $\mathrm{pH}$ and $\mathrm{p}K_\text{a}$ are equal. If the $\mathrm{pH}$ is higher (that is to say, more basic), then you'll find that your conjugate base (in this case, $\ce{NH3}$) will predominate by factor of $10^{\mathrm{pH} - \mathrm{p}K_\text{a}}$. Conversely, if the $\mathrm{p}K_\text{a}$ is higher, then the conjugate acid ($\ce{NH4+}$) will predominate by a factor of $10^{\mathrm{p}K_\text{a} - \mathrm{pH}}$.

If you prefer to use an inequality and some generic conjugate acid/base pair $\ce{HA/A- }$, then the logic goes as follows.

If $\mathrm{pH} > \mathrm{p}K_\text{a}$:

$$\mathrm{pH} > \mathrm{p}K_\text{a}\\ -\log\ce{[H+]} > -\log\frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\ \ce{[H+]} < \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\ 1 < \frac{\ce{[A- ]}}{\ce{[HA]}}\\ \ce{[HA]} < \ce{[A- ]}$$

If $\mathrm{pH} = \mathrm{p}K_\text{a}$:

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

If $\mathrm{pH} < \mathrm{p}K_\text{a}$:

$$\mathrm{pH} < \mathrm{p}K_\text{a}\\ -\log\ce{[H+]} < -\log\frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\ \ce{[H+]} > \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\ 1 > \frac{\ce{[A- ]}}{\ce{[HA]}}\\ \ce{[HA]} > \ce{[A- ]}$$

If you rearrange the equation to $$K_a[HA] = [A^-][H^+]$$ the relationship becomes more obvious. If $[H^+]>K_a$, then $[HA]>[A^-]$ in order to keep the two products equal. otherwise $[H^+]>K_a$ and $[A^-]>[HA]$, then the relationship would be an inequality: $$[A^-][H^+] > K_a[HA].$$