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The Henderson-Hasselbalch relationship describing each ionizable group is:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[A-]}}{\ce{[AH]}}$$
We can solve for the ratio:
$$10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})}= \frac{\ce{[A-]}}{\ce{[AH]}}$$
However, we really want the fraction of protonated among the total (not the ratio of deprotonated ...
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Below a more general approach.
Suppose that we have two weak acids $\ce{HA}$ and $\ce{HB}$.
The initial concentrations are $C^0_\ce{HA}$ and $C^0_\ce{HB}$, and their constants are ${K_\ce{a}}_\ce{(HA)}$ and ${K_\ce{a}}_\ce{(HB)}$.
Suppose yet that volumes, $V_\ce{HA}$ and $V_\ce{HB}$, are additives.
So we have:
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$$\ce{HA + H2O <=> H3O+ + A-}...
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