# What is the dissociation percentage of a polybasic organic acid at a certain pH?

For my Master's Thesis I am working with the dicarboxylic acid muconic acid $$(\ce{H2MA},$$ $$\mathrm{p}K_\mathrm{a1} = 2.9,$$ $$\mathrm{p}K_\mathrm{a2} = 4.0).$$ It thus dissociates into $$\ce{HMA-}$$ and $$\ce{MA^2-},$$ respectively.

I am aware of how to calculate the $$\mathrm{pH}$$ and the subsequent dissociation percentage when dissolving a weak acid in water. But I am wondering how you can calculate it the other way around. Say I dissolve $$\pu{1 g}$$ of the acid in water and adjust the $$\mathrm{pH}$$ to 4 with a strong acid such as $$\ce{H2SO4},$$ how do I know what the concentration is of each species $$(\ce{H2MA},$$ $$\ce{HMA-}$$ and $$\ce{MA^2-})$$ at that given $$\mathrm{pH}?$$

In general, how can I calculate the dissociation percentage of a weak polybasic organic acid when the pH is given?

• Hi, Welcome to Chem.SE! Please note the edits. About the question, if the pH is fixed, then you have $\ce{[H+]}$. You have the initial concentration of the acid which will be the total of $\ce{[H2MA] +[HMA-] +[MA^2-]}$ in the solution. Write the equations for protonation/deprotonation equilibria and then use the above, and there should be a system of equations you can solve. May 20, 2021 at 14:03
• Following on from the detailed comment by @ShoubhikRMaiti, you just compute the dissociation fractions at pH 4 or for a range of pH values. Here is what they look like for phosphoric acid: chemistry.stackexchange.com/a/118933/79678.
– Ed V
May 20, 2021 at 14:13
• I remember posting such calculation formulas, perhaps few months ago. See the answer chemistry.stackexchange.com/questions/149282/… May 20, 2021 at 14:21
• I have read them few times somewhere and derived them once for myself many years ago. Since then I remember them. Just taylor it for just 2 dissociation degrees. May 20, 2021 at 14:28

Reused my answer to how-to-numerically-model-a-phosphoric-acid-titration-curve, edited for diprotic cases:

First, calculate the common denominator $$CD$$:

$$a_1 = [\ce{H+}]^2$$ $$a_2 = [\ce{H+}] \cdot K_\mathrm{a1} = a_1 \cdot \frac {K_\mathrm{a1}}{[\ce{H+}]}$$ $$a_3 = K_\mathrm{a1} \cdot K_\mathrm{a2} = a_2 \cdot \frac {K_\mathrm{a2}}{[\ce{H+}]}$$

$$CD = a_1 + a_2 + a_3$$

.. and then reuse it and its additive terms:

$$[\ce{H2A}]=c_0 \cdot \frac {a_1}{CD}= c_0 \cdot\frac{[\ce{H+}]^2 }{[\ce{H+}]^2 + [\ce{H+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$

$$[\ce{HA-}]=c_0 \cdot \frac {a_2}{CD}= c_0 \cdot\frac{ [\ce{H+}] \cdot K_\mathrm{a1} }{[\ce{H+}]^2 + [\ce{H+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$

$$[\ce{A^2-}]=c_0 \cdot \frac {a_3}{CD}=c_0 \cdot \frac{K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H+}]^2 + [\ce{H+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$

$$c_0$$ is the total concentration of respective particular acid forms.

For the general degree of "N-proticity", the expression terms span from $$\ce{[H+]}^N$$ to $$K_1 \cdot ... \cdot K_\mathrm{aN}$$