59 words

Isoelectric Point Calculation

By rearranging the Henderson-Hasselbalch equation:

\[ \text{pH}=\text{pK}_\text{a} + \log\left(\dfrac{\text{[A}^{-}\text{]}}{\text{[HA]}}\right) \]

we can get the ratio between an acid and its conjugate base:

\[ \frac{\text{[HA]}}{\text{[A}^{-}\text{]}}=10^{(\text{pK}_\text{a}-\text{pH})} \]

and between an base and its conjugate acid:

\[ \frac{\text{[B]}}{\text{[BH}^+\text{]}} = 10^{(\text{pH}-\text{pK}_\text{a})} \]

Thus, the proportion of deprotonated acid is calculated as follows:

\[ \dfrac{\text{[A}^{-}\text{]}}{\text{[A]}_\text{total}}=\dfrac{\text{[A}^{-}\text{]}}{\text{[HA]}+\text{[A}^{-}\text{]}}=\dfrac{1}{1+\frac{\text{[HA]}}{\text{[A}^{-}\text{]}}}=\dfrac{1}{1+10^{(\text{pK}_\text{a}-\text{pH})}} \]

Similarly, for basic species:

\[ \dfrac{\text{[BH}^+\text{]}}{\text{[B]}_\text{total}}=\dfrac{\text{[BH}^+\text{]}}{\text{[B]}+\text{[BH}^+\text{]}}=\dfrac{1}{1+\frac{\text{[B]}}{\text{[BH}^+\text{]}}}=\dfrac{1}{1+10^{(\text{pH}-\text{pK}_\text{a})}} \]

http://fields.scripps.edu/DTASelect/20010710-pI-Algorithm.pdf