Calculate the isoelectric and isoionicpHof 0.010Mthreonine.

• The pH forwhich a molecule essentially has almost no net electrical charge or is electrically neutral is characterizedas the isoelectric point (pI, pH (I), IEP).
• The standard abbreviation for the isoelectric point is pH (I), though pI is also used and is used in this article for brevity.
• When a pure, neutral polyprotic acid is dissolved in water, the isoionic point or isoionic pH is obtained. The isoelectric point is the pH at which the polyprotic acid's average charge is zero (0).

First, take the values of and of threonine from P g 900 :

${\mathrm{K}}_1=8.17\times {10}^{-3}{\mathrm{K}}_2=7.94\times {10}^{-10}$

Next, calculate the isoionic $\left[{\mathrm{H}}^{+}\right]$via the equation

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=\sqrt{\left({\mathrm{K}}_1{\mathrm{K}}_2\left(0.01\right)+{\mathrm{K}}_2{\mathrm{K}}_{\mathrm{w}}\right)/\left({\mathrm{K}}_1+0.01\right)} \\ \left[{\mathrm{H}}^{+}\right]=\sqrt{\left(\left(8.17\times {10}^{-3}\right)\times \left(7.94\times {10}^{-10}\right)\right)\times 0.01+\left(7.94\times {10}^{-10}\right)\times \left({10}^{-14}\right)/\left(8.17\times {10}^{-3}\right)}\left[{\mathrm{H}}^{+}\right]=1.891\times {10}^{-6}\mathrm{M} \\ \left[{\mathrm{H}}^{+}\right]=1.891\times {10}^{-6}\mathrm{M} \end{gathered}$$

Then calculate the isoionicby the following:

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \mathrm{pH}=-\log \left(1.891\times {10}^{-6}\right)=5.72 \end{gathered}$$

Next, we can calculate the isoelectric pH by the following values of ${\mathrm{pK}}_{\mathrm{a}}$also taken from Pg900

$$\begin{gathered} \mathrm{pH}=\frac{{\mathrm{pK}}_1+{\mathrm{pK}}_2}{2} \\ \mathrm{pH}=\frac{2.088+9.100}{2}=5.59 \end{gathered}$$

The isoelectric and isoionic pH of 0.010M threonine is 5.72 and 5.59.