Question

Calculate the pH of a$\mathbf{0}\mathbf{.}\mathbf{010}\mathbf{M}$solution of each amino acid in the form drawn here

Short answer

The isoionic Ph is $=10.37$

Step-by-step solution

Step 1: Define the term isoionic ph.      

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).

Calculate the isoionic PH

The pH of a neutral, polyprotic molecule in pure solution.

The only ions present are${\mathrm{H}}^{+},{\mathrm{OH}}^{-}$

Isoionic is the name given to polyprotic species.

$\left[\mathrm{H}+\right]=\sqrt{\frac{{\mathrm{K}}_2{\mathrm{K}}_3\mathrm{F}+{\mathrm{K}}_2{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_2+\mathrm{F}}}$

Where F is the formal concentration.

Here Arginine has four forms ${\mathrm{H}}_3{\mathrm{A}}^{2+},{\mathrm{H}}_2{\mathrm{A}}^{+},\mathrm{HA}$and$A^{-}$

HA form shown as

$\left[\mathrm{H}+\right]=\sqrt{\frac{{\mathrm{K}}_2{\mathrm{K}}_3\left(0.010\right)+{\mathrm{K}}_2{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_2+0.010}}$

$$\begin{gathered} =4.28\times {10}^{-11}\mathrm{MpH} \\ =10.37 \end{gathered}$$

Thus the isoionic Ph is$=10.37$