# pH of amphoteric species

For a solution of $$\ce{HA-}$$, I have seen the following approximations for the $$\mathrm{pH}:$$

$$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2})\tag{1}$$

$$\mathrm{pH} = \sqrt{\frac{K_1K_2[\ce{HA-}] + K_1K_\mathrm{w}}{K_1 + [\ce{HA-}]}}\tag{2}$$

How do they differ? Is the second approximation always more accurate than the first? I know that the second is derived from the mass and charge balance equations and the first is the same as the second, with several assumptions.

• The second equation should be for $\ce{[H+]}$ not $\pu{pH}$.
– MaxW
Dec 25, 2019 at 5:15

The correct expressions are:

$$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2})\tag{1}$$

$$\ce{[H+]} = \sqrt{\frac{K_1K_2[\ce{HA-}] + K_1K_\mathrm{w}}{K_1 + [\ce{HA-}]}}\tag{2}$$

Equation 2 is an exact expression (neglecting activities vs. concentrations), but Expression 1 is an approximation.

To dervice the approximate expression start by taking the -log of the second equation:

$$-\log{\ce{[H+]}} = \mathrm{pH} = -\log\left(\sqrt{\frac{K_1K_2[\ce{HA-}] + K_1K_\mathrm{w}}{K_1 + [\ce{HA-}]}}\right)$$

removing $$K_1$$ from numerator and simplifying...

$$\mathrm{pH} = \frac{1}{2}\left(\mathrm{pK}_1 -\log\left(\frac{K_2[\ce{HA-}] + K_\mathrm{w}}{K_1 + [\ce{HA-}]}\right)\right)$$

Now to reduce $$\dfrac{K_2[\ce{HA-}] + K_\mathrm{w}}{K_1 + [\ce{HA-}]}$$ to $$K_2$$ two conditions must be met:

• $$K_2[\ce{HA-}] \gg K_\mathrm{w}$$
• $$K_1 \ll [\ce{HA-}]$$
• Derivation of first equation is where Sir ? From where i can see? Apr 24, 2022 at 20:32

The former equation assumes $$[\ce{H2A}]\simeq [\ce{A^2-}]$$

due reaction

$$\ce{ 2 HA- <=> H2A + A^2-}$$

The is possible with 2 simplifying conditions:

The concentration of oxonium resp. hydroxide ions originated from water dissociation is much lower than concentration of the basic resp. acidic ampholyte form.

$$[\ce{H2A}] \gg \sqrt{K_\mathrm{w}}$$ $$[\ce{A^2-}] \gg \sqrt{K_\mathrm{w}}$$

That assures the effect of ampholyte reaction with water dissociation products

$$\ce{ A^2- + H3O+ -> HA- + H2O}$$ $$\ce{ H2A + OH- -> HA- + H2O}$$

is negligible for the ampholyte form concentration ratios and you can ignore water dissociation. This condition is critical for low ampholyte concentration near neutral $$\mathrm{pH}$$.

The concentration of oxonium resp. hydroxide ions is much lower than concentration of acidic resp. basic ampholyte forms.

$$[\ce{H2A}] \gg [\ce{H+}]$$ $$[\ce{A^2-}] \gg [\ce{OH-}]$$

That assures the effect of production of oxonium resp. hydroxide ions to reach the final $$\mathrm{pH}$$

$$\ce{ HA- + H2O -> A^2- + H3O+}$$ resp. $$\ce{ HA- + H2O -> H2A + OH-}$$

does not compete with creation of acidic/basic ampholyte forms.

$$\ce{ 2 HA- + -> A^2- + H2A }$$

This condition is not valid for very low /very high target $$\mathrm{pH}$$, as respective ions are produced in expense of the deviation from the same ratio of ampholyte form concentration.