Question

How do pH and pKa relate?

Answer 1

The `"p"H` of a solution is directly related to the `"p"K_"a"` of a solution via the Henderson-Hasselbach equation,

`"p"H = "p"K_"a" + log(([A^-])/([HA]))`

Let's do an example:

What is the `"p"H` of a `"1-L"` solution of `0.12"M"` of `NH_4Cl` to which `"1 L"` of `0.03"M"` of `NaOH` was added (`"p"K_"a"` of `NH_4^(+)` is `9.25^([1])`)?

Consider the equilibrium,

`NH_4^(+) + OH^(-) rightleftharpoons H_2O + NH_3`

It is safe to assume that the hydroxide ion will consume one equivalent ammonium's protons, leaving `0.09"mol"` `NH_4^+` ions and `0.03"mol"` `NH_3`.

Since the total volume cancels out in the ratio of concentrations, we can translate these concentrations into mols and proceed.

To be sure, the hydroxide is treated as a strong base, and the ammonium as a weak acid.

Hence,

`"p"H = "p"K_"a" + log(([NH_3]_"eq")/([NH_4^+]_"eq"))`

`= 9.25 + log("0.03 mols"/("0.12 mols" - "0.03 mols")) approx 8.77`

[1]: Nelson, D. L., Cox, M. M., & Lehninger, A. L. (2017). Lehninger Principles of Biochemistry. New York, NY: W.H. Freeman and Company.