Question

How do we calculate the `pH` of a buffer that is composed of `HPO_4^(2-)` and `H_2PO_4^(-)`?

Answer 1

The buffer equation tells us that `pH=pK_a+log_10([[HPO_4^(2-)]]/[[H_2PO_4^(-)]])`, but here `pK_a=7.20` for `H_2PO_4^(-)`.

Explanation:

A buffer is formed by mixing appreciable quantities of a weak acid and its conjugate base. Such a mixture keeps the `pH` of the solution tolerably close to the `pK_a` of the weak acid.

Now this site gives `pK_a=7.20` for the following reaction:

`H_2PO_4^(-) + H_2O(l) rightleftharpoonsHPO_4^(2-)`,

And thus solutions which contain tolerably equal concentrations of dihydrogen phosphate, and biphosphate, should have `pH` reasonably close to `7.20`.