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

Apr 13, 2017

The buffer equation tells us that $p H = p {K}_{a} + {\log}_{10} \left(\frac{\left[H P {O}_{4}^{2 -}\right]}{\left[{H}_{2} P {O}_{4}^{-}\right]}\right)$, but here $p {K}_{a} = 7.20$ for ${H}_{2} P {O}_{4}^{-}$.

#### Explanation:

A buffer is formed by mixing appreciable quantities of a weak acid and its conjugate base. Such a mixture keeps the $p H$ of the solution tolerably close to the $p {K}_{a}$ of the weak acid.

Now this site gives $p {K}_{a} = 7.20$ for the following reaction:

${H}_{2} P {O}_{4}^{-} + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s H P {O}_{4}^{2 -}$,

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