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

1 Answer
Apr 13, 2017

Answer:

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#.