Question

Calculate the pH of a 0.1620 M aqueous solution of sodium dihydrogen phosphate, NaH₂PO₄.

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pH =

Answer 1

Let's simplify this situation assuming that water ionization can be neglected, and all equations we have to use are these for Ka1, Ka2 and mass balance:

12.1

12.2

12.3

CHA- is the concentration of the source of the amphiprotic substance - for example of the NaHA salt.

H+ ions are produced in the dissociation reaction (together with A2-) and consumed in the hydrolysis (yielding H2A), so their concentration is:

12.4

Using 12.1 and 12.2 we can rewrite 12.4 in the form:

12.5

or, after rearranging:

12.6

It is time for two more assumptions. First of all, let's assume that neither dissociation nor hydrolysis goes too far, and [HA-] = CHA-. If so

12.7

Now, let's look at the denominator - if the CHA- is sufficiently larger than Ka1 we can neglect Ka1 and whole equation takes form

12.8

or

12.9