Question

A diprotic acid, H2A, has the following ionization constants: Ka1 = 1.1 10-3 and Ka2 = 2.5 10-6. In order to make up a buffer solution of pH 5.80, which combination would you choose, NaHA/H2A or Na2A/NaHA?

NaHA/H2A

Na2A/NaHA

pKa of the acid component

Answer 1

K_a = [H⁺][Acid^-]/[HAcid]

Note that in your case, the HAcid/Acid^- pairs are H₂A and HA^- in the first case, and HA^- and A^2- in the second case. Anyway, the equation can be rearranged to:

[H⁺] = K_a[HA]/[A^-]

This is very useful for buffer solutions because we can calculate [H⁺] (and hence pH) or

First calculate the value of pka : - log K_a = pK_a

log (1.1*10^-3)= -pKa

pka=log (1.1*10^-3)= -(-3)= 3

Applying Henderson–Hasselbalch equation

pH = pK_a + log [Base]/[Acid]

5.80 = 3 + log[NaHA]/[H₂A]

log[NaHA]/[H₂A] = 2.8

[NaHA]/[H₂A] = 10^-2.8 = 0.0015

The second one, because roughly equal concentrations of NaHA and Na₂A will give a pH nearer to 5.8 than the other one - a more effective buffer will result.

HA^-<---> A^2-
pKa = - log (2.5 x 10^-6 ) = 5.60

5.80 = 5.60 + log [A^2-] / [HA^-]
10^-0.2 = 0.63 = [A^2-] / [HA^-]