A diprotic acid, H2A, has the following ionization constants: Ka1 = 1.1 10-3 and

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
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
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

Explanation / Answer

Ka = [H+][Acid-]/[HAcid]

Note that in your case, the HAcid/Acid- pairs are H2A and HA- in the first case, and HA- and A2- in the second case. Anyway, the equation can be rearranged to:

[H+] = Ka[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 Ka = pKa

log (1.1*10-3)= -pKa

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

Applying Henderson–Hasselbalch equation

pH = pKa + 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-<---> A2-
pKa = - log (2.5 x 10-6 ) = 5.60

5.80 = 5.60 + log [A2-] / [HA-]
10-0.2 = 0.63 = [A2-] / [HA-]

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