Question

If pKa for phosphoric acid (h3po4) is 2.15, what is Kb for H2po4-?

Answer 1

First, identify the ionization equations for H3PO4, phosphoric acid:

H3PO4(aq) <--> H2PO4-(aq) + H+(aq); Ka1

H2PO4-(aq) <--> HPO4-2(aq) + H+(aq); Ka2

HPO4-2(aq) <--> PO4-3(aq) + H+(aq); Ka3

clearly, we need only

H3PO4(aq) <--> H2PO4-(aq) + H+(aq); Ka1

since H3PO4 and H2PO4- are present in such equation

if pKa = 2.15

then

at T = 25°C, expect

14 = pKa + pKb

to be valid

pKb = 14-pKa = 14-2.15 = 11.85

pKb = 11.85

we need Kb only so

pKb = -log(Kb)

solve for Kb

11.85 = -log(Kb)

Kb = 10^- 11.85

Kb = 1.412*10^-12