Question

Find the value of the equilibrium constant for the following equilibrium HAsO42–(aq) + H2O(l) ⇌ H2AsO4–(aq) + OH–(aq) given the acid ionization constants for H3AsO4: pKa1=2.26, pKa2=6.77, pKa3=11.29.

Answer 1

Consider reaction, HAsO42–( aq) + H2O (l) H2AsO4– (aq) + OH–(aq)

Equilibrium constant for above reaction is K b = [ H2AsO4– ] [ OH– ] / [ HAsO42– ]

HAsO42– is conjugate base of acid H2AsO4–.

For conjugate acid base pair, we have relation K a K b = 1.00 10 -14

K b = 1.00 10 -14 / K a

We have relation, pKa = - log Ka

Ka = 10 - pKa

For H2AsO4– pKa is 6.77

Ka of  H2AsO4– = 10 -6.77 = 1.7 10 -7

Substituting Ka = 1.7 10 -7 in  K b = 1.00 10 -14 / K a , we get

K b = 1.00 10 -14 / 1.7 10 -7

K b = 5.9 10 -8

ANSWER :   K b = 5.9 10 -8