Question

Phosphoric acid has a formula of H₃PO₄, and has a K_a1 of 7.5×10^–3 , K_a2 of 6.2×10^–8 , and K_a3 of 4.2×10^–13 at 25 ºC.

What is the equilibrium constant for the following reaction at 25°C?

HPO₄^2–_(aq) + H₂O_(l) ⇄ H₂PO₄^–_(aq) + OH^–_(aq)

A.) 4.2×10^–13

B.) 1.3×10^–12

C.) 2.4×10^–2

D.) 7.5×10^–3

E.) 1.6×10^–7

Answer 1

H3PO4 is triprotic acid the balanced equation are as follows

H3PO4 (aq) + H2O(l) -------------------------------> H2PO4-(aq) + H3O+(aq) => Ka1

H2PO4- (aq) + H2O(l) -------------------------------> HPO4-2(aq) + H3O+(aq) => Ka2

HPO4-2 (aq) + H2O(l) -------------------------------> PO4-3(aq) + H3O+(aq) => Ka3

but in the reaction the given equation is

HPO4-2 (aq) + H2O(l) -------------------------------> H2PO4-(aq) +OH-(aq) => Kb2

because here OH- formed in the product side so here the value is as follows

Kw = Ka * Kb

=>1.0*10^-14 = 6.2*10^-8 * Kb

=> Kb = 1.6129*10^-7

answer => E