Question

Consider a diprotic acid H2A with ka1=2.5*10^-6 ka2 2.8*10^-12
A) what will be the concentrations of all species present in a 0.125 M solution of H2A
B) what will be the concentrations of all species present in a 0.125 M solution of NaHA

Answer 1

(A) For the first dissociation of H2A

--------------------- H2A + H2O -----> H3O⁺ + HA^-(aq), Ka1 = 2.5x10^-6

Initial conc(M):0.125, ---------------- 0 --------- 0

Eqm.conc(M): (0.125 - y₁) -------- y₁ -------- y₁

Ka1 = 2.5x10^-6 = [H3O⁺] x [HA^-(aq)] / [H2A] = y₁² / (0.125 - y₁)

=> y₁ = 5.58x10^-4 M

Hence [H2A] = (0.125 - y₁) = (0.125 - 5.58x10^-4 M) = 0.12444 M (answer)

For the second dissociation of H2A

--------------------- HA^-(aq) + H2O -----> H3O⁺ + A^2-(aq), Ka2 = 2.8x10^-12

Initial conc(M):5.58x10^-4 ,--------------- 0 ------- 0

Eqm.conc(M):(5.58x10^-4 - y₂) -------- (y₁+y₂), y₂

Ka2 = 2.8x10^-12 = [H3O⁺] x [A^2-(aq)] / [HA^-(aq)] = [(y₁+y₂) x y₂] / (5.58x10^-4 - y₂)

=> Ka2 = 2.8x10^-12 = [(5.58x10^-4 +y₂) x y₂] / (5.58x10^-4 - y₂)

=>  y₂ = 2.80 x 10^-12 M

Hence [HA^-(aq)] = (5.58x10^-4 - y₂) 5.58x10^-4 M (answer)

[A^2-(aq)] = y₂ =  2.80 x 10^-12 M (answer)

(B) HA^-(aq) can either dissociated to form A^2-(aq) or hydrolysed to form H2A

For dissociation of HA^-(aq)

--------------------- HA^-(aq) + H2O -----> H3O⁺ + A^2-(aq), Ka2 = 2.8x10^-12

Initial conc(M):0.125,---------------------- 0 ------- 0

Eqm.conc(M):(0.125- y₂) -------------   y₂ -----, y₂

Ka2 = 2.8x10^-12 = [H3O⁺] x [A^2-(aq)] / [HA^-(aq)] = [y₂ x y₂] / (0.125- y₂) =  y₂² / (0.125- y₂)

=>  y₂ = 5.92x10^-6 M

Hence  [A^2-(aq)] = y₂ = 5.92x10^-6 M (answer)

For the hydrolysis of HA^-(aq)

----------------------HA^-(aq) + H2O ---- > H2A + OH^- ; Kb1 = Kw / Ka1 = 10^-14 / 2.5x10^-6 = 4x10^-9

Initial conc(M): (0.125 - y₂) ------------- 0 ------ 0

Eqm.conc(M): (0.125 - y₂ - y₃), ------- y₃ --- y₃

Kb1 = 4x10^-9 = [H2A] x [OH^-] / [HA^-(aq)] = y₃² / (0.125 - y₂ - y₃)

=> y₃ = 2.24x10^-5

Hence [H2A] = y3 = 2.24x10^-5 M (answer)

[HA^-(aq)] = (0.125 - y₂ - y₃) = (0.125 - 5.92x10^-6  - 2.24x10^-5 ) 0.125 M (answer)