Question

For the diprotic weak acid H2A, a1=3.2×10−6 and a2=6.1×10−9 .

Answer 1

In the first dissociation of H2A:

molarity    H2A(aq)↔ (HA)^-(aq) + H^+(aq)

initial                0.05 m          0 m           0 m

change               -x                 +x               +x

equilibrium    0.05-x               x                 x

we can neglect X in [H2A] as it so small compared to the 0.05

so by substitution in Ka equation:

Ka1 = [HA][H] / [H2A]

2.2x10^-6 = X^2/0.05

X = √(2.2x10^-6)*(0.05)= 1.1x10^-7

X=   3.32x10^-4 m

∴ [H2A] = 0.05 - 3.32x10^-4 = 0.0497 m

[HA] = 3.32x10^-4 m

[H] = 3.32x10^-4 m

the second dissociation of H2A: when ka2 = 8.2x10^-9

                          HA-(aq)     ↔ A^2- (aq) + H+(aq)

at equilibrium   3.32x10^-4        y              3.32x10^-4

Ka2           = [H+][A^2-] / [HA]

8.2x10^-9 = Y(3.32x10^-4)/(3.32x10^-4)

∴y = 8.2x10^-9 m

∴[A] = 8.2x10^-9 m

PH= -㏒[H+]

   = -㏒(3.32x10^-4)= 3.479  

[A]=8.2x10^-9 m

[H2A] = 0.0497 ≈ 0.05 m