For the diprotic weak acid h2a, ka1 = 2.2 * 10-6 m and ka2 = 8.2 * 10-9 m. what is the ph of a 0.0500 m soluti? for the diprotic weak acid h2a, ka1 = 2.2 * 10-6 m and ka2 = 8.2 * 10-9 m. what is the ph of a 0.0500 m solution of h2a? what are the equilibrium concentrations of h2a and a2 - in this solution?

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