Chemistry Question

Question

Given a diprotic acid, H2A, with two ionization constants of Ka1=4.6×10^− 4 and Ka2=5.0×10^−11, calculate the pH for a 0.168M solution of NaHA.

Answer 1

Since H2A is a weak acid, NaHA reacts with H2O as follows:

NaHA + H2O = H2A + OH- + Na+ The equal sign represents the double arrow of equilibrium.

As HA- acts as a base, the equilibrium expressing can be written

Kb = [OH-] [H2A] / [HA-] – [OH-] We will use the approximation that [OH-] << [HA-] and since

[OH-] = [H2A]

Kb = [OH-]² / [HA-]. Solving for [OH-], we get

[OH-] = SqRt {Kb x [HA-]}

 Kb = Kw / Ka = 1.0 x 10^-14 / 4.6 x 10^-4 = 2.17 x 10^11

So [OH-] = SqRt {2.17 x 10^-11 x 0.168} = 1.91 x 10^-6.

[H+] =  Kw/[OH-] = 1.0 x 10^-14/1.91 x 10^-6 = 5.2 x 10^-9

pH = - log [H+] = 8.3

Since the [OH-] calculated is << 0.168, the approximation used is justified.

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