Question

Use the method of successive approximations to determine the pH and concentrations of H2A, HA–, and A2– in a solution of 0.00273 M monopotassium fumarate (KHA). The pKa values for fumaric acid are 3.02 (pKa1) and 4.48 (pKa2).

Answer 1

Monopotassiu Fumarate (KHA) is a strong electrolyte and hence ionizes completely in an aqueous phase.

Hence we write,

KHA (aq.) ------------> K⁺ (aq.) + HA^- (aq.)

[HA^-] = [KHA] = 0.00273 M

The HA^- species thus generated in aqueous medium will undergo reprotonation to H2A (Fumaric acid) or further

deprotonation to A^2- at equilibrium.

1) For pH determination,

HA^- (aq.) -------------> A^2- (aq.) + H⁺ (aq.)

Let 'X' M/L of HA^- ionizes at equilibrium and hence we write ICE table as,

HA^- (aq.) -------------> A^2- (aq.) + H⁺ (aq.)

Initial 0.00273 M 0 0

Change - X + X + X

Equilibrium (0.00273 - X) X X

Ka2 = [A^2-][H⁺] / [HA^-]

Ka2 = (X) x (X) / (0.00273 - X)

HA- being weakly acidic species we expect X <<< 0.00273 M. Small concentration approximation says, 0.00273 - X = 0.00273 only.

Ka2 = X² / 0.00273

X² = 0.00273 x Ka2

X = (0.00273 x Ka2)^1/2.

Here we have to use pKa2 = 4.48 as ionizing species is HA- which ionizes second proton.

pKa = - log(Ka)

Ka = 10^-pKa.

For pKa2 = 4.48

Ka2 = 10^-4.48.

Ka2 = 3.31 x 10^-5.

Using this value in eq.(1)

X = (0.00273 x 3.31 x 10^-5)^1/2.

X = 3.01 x 10^-4 M

I.e.

[A^2-] = [H⁺] = 3.01 x 10^-4 M

Then,

pH = -log[H⁺]

pH = -log(3.01 x 10^-4)

pH = 3.52.

2) The species H2A would form as follows,

HA^- (aq.) + H⁺ (aq.) -----------> H₂A.

Reverse of this reaction as,

H₂A.----------------> HA^- (aq.) + H⁺ (aq.)

The ICE table is given as,

H₂A.----------------> HA^- (aq.) + H⁺ (aq.)

Initial 0 0.00273 3.01 x 10^-4.

Change +Y -Y -Y

At Eq^m. +Y (0.00273 - Y) (3.01 x 10^-4 - Y)

Equilibrium constant Ka1 is given as,

Ka1 = [HA^-][H⁺] / [H₂A]

Using ICE table,

Ka1 = (0.00273 - Y)(3.01 x 10^-4 - Y) / (Y)

Small concentration approximation states,

(0.00273 - Y) = 0.00273 and (3.01 x 10^-4 - Y) = 3.01 x 10^-4.

With we write,

Ka1 = 0.00273 x 3.01 x 10^-4 / Y

Y = (0.00273 x 3.01 x 10^-4) / Ka1 .......(2)

pKa1 = 3.02

Ka1 = 10^-Ka1.

Ka1 = 10^-3.02.

Ka1 = 9.55 x 10^-4.

Using this value in eq.(2) we get,

Y = (0.00273 x 3.01 x 10^-4) / (9.55 x 10^-4)

Y = 8.60 x 10^-4 M

I.e. [H₂A] = 8.60 x 10^-4 M

Finally,

[A^2-] = [H⁺] = 3.01 x 10^-4 M

pH = 3.52

[H₂A] = 8.60 x 10^-4 M

===================XXXXXXXXXXXXXXXXXXXXXXX==========================