Question

A diprotic acid, H2A, has acid dissociation constants of Ka1 = 1.63× 10–4 and Ka2 = 4.71× 10–11. Calculate the pH and molar concentrations of H2A, HA–, and A2– at equilibrium for each of the solutions below.

(a) a 0.184 M solution of H2A

(b) a 0.184 M solution of NaHA

(c) a 0.184 M solution of Na2A

Answer 1

(a) 0.184 M H₂A.

Ionization of H₂A is given as

H₂A. <--------> HA^- (aq) + H⁺ (aq) Ka1 = 1.63 x 10^-4.

Ka1 = [HA^-][H⁺] / [H₂A] = 1.63 x 10^-4 . ---------------- (1)

Initially, [H₂A] = 0.184 M, let at equilibrium 'X' M of H₂A ionizes as per above first ionization equation.

Accordingly we write ICE table as,

H₂A. <--------> HA^- (aq) + H⁺ (aq)

Initially 0.184 0 0

Change -X +X +X

At eq^m. (0.184-X) X X

Using equilibrium concentrations in eq.(1) we have,

(X)(X) / (0.184-X) = 1.63 x 10^-4.

From dissociation constant values it is clear that H₂A is a weak acid and ionizes to very small extent hence, "small concentration assumption" holds true. So, (0.184-X) = 0.184 (apprx.) with this abve eq. takes form,

X² / 0.184 = 1.63 x 10^-4.

X² = 0.184 x 1.63 x 10^-4.

X² = 29.99 x 10^-6.

X = 5.48 x 10^-3.

From ICE table,

[HA^-] = [H⁺] = X = 5.48 x 10^-3 M

And [H₂A] = 0.184 - X = 0.184 - 5.48 x 10^-3 = 0.179 M.

By definition of pH,

pH = -log[H⁺]

pH = -log(5.48 x 10^-3)

pH = 2.26

Second dissociation can be neglected for as ionizing species at second step HA^- is very small in concentration and dissociation constant Ka2 also is very much small.

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b) 0.184 M NaHA.

NaHA is a strong electrolyte.

NaHA ------> Na⁺ (aq) + HA^- (aq)

So, [HA^-] = [NaHA] = 0.184 M.

(note, here [HA^-] is considerable and hence ionization is also considerable unlike in case (a).

HA^- undergo ionization according to the equation,

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

Ka2 = [A^2-][H⁺] / [HA^-] =

Initially, [HA^-] = 0.184 M and let at equilibrium ‘X’ M HA^- ionizes.

ICE table is

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

Initial                                   0.184                   0               0

Change                                 -X                      +X             +X

At Eq^m.                                (0.184-X)               X               X

Using equilibrium concentrations in eq.(2)

(X)(X) / (0.184-X) = 4.71 x 10^-11.

Being HA^- weak acid, “small concentration assumption” holds true and so 0.184-X = 0.184 (apprx.)

X² / 0.184 = 4.71 x 10^-11.

X² = 0.184 x 4.71 x 10^-11.

X² = 8.67 x 10^-12.

X = 2.95 x 10^-6.

By ICE table,

[A^2-] = [H⁺] = X = 2.95 x 10^-6M

[HA^-] = 0.184 – X = 0.184 - 2.95 x 10^-6. = 0.183 M (appx.)

By definition of pH

pH = -log[H⁺]

pH = -log(2.95 x 10^-6)

pH = 5.53

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c) 0.184 M Na2A.

Na2A is a strong electrolyte and hence,

[A^2-] = [Na₂A] = 0.184 M

A^2- is a strong conjugate base and hence abstract proton from H2O and gives HA^- and HO^- in solution.

Kb1 for A^2- = Kw / Ka2 = (1x10^-14) /(4.71x10^-11) = 2.12 x 10^-4.

Corresponding hydrolysis reaction of A^2- is given as,

A^2- (aq) + H₂O <----------> HA^- (aq) + HO^- (aq).

Kb1 = [HA^-][HO^-] / [A^2-] = 2.12 x 10^-4. ------------- (3)

Initially, [A^2-] = 0.184 M and let ‘X’ M A^2- hydrolyzes to HA^- and HO^-.

Accordingly we write ICE table as,

                                        A^2- (aq) + H₂O <----------> HA^- (aq) + HO^- (aq).

Initially                             0.184                                  0             0

Change                              -X                                   +X              +X

At Eq^m.                             (0.184-X)                         X                  X

Using equilibrium concentrations in eq.(3) we get,

(X)(X)/(0.184-X) = 2.21 x 10^-4.

Small concentration assumption holds tru and hence 0.184-X = 0.184.

X² = 0.184 x 2.1 x 10^-4.

X² = 40.66 x 10^-6.

X = 6.38 x10^-3.

By ICE table,

[HA^-] = [HO^-] = X = 6.38 x 10^-3 M

[A^2-] = 0.184 – X = 0.184 – 6.38 x 10^-3 = 0.178 M. (apprx.)

By definition of pOH,

pOH = -log[HO^-]

pOH = -log(6.38 x 10^-3)

pOH = 2.2

pH + pOH = 14

pH = 14 – pOH = 14 – 2.2 = 11.8

pH= 11.8

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