A diprotic acid, H2A, has acid dissociation constants of Ka1 = 1.78× 10–4 and Ka

Question

A diprotic acid, H2A, has acid dissociation constants of Ka1 = 1.78× 10–4 and Ka2 = 4.03× 10–12.

Calculate the pH and molar concentrations of H2A, HA–, and A2– at equilibrium for each of the solutions below.

(a) a 0.137 M solution of H2A

(b) a 0.137 M solution of NaHA

(c) a 0.137 M solution of Na2A

(a) a 0.137 M solution of H2A pH H, A Number Number (b) a 0.137 M solution of NaHA pH H,A Number Number (c) a 0.137 M solution of Na2A pH H,A Number Number M M M HA Number HA Number HA Number M Number Number Number M M M

Explanation / Answer

Ka1 = 1.78 x 10-4 ; pKa1 = 3.75

Ka2 = 4.03 x 10-12 ; pKa2 = 11.39

a)

C = 0.137 M

H2A -> HA- + H+

(C – x) -> (x - y) +(x + y)

HA- -> A2- + H+

(x - y) -> y + (x + y)

Ka2 = y (x + y) / (x – y)

Assuming y << x, we get

Ka2 = y = 4.03 x 10-12

Ka1 = (x – y)(x + y)/(C – x)

Ka1 = x2 / (C – x) = 1.78 x 10-4

x2 / (0.137 – x) = 1.78 x 10-4

Solving we get x = 4.85 x 10-3 M >> y

[H+] = x

pH = -log(x) = 2.31

[H2A] = C – x = 0.132 M

[HA-] = x = 4.85 x 10-3 M

[A2-] = y = 4.03 x 10-12 M ~ 0

b)

C = 0.137 M

Since C >> Ka1

pH = (pKa1 + pKa2) / 2

= (3.75 + 11.39) / 2

= 7.57

[H+] = 10-7.57 = 2.68 x 10-8 M

[OH-] = Kw / [H+] = 3.73 x 10-7 M

HA- -> A2- + H+

(C – x - y) -> x + [H+]

Ka2 = x * [H+] / (C – x – y)

Assuming C >> x + y, we get

x = Ka2 * C / [H+] = 2.06 x 10-5 M

HA- + H2O -> H2A + OH-

(C – x - y) -> y + [OH-]

Kb1 = Kw/Ka1 = 5.62 x 10-11 = y [OH-] / (C – x – y)

y = 5.62 x 10-11 * C / [OH-] = 2.06 x 10-5 M

pH = -log([H+]) = 7.57

[H2A] = y = 2.06 x 10-5 M

[HA-] = C – x - y = 0.13696 M

[A2-] = x = 2.06 x 10-5 M

c)

C = 0.137 M

A2- + H2O -> HA- + OH- ; Kb2 = Kw/Ka2 = 2.48 x 10-3

(C – x - y) -> (x – y) + (x + y)

HA- + H2O -> H2A + OH- ; Kb1 = Kw/Ka1 = 5.62 x 10-11

(x - y) -> y + (x + y)

Kb1 = y (x + y) / (x – y)

Assuming y << x, we get

Kb1 = y = 5.62 x 10-11

Kb2 = (x – y)(x + y)/(C – x)

Kb2 = x2 / (C – x) = 2.48 x 10-3

x2 / (0.137 – x) = 2.48 x 10-3

Solving we get x = 0.017 M >> y

[OH-] = x

pOH = -log (x) = 1.76

pH = 14 - 1.76 = 12.24

[H2A] = y = 5.62 x 10-11 M

[HA-] = x = 0.017 M

[A2-] = C - x = 0.12 M

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