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

At equilibrium, the pH of the solution is 3.48; [H2A] = 0.0496 and [A2-] = 8.2 × 10^-9 M

First ionization

H2A <> H+ + HA- ;Ka1 = 2.2 × 10^-6

Ka1 = 2.2x10^-6 = [H+][HA-]/[H2A]

At equilibrium:

[H+] = X[HA-] = X[H2A] = 0.05 - X

Substituting in the equation of Ka1

2.2 × 10^-6 = X^2/0.050 - X

Assuming X is small relative to 0.050, 0.050 - X simplifies to 0.050

X^2 = 1.1 × 10^-7

X = 3.31 × 10^-4

Since Ka1 <<< Ka2 and [H+] = X

[H+] = 3.31 × 10^-4

pH = - log 3.31 × 10^-4

pH = 3.48

Equilibrium concentration of H2A

[H2A] = 0.05 - 3.31 × 10^-4

[H2A] = 0.0496

Second ionization:

HA- <> H+ + A- Ka2 = 8.2 × 10^-9

Ka2 = 8.2 × 10^-9 = [H+][A2-]/[HA-]

HA- > H+ + A2-

3.31 × 10^-4 3.31 × 10^-4 0 Initial

-y +y +y Change

{3.31 ×10^-4 -y} {3.31 x10^-4+y} {+y} Equilibrium

Assuming y is very small:

{3.31 ×10^-4 - y} simplifies to 3.31 ×10^-4 {3.31 ×10^-4 + y} simplifies to 3.31 ×10^-4

Substituting in the equation of Ka2

8.2 × 10^-9 = (3.31x10^-4) × y /3.31 ×10^-4

y = 8.2 × 10^-9

Equilibrium concentration of A2-

Since [A2-] = y

[A2-] = 8.2 × 10^-9 M

Therefore, at equilibrium, the pH of the solution is 3.48; [H2A] = 0.0496 and [A2-] = 8.2 × 10^-9 M

Learn more about pH and equilibrium concentration of acid-base conjugates of weak acid at: link

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