Answer:
[H₂A] = 5.0409x10⁻⁷M
[HA⁻] = 0.001951M
[A²⁻] = 0.234
11.29 = pH
Explanation:
When Na₂A = A²⁻is in equilibrium with water, the reactions that occurs are:
A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + OH⁻(aq)
By definition of Ka of reaction, you can find Kb (the inverse basic reaction), thus:
Kb1 = KwₓKa2 =
Where Kw is K of equilibrium of water, 1x10⁻¹⁴
1x10⁻¹⁴/ 6.19x10⁻⁹ =
1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]
And HA⁻ will be in equilibrium as follows:
HA⁻(aq) + H₂O(l) ⇄ H₂A(aq) + OH⁻(aq)
Kb2 = KwₓKa1 = 1x10⁻¹⁴/ 7.68x10⁻⁵ =
1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]
Thus, the Na₂A has 2 equilibriums, for the first reaction, concentrations wil be:
[HA⁻] = X
[OH⁻] = X
[A²⁻] = 0.236M - X
X as how much will react, Reaction coordinate
Replacing in Kb1:
1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]
1.6155x10⁻⁶ = [X] [X] / [0.236-X]
3.8126x10⁻⁶ - 1.6155x10⁻⁶X = X²
3.8126x10⁻⁶ - 1.6155x10⁻⁶X - X² = 0
Solving for X :
X = -0.00195 → False solution. There is no negative concentrations
X = 0.001952.
Replacing, concentrations for the first equilibrium are:
Now, in the second equilibrium:
[HA⁻] = 0.001952 - X
[OH⁻] = X
[H₂A] = X
Replacing in Kb1:
1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]
1.3021x10⁻¹⁰ = [X] [X] / [0.001952 - X]
2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X = X²
2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X - X² = 0
Solving for X :
X = -5.04x10⁻⁷ → False solution. There is no negative concentrations
X = 5.0409x10⁻⁷
Replacing, concentrations for the second equilibrium are:
Thus, you have concentrations of H2A, HA−, and A2−
Now, for pH, you need [OH⁻] concentration that is:
[OH⁻] = 0.0019525
pOH = -log[OH⁻] = 2.709
As 14 = pH+ pOH
14-2.709=pH
