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juicemankinnie95
15.07.2020 •
Chemistry
Calculate the pH and concentrations of H2A, HA−, and A2−, at equilibrium for a 0.236 M solution of Na2A. The acid dissociation constants for H2A are Ka1=7.68×10−5 and Ka2=6.19×10−9.
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Ответ:
[H₂A] = 5.0409x10⁻⁷M
[HA⁻] = 0.001951M
[A²⁻] = 0.234
11.29 = pH
Explanation:
When Na₂A is in equilibrium with water, the reactions that occurs are:
2Na⁺ + A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + 2Na⁺(aq) + OH⁻(aq)
As sodium ion doesn't react:
A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + OH⁻(aq)
Kb1 = KwₓKa2 = 1x10⁻¹⁴/ 6.19x10⁻⁹ = 1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]
And HA⁻ will be in equilibrium:
HA⁻(aq) + H₂O(l) ⇄ H₂A(aq) + OH⁻(aq)
Kb2 = KwₓKa1 = 1x10⁻¹⁴/ 7.68x10⁻⁵ = 1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]
In the reaction, you have 2 equilibriums, for the first reaction, concentrations in equilibrium are:
[HA⁻] = X
[OH⁻] = X
[A²⁻] = 0.236M - X
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:
[HA⁻] = 0.001952
[OH⁻] = 0.001952
[A²⁻] = 0.234
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:
[HA⁻] = 0.001951M
[OH⁻] = 5.0409x10⁻⁷M
[H₂A] = 5.0409x10⁻⁷M
Thus, you have concentrations of H2A, HA−, and A2−
Now, for pH, the sum of both productions of [OH⁻] is:
[OH⁻] = 0.0019525
pOH = -log[OH⁻] = 2.709
As 14 = pH+ pOH
11.29 = pH
Ответ:
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