Answer:
0.0034M = [C2O4²⁻
Explanation:
When H2C2O4 is in water, the equilibrium occurs as follows:
H2C2O4(aq) ⇄ HC2O4⁻(aq) + H⁺(aq)
The equilbrium constant, Ka1 (10^-pKa1 = 0.056234) is defined as:
Ka1 = 0.056234 = [HC2O4⁻] [H⁺] / [H2C2O4]
As both HC2O4⁻ and H⁺ comes from the same equilibrium, [HC2O4⁻] = [H⁺]. We can say:
[HC2O4⁻] = X
[H⁺] = X
[H2C2O4] = 0.19-X
<em>Where X is reaction coordinate.</em>
Replacing:
0.056234 = [X] [X] / [0.19-X]
0.010684 - 0.056234X = X²
0.010684 - 0.056234X - X² = 0
Solving for X:
X = -0.135. False solution. There are no negative concentrations.
X = 0.0790M. Right solution.
That means: [HC2O4⁻] = 0.0790M
Now, the HC2O4⁻ is in equilibrium as follows:
HC2O4⁻ ⇄ C2O4²⁻ + H⁺
The equilibrium constant of this reaction, Ka2, is:
Ka2 = 10^-3.81 = 1.5488x10⁻⁴ = [C2O4²⁻] [H⁺] / [HC2O4⁻]
Based on the same of the first equilibrium:
1.5488x10⁻⁴ = [X] [X] / [0.0790M - X]
1.2236x10⁻⁵ - 1.5488x10⁻⁴X - X² = 0
Solving for X:
X = -0.0036M
X = 0.0034M = [C2O4²⁻]
