Question

Calculate the concentrations of H₂C₂O₄, HC₂O₄⁻, C₂O₄²⁻, and H⁺ ions in a 0.193 M oxalic acid solution at 25

°C. (K_a1 = 6.5 × 10⁻² and K_a2 = 6.1 ×10⁻⁵ for oxalic acid.) (Enter your answer in scientific notation.)

Answer 1

H₂C₂O₄ ---------> H⁺ + HC₂O₄^- K_a1 = 6.5*10^-2
At equilibrium concentrations of reactant and products will be,
[H₂C₂O₄] = 0.193-X ; [H⁺] = +X ; [HC₂O₄^-] = +X
K_a1 = [H⁺] [HC₂O₄^-] / [H₂C₂O₄] = X² / (0.193-X)
6.5*10^-2 = X² / (0.193-X)
0.012545 - 6.5*10^-2X = X²
X² + 6.5*10^-2X - 0.012545 = 0
solve this expression for X
X = 0.08412 M = [H⁺] = [HC₂O₄^-] ; [H₂C₂O₄] = 0.193-0.08412 = 0.10888M
Now consider,
HC₂O₄^- ---------> H⁺ + C₂O₄^-
at equilibrium concentrations will be
[HC₂O₄^-] = 0.08412-Y ; [H⁺] = Y ; [C₂O₄^-] = Y
K_a2 = [H⁺] [C₂O₄^-] / [HC₂O₄^-] = Y² / (0.08412-Y) = 6.1*10^-5
5.1313*10^-6- 6.1*10^-5Y = Y
Y+6.1*10^-5Y-5.1313*10^-6 = 0
After solving the above quadratic expression we get Y = 2.2349*10^-3
Therefore the concentrations will be,
[H⁺] = [C₂O₄^-] = 2.2349*10^-3 M
[HC₂O₄^-] = 0.08412 - 2.2349*10^-3 = 0.08188 M
[H₂C₂O₄] = 0.10888 M