What is the equilibrium concentration of H₂C₂O₄ in a 0.370 M oxalic acid,H₂C₂O₄,solution? For oxalic acid,Ka₁ = 5.6 × 10^-2 and Ka₂ = 5.1 × 10^-5. A)2.5 × 10^-1 M B)1.4 × 10^-1 M C)3.7 × 10^-1 M D)5.1 × 10^-5 M E)1.2 × 10^-1 M

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The correct answer is A because the equilibrium concentration of H2C2O4 in a 0.370 M oxalic acid, H2C2O4, solution is 2.5 × 10-1 M. This is because the first dissociation constant of oxalic acid is Ka1 = 5.6 × 10-2 and the second dissociation constant is Ka2 = 5.1 × 10-5. The equilibrium constant for the first dissociation is K1 = [H+][HC2O4-]/[H2C2O4] = 5.6 × 10-2. The equilibrium constant for the second dissociation is K2 = [H+][C2O42-]/[HC2O4-] = 5.1 × 10-5. The equilibrium concentration of H+ can be calculated using the first dissociation constant: [H+] = (Ka1[H2C2O4])/[HC2O4-] = (5.6 × 10-2[0.370 M])/x = 2.1 × 10-2 M. The equilibrium concentration of HC2O4- can be calculated using the second dissociation constant: [HC2O4-] = (Ka2[C2O42-])/[H+] = (5.1 × 10-5[x])/(2.1 × 10-2 M) = 2.4 × 10-3 M. The equilibrium concentration of H2C2O4 can be calculated using the first dissociation constant: [H2C2O4] = [H+][HC2O4-]/Ka1 = (2.1 × 10-2 M)(2.4 × 10-3 M)/5.6 × 10-2 = 2.5 × 10-1 M.

What Is the Equilibrium Concentration of H2C2O4 in a 0

What Is the Equilibrium Concentration of H₂C₂O₄ in a 0