What is the hydroxide-ion concentration of a 0.190 M sodium oxalate (Na₂C₂O₄)solution? For oxalic acid (H₂C₂O₄),Ka₁ = 5.6 × 10^-2 and Ka₂ = 5.1 × 10^-5.

Question

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First, write the dissociation equation for sodium oxalate: Na₂C₂O₄ -> Na+ + S2O42- This compound does not contribute to the hydroxide-ion concentration, so we can ignore it for the purposes of this question. The dissociation equation for oxalic acid is: H₂C₂O₄ -> H+ + S1O42- Using the first acid dissociation constant (Ka1= 5.6 × 10^-2), we can find the equilibrium concentrations of H+ and S1O42-: Ka1 = [H+][S1O42-]/[H₂C₂O₄] 5.6 × 10^-2 = [H+]^2 / (0.190 - [H+]) Because the concentration of oxalic acid is much smaller than 0.190, we can approximate 0.190 - [H+] as 0.190: 5.6 × 10^-2 = [H+]^2 / 0.190 [H+] = 3.1 × 10^-3 M Now we can use the second acid dissociation constant (Ka2 = 5.1 × 10^-5) to find the concentration of S2O42-: Ka2 = [H+][S2O42-]/[H₂C₂O₄] 5.1 × 10^-5 = (3.1 × 10^-3)^2 / [S2O42-] [S2O42-] = 6.1 × 10^-6 M Because each mole of oxalic acid produces two moles of hydroxide ions when it fully dissociates, the hydroxide-ion concentration is: 2[H+] = 6.1 × 10^-6 M Therefore, the answer is A.

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