What is the hydroxide-ion concentration of a 0.190 M sodium oxalate (Na₂C₂O₄)solution? For oxalic acid (H₂C₂O₄),K_a1 = 5.6 × 10^-2 and K_a2 = 5.1 × 10^-5.(K_w = 1.01 × 10^-14) A) 6.1 × 10^-6 M B) 1.0 × 10^-7 M C) 7.9 × 10^-2 M D) 3.1 × 10^-3 M E) 1.8 × 10^-7 M

Question

Quizplus · Accepted Answer

The balanced equation for the reaction between oxalic acid and hydroxide ion is: 
HS1U1B12S1U1B0CS1U1B12S1U1B0OS1U1B14S1U1B0 + 2OH1U1B1-S1U1B0 → S1U1B12S1U1B0CS1U1B12S1U1B0O4S1U1B1-2S1U1B0 + H2O
The equilibrium expression for this reaction is:
KS1U1B1wS1U1B0 = [S1U1B12S1U1B0CS1U1B12S1U1B0O4S1U1B1-2S1U1B0][OH1U1B1-S1U1B0]2/[HS1U1B12S1U1B0CS1U1B12S1U1B0OS1U1B14S1U1B0]
At equilibrium, the concentration of [HS1U1B12S1U1B0CS1U1B12S1U1B0OS1U1B14S1U1B0] is equal to the initial concentration of sodium oxalate because they are in a 1:1 stoichiometric ratio. Therefore, [HS1U1B12S1U1B0CS1U1B12S1U1B0OS1U1B14S1U1B0] = 0.190 M. 
Let x be the concentration of hydroxide ion at equilibrium. 
Then, using the equilibrium expression and the given equilibrium constants: 
1.01 × 10S1U1P1-14S1S1P0 = [S1U1B12S1U1B0CS1U1B12S1U1B0O4S1U1B1-2S1U1B0](x2)/(0.190)
x = 6.1 × 10S1U1P1-6S1S1P0 M, which is the hydroxide-ion concentration at equilibrium. Therefore, the answer is A.

What Is the Hydroxide-Ion Concentration of a 0