What is the hydroxide-ion concentration of a 0.150 M oxalic acid,H₂C₂O₄,solution? For oxalic acid,Ka₁ = 5.6 × 10^-2 and Ka₂ = 5.1 × 10^-5. A)3.6 × 10^-12 M B)1.1 × 10^-13 M C)2.7 × 10^-3 M D)1.5 × 10^-13 M E)1.0 × 10^-7 M

Question

Quizplus · Accepted Answer

The balanced equation of the dissociation of oxalic acid is:

HS1U1B12S1U1B0CS1U1B12S1U1B0OS1U1B14S1U1B0(aq) + H2O(l) ↔ H3O+(aq) + HC2O4-S1U1B12S1U1B0O4S1U1B0-(aq)

Let x be the concentration of H3O+ and HC2O4-S1U1B12S1U1B0O4S1U1B0- ions in the solution then [H3O+] = [HC2O4-S1U1B12S1U1B0O4S1U1B0-] = x.

Using the Ka for the dissociation of HC2O4-S1U1B12S1U1B0O4S1U1B0-:

KaS1U1B11S1U1B0 = [H3O+][C2O4-S1U1B12S1U1B0O4S1U1B0-]/[HC2O4-S1U1B12S1U1B0O4S1U1B0-]

5.6 × 10S1U1P1-2S1S1P0 = x2/0.150 - x

Since the acid is diprotic, use KaS1U1B12S1U1B0, which gives:

KaS1U1B12S1U1B0 = [H3O+][HC2O4-S1U1B12S1U1B0O4S1U1B0-]/[C2O4-S1U1B12S1U1B0O4S1U1B0-]

5.1 × 10S1U1P1-5S1S1P0 = x2/(2x + 0.150)

Solving the two equations simultaneously yields x = 1.5 × 10S1U1P1-13S1S1P0 M, which is the concentration of hydroxide ions in the solution. 
Therefore, the answer is D) 1.5 × 10S1U1P1-13S1S1P0 M.

What Is the Hydroxide-Ion Concentration of a 0