The Ka1 value for oxalic acid is 5.9 x10-2 , and the Ka2 value is 4.6 x 10-5 . What are the values of Kb1 and Kb2 of the oxalate ion

2.17x10⁻¹⁰ = Kb1

1.69x10⁻¹³ = Kb2

Explanation:

Oxalic acid, C₂O₄H₂, has two intercambiable protons, its equilibriums are:

C₂O₄H₂ ⇄ C₂O₄H⁻ + H⁺ Ka1 = 5.9x10⁻²

C₂O₄H⁻ ⇄ C₂O₄²⁻ + H⁺ Ka2 = 4.6x10⁻⁵

Oxalate ion, C₂O₄²⁻, has as equilibriums:

C₂O₄²⁻ + H₂O ⇄ C₂O₄H⁻ + OH⁻ Kb1

C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2

Also, you can know: KaₓKb = Kw

Where Kw is 1x10⁻¹⁴

Thus:

Kw = Kb2ₓKa1

1x10⁻¹⁴ =Kb2ₓ4.6x10⁻⁵

And:

Kw = Kb1ₓKa2

1x10⁻¹⁴ =Kb1ₓ5.9x10⁻²

That is because the inverse reaction of, for example, Ka1:

C₂O₄H⁻ + H⁺ ⇄ C₂O₄H₂ K = 1 / Ka1

+ H₂O ⇄ H⁺ + OH⁻ K = Kw = 1x10⁻¹⁴

=

C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2 = Kw × 1/Ka1

the correct answer is boyle

i'm assuming that it's c.) red bank..