Question

From the equilibrium constants below, calculate the equilibrium constant for the overall reaction?

From the equilibrium constants below, calculate the equilibrium constant for the overall reaction: H02CC02H ↔ 2H(+) + C2O4(2-)

Given that,
Oxalic ↔ H(+) + HOOCCOO(-) K= `5.6*10^-3`

and,
HOOCCOO(-) ↔ H(+) + Oxalate K= `5.4*10^-3`

Answer 1

The `"pK"_a` values of oxalic acid are given right here:

https://pubchem.ncbi.nlm.nih.gov/compound/oxalic_acid#section=Environmental-Fate

and they are `1.25` and `4.28`, respectively. That gives

`K_(a1) = 10^(-"pK"_(a1)) = 5.62 xx 10^(-2)`

`K_(a2) = 10^(-"pK"_(a2)) = 5.25 xx 10^(-5)`

That makes more physical sense. Oxalic acid has one somewhat weak proton, and one very weak proton. They can't both be identical.

`"H"_2"A"(aq) rightleftharpoons "H"^(+)(aq) + "HA"^(-)(aq)`,

`K_(a1) = 5.62 xx 10^(-2) = (["H"^(+)]["HA"^(-)])/(["H"_2"A"])`

`"HA"^(-)(aq) rightleftharpoons "H"^(+)(aq) + "A"^(2-)(aq)`,

`K_(a2) = 5.25 xx 10^(-5) = (["H"^(+)]["A"^(2-)])/(["HA"^(-)])`

The sum of two reactions must lead to a product of equilibrium constants...

`"H"_2"A"(aq) rightleftharpoons "H"^(+)(aq) + cancel("HA"^(-)(aq))`
`ul(cancel("HA"^(-)(aq)) rightleftharpoons "H"^(+)(aq) + "A"^(2-)(aq))`
`"H"_2"A"(aq) rightleftharpoons 2"H"^(+)(aq) + "A"^(2-)(aq)`

The net equilibrium constant is...

`color(blue)(K_a) = (["H"^(+)]^2["A"^(2-)])/(["H"_2"A"])`

`= underbrace((["H"^(+)]["A"^(2-)])/(cancel(["HA"^(-)])))_(K_(a1))underbrace((["H"^(+)]cancel(["HA"^(-)]))/(["H"_2"A"]))_(K_(a2))`

`= K_(a1)K_(a2)`

`= 5.62 xx 10^(-2) cdot 5.25 xx 10^(-5)`

`= color(blue)ul(2.95 xx 10^(-6))`