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^-35.6103

and,
HOOCCOO(-) ↔ H(+) + Oxalate K= 5.4*10^-35.4103

1 Answer
Feb 4, 2018

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

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

and they are 1.251.25 and 4.284.28, respectively. That gives

K_(a1) = 10^(-"pK"_(a1)) = 5.62 xx 10^(-2)Ka1=10pKa1=5.62×102

K_(a2) = 10^(-"pK"_(a2)) = 5.25 xx 10^(-5)Ka2=10pKa2=5.25×105

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)H2A(aq)H+(aq)+HA(aq),

K_(a1) = 5.62 xx 10^(-2) = (["H"^(+)]["HA"^(-)])/(["H"_2"A"])Ka1=5.62×102=[H+][HA][H2A]

"HA"^(-)(aq) rightleftharpoons "H"^(+)(aq) + "A"^(2-)(aq)HA(aq)H+(aq)+A2(aq),

K_(a2) = 5.25 xx 10^(-5) = (["H"^(+)]["A"^(2-)])/(["HA"^(-)])Ka2=5.25×105=[H+][A2][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))