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#

1 Answer
Feb 4, 2018

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))#