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#
From the equilibrium constants below, calculate the equilibrium constant for the overall reaction: H02CC02H ↔ 2H(+) + C2O4(2-)
Given that,
Oxalic ↔ H(+) + HOOCCOO(-) K=
and,
HOOCCOO(-) ↔ H(+) + Oxalate K=
1 Answer
The
https://pubchem.ncbi.nlm.nih.gov/compound/oxalic_acid#section=Environmental-Fate
and they are
#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))#
