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^-35.6⋅10−3
and,
HOOCCOO(-) ↔ H(+) + Oxalate K= 5.4*10^-35.4⋅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)Ka1=10−pKa1=5.62×10−2
K_(a2) = 10^(-"pK"_(a2)) = 5.25 xx 10^(-5)Ka2=10−pKa2=5.25×10−5
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)H2A(aq)⇌H+(aq)+HA−(aq) ,
K_(a1) = 5.62 xx 10^(-2) = (["H"^(+)]["HA"^(-)])/(["H"_2"A"])Ka1=5.62×10−2=[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=5.25×10−5=[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))