Question #25ef1

1 Answer
Ernest Z.
Apr 9, 2017

Yes, this is correct.

Explanation:

The equations are

"H"_2"CO"_3 + "H"_2"O" ⇌ "HCO"_3^"-" + "H"_3"O"^"+";color(white)(l) K_1 = X1H2CO3+H2O⇌HCO-3+H3O+;lK1=X1
"HCO"_3^"-" + "H"_2"O" ⇌ "CO"_3^"2-" + "H"_3"O"^"+"; color(white)(ml)K_2 = X2HCO-3+H2O⇌CO2-3+H3O+;mlK2=X2
stackrel(————————————————)("H"_2"CO"_3 + "2H"_2"O" ⇌ "CO"_3^"2-" + "2H"_3"O"^"+"); K_3 = X3

K_1 = (["HCO"_3^"-"]["H"_3"O"^"+"])/(["H"_2"CO"_3]) = X1

K_2 = (["CO"_3^"2-"]["H"_3"O"^"+"])/(["HCO"_3^"-"]) = X2

K_3 = (["CO"_3^"2-"]["H"_3"O"^"+"]^2)/(["H"_2"CO"_3]) = X

Multiply K_text(3) by (["HCO"_3^"-"])/(["HCO"_3^"-"])

Then

K_3 = (["CO"_3^"2-"]["H"_3"O"^"+"]^2)/(["H"_2"CO"_3])× (["HCO"_3^"-"])/(["HCO"_3^"-"]) = (["HCO"_3^"-"]["H"_3"O"^"+"])/(["H"_2"CO"_3]) × (["CO"_3^"2-"]["H"_3"O"^"+"])/(["HCO"_3^"-"]) = X1 × X2 = X

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