Question

Question #25ef1

Answer 1

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 = X1`
`"HCO"_3^"-" + "H"_2"O" ⇌ "CO"_3^"2-" + "H"_3"O"^"+"; color(white)(ml)K_2 = 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`