Question #25ef1

1 Answer
Apr 9, 2017

Answer:

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#