Question #e6c92

1 Answer
Feb 28, 2017

Answer:

One way is to use the #K_"a"# values of the acids.

Explanation:

This is sort of a Hess's Law problem.

Our target equation is:

#"H"_3"PO"_3 "(aq)" + "HSO"_3^"-""(aq)" ⇌ "H"_2"PO"_3^"-""(aq)" + "H"_2"SO"_3"(aq)"#; #K = ?#

Our two given equations are:

#bb"(a)"color(white)(l)"H"_3"PO"_3"(aq)" + "H"_2"O(l)" ⇌ "H"_3"O"^"+""(aq)" + "H"_2"PO"_3^"-""(aq)"#; #K_1 = 5×10^"-2"#
#bb"(b)"color(white)(l) "H"_2"SO"_3"(aq)"+ "H"_2"O(l)" ⇌ "H"_3"O"^"+""(aq)" +"HSO"_3^"-""(aq)"#; #color(white)(l)K_2 = 1.6×10^"-2"#

Now, we combine equations #bb"(a)"# and #bb"(b)"# to get the target equation.

#"H"_3"PO"_3"(aq)" + color(red)(cancel(color(black)("H"_2"O(l)"))) ⇌ color(red)(cancel(color(black)("H"_3"O"^"+""(aq)"))) + "H"_2"PO"_3^"-""(aq)"#; #K_1 = 5×10^"-2"#
#"HSO"_3^"-""(aq)" + color(red)(cancel(color(black)("H"_3"O"^"+""(aq)"))) ⇌ "H"_2"SO"_3"(aq)"+ color(red)(cancel(color(black)("H"_2"O(l)")))#; #K_3 = 1/K_2#

#bb("H"_3"PO"_3"(aq)" + "HSO"_3^"-""(aq)" ⇌ "H"_2"PO"_3^"-""(aq)" + "H"_2"SO"_3"(aq)")#; #bb(K = K_1K_3)#

#K_"c" = K_1K_3 = K_1/K_2 = (5×10^"-2")/(1.6×10^"-2") = 3#