Question

Question #e6c92

Answer 1

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`