# Question e6c92

Feb 28, 2017

One way is to use the ${K}_{\text{a}}$ values of the acids.

#### Explanation:

This is sort of a Hess's Law problem.

Our target equation is:

$\text{H"_3"PO"_3 "(aq)" + "HSO"_3^"-""(aq)" ⇌ "H"_2"PO"_3^"-""(aq)" + "H"_2"SO"_3"(aq)}$; K = ?

Our two given equations are:

$\boldsymbol{\text{(a)"color(white)(l)"H"_3"PO"_3"(aq)" + "H"_2"O(l)" ⇌ "H"_3"O"^"+""(aq)" + "H"_2"PO"_3^"-""(aq)}}$; K_1 = 5×10^"-2"
$\boldsymbol{\text{(b)"color(white)(l) "H"_2"SO"_3"(aq)"+ "H"_2"O(l)" ⇌ "H"_3"O"^"+""(aq)" +"HSO"_3^"-""(aq)}}$; color(white)(l)K_2 = 1.6×10^"-2"

Now, we combine equations $\boldsymbol{\text{(a)}}$ and $\boldsymbol{\text{(b)}}$ to get the target equation.

$\text{H"_3"PO"_3"(aq)" + color(red)(cancel(color(black)("H"_2"O(l)"))) ⇌ color(red)(cancel(color(black)("H"_3"O"^"+""(aq)"))) + "H"_2"PO"_3^"-""(aq)}$; 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)"))); ${K}_{3} = \frac{1}{K} _ 2$

$\boldsymbol{\text{H"_3"PO"_3"(aq)" + "HSO"_3^"-""(aq)" ⇌ "H"_2"PO"_3^"-""(aq)" + "H"_2"SO"_3"(aq)}}$; $\boldsymbol{K = {K}_{1} {K}_{3}}$

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