Chapter 16, Problem 101GQ

Interpretation Introduction

Interpretation:

The equilibrium constant has to be calculated for the following rection.

${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{(aq) + 2H}}_{\text{2}}\text{O}\rightleftharpoons {\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}{\text{(aq) + 2H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}$

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

${\text{K}}_{\text{a}}$ is an acid constant for equilibrium reactions.

$\begin{array}{l}{\text{HA + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+ A}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

${\text{K}}_{\text{b}}$ is base constant for equilibrium reaction.

$\begin{array}{l}{\text{BOH + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{B}}^{\text{+}}{\text{+ OH}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[B}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\end{array}$