Chapter 5.3, Problem 2E

Interpretation Introduction

Interpretation : The mathematical conversion of below equation (1) to (2) needs to be explained.

  $K_{eq}=\frac{\left[{\text{A}}^{-}\right]\left[\text{HB}\right]}{\left[\text{HA}\right]\left[{\text{B}}^{-}\right]}\to \left(1\right)$

  $K_{eq}=\frac{K_a(\text{HA)}}{K_a(\text{HB)}}\to \left(2\right)$

Concept Introduction :

The equilibrium expression is defined as the ratio of concentration of product to reactant.

  $K_{eq}=\frac{\left[{\text{A}}^{-}\right]\left[\text{HB}\right]}{\left[\text{HA}\right]\left[{\text{B}}^{-}\right]}\to \left(1\right)$

Here, K_eq is the equilibrium constant while [ ] indicates the concentration of each species.

The equilibrium equation can also be written as ratio of acid dissociation constant and base dissociation constant.

  $K_{eq}=\frac{K_a(\text{HA)}}{K_a(\text{HB)}}\to \left(2\right)$

K_a is the dissociation constant.