Chapter 18, Problem 18.72P

Interpretation Introduction

Interpretation:

${\text{K}}_{\text{a}}$ of $\text{HY}$ has to be calculated when the solution of has $\text{pH 2}\text{.68}$.

Concept introduction:

An equilibrium constant $\left(K\right)$ is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

For the general acid HA,

  $\text{HA}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{+}\left(\text{aq}\right)+{\text{A}}^{-}\left(\text{aq}\right)$

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

$K_a=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$                                                                                                 $\left(1\right)$

An equilibrium constant $\left(K\right)$ with subscript $a$ indicate that it is an equilibrium constant of an acid in water.

The formula is given below for calculating concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$,

${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] = 10}}^{\text{-pH}}$

The molarity can be calculated by using following formula,

$\text{Molarity}\text{= }\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{the}\text{solution}}$