Chapter EQ, Problem 28PQ

Interpretation Introduction

Interpretation:

0.15 M solution of a weak acid is found to be 1.3% ionized. ${\text{K}}_{\text{a}}$ has to be calculated.

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.

Percent dissociation can be calculated by using following formula,

  $\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}$

Explanation of Solution

Given,

0.15 M solution of a weak acid is found to be 1.3% ionized.

Reason for the correct option.

ICE table:

  $\text{HY(aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{Y}}^{\text{-}}\text{(aq) }\text{+}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}$

Initial concentration	0.15 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.15 - x		x	x

The initial concentration is $\text{0}\text{.15 M}$ acid, hence each mole of acid completely dissociates to form one mole of $Y^{\text{-}}$.

$\begin{array}{l}\text{HY(aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{Y}}^{\text{-}}\text{(aq) }\text{+}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\\ {\text{K}}_{\text{a}}\text{ =}\frac{\left[{\text{Y}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[\text{HY}\right]}\\ \text{given,}\\ {\text{K}}_{\text{a}}\text{ =}\frac{x^2}{(0.15-x)}\end{array}$

Percent dissociation is calculated as follows,

  $\begin{array}{l}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \text{1}\text{.3 = }\frac{x}{0.15}\text{×100}\\ x=1.95\times {10}^{-3}\end{array}$

  $\begin{array}{l}\text{therefore,}\\ \\ {\text{K}}_{\text{a}}\text{ =}\frac{{\text{(}1.95{\text{×10}}^{\text{-3}}\text{)}}^{\text{2}}}{\text{(}\text{0}\text{.15}\text{-}1.95{\text{×10}}^{\text{-3}}\text{)}}\\ {\text{K}}_{\text{a}}\text{ =}\frac{\text{(}3.8025{\text{×10}}^{\text{-6}}\text{)}}{\text{(}0.14805\text{)}}\\ {\text{K}}_{\text{a}}\text{ = 2}{\text{.568×10}}^{\text{-5}}\\ {\text{K}}_{\text{a}}\text{ = 2}{\text{.6×10}}^{\text{-5}}\end{array}$

According to the calculation option (d) is correct answer.

Reason for the incorrect option:

On the basis of equilibrium constant calculation, the rest of the options (a), (b) and (c) are wrong answer.