Chapter EQ, Problem 26PQ

Interpretation Introduction

Interpretation:

In 0.050 M solution of a weak acid, [H⁺] = 1.8 × 10^-3. ${\text{K}}_{\text{a}}$ 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.

Explanation of Solution

Given,

In 0.050 M solution of a weak acid, [H⁺] = 1.8 × 10^-3.

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 $\text{0}\text{.05 M}$ acid, hence one mole of acid completely dissociates to form one mole of $Y^{\text{-}}$.

$\begin{array}{l}\text{HY(aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{Y}}^{\text{-}}\text{(aq) }\text{+}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\\ {\text{K}}_{\text{a}}\text{ =}\frac{\left[{\text{Y}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[\text{HY}\right]}\\ {\text{K}}_{\text{a}}\text{ =}\frac{x^2}{(0.15-x)}\\ \text{given,}\\ {\text{K}}_{\text{a}}\text{=}\text{0}\text{.15}\text{M}\\ \text{x}\text{=}\left[{\text{Y}}^{\text{-}}\right]\text{=}\left[{\text{H}}^{\text{+}}\right]\text{=}1.8\times {10}^{-3}\text{M}\\ \text{x}\text{=}1.8\times {10}^{-3}\text{M}\end{array}$

  $\begin{array}{l}\text{therefore,}\\ {\text{K}}_{\text{a}}\text{ =}\frac{\left[{\text{Y}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[\text{HY}\right]}\\ {\text{K}}_{\text{a}}\text{ =}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.05-x)}}\\ {\text{K}}_{\text{a}}\text{ =}\frac{\text{(}1.8\times {10}^{-3}\text{M)(}1.8\times {10}^{-3}\text{M)}}{\text{(0}\text{.05-}1.8\times {10}^{-3}\text{M)}}\text{,}\\ {\text{K}}_{\text{a}}\text{ =}\frac{\text{(3}\text{.24}\times {10}^{-6}\text{M)}}{\text{(}0.0482\text{M)}}\\ {\text{K}}_{\text{a}}\text{ = 6}{\text{.72×10}}^{\text{-5}}\\ {\text{K}}_{\text{a}}\text{ = 6}{\text{.7×10}}^{\text{-5}}\end{array}$

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

Reason for the incorrect option:

According to the equilibrium constant calculation, the rest of the options (a), (b) and (d) are wrong answer.