Chapter 8, Problem 104P

8-112 Consider an initial 0.040 M hypobromous acid (HOBr) solution at a certain temperature.

At equilibrium after partial dissociation, its pH is found to be 5.05. What is the acid ionization constant, K_a, for hypobromous acid at this temperature?

Interpretation Introduction

Interpretation:

The acid dissociation constant of hypobromous acid is to be calculated.

Concept Introduction:

Weak acids do not dissociate completely. Let HA be a weak acid. The dissociation of the weak acid can be represented by the chemical equation,

$\text{HA}(\text{a}q)\underset{}{\rightleftharpoons }{\text{H}}^{+}(\text{a}q)+{\text{A}}^{-}(\text{a}q)$

The equation for acid dissociation constant can be written from this chemical equation.

$K_{\text{a}}=\frac{[{\text{H}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}$

Here, $\left[{\text{H}}^{+}\right]$ is concentration of hydrogen ion, $\left[{\text{A}}^{-}\right]$ is concentration of conjugate base and $\left[\mathrm{HA}\right]$ is concentration of acid.

Answer to Problem 104P

The acid dissociation constant of hypobromous acid is $K_{\text{a}}=1.9858\times {10}^{-9}$.

Explanation of Solution

Hypobromous acid is a weak acid. Hence, it do not dissociate completely. The dissociation of the given weak acid can be represented by the chemical equation,

$\text{HOBr}(\text{a}q)\underset{}{\rightleftharpoons }{\text{H}}^{+}(\text{a}q)+{\text{OBr}}^{-}(\text{a}q)$

The equation for acid dissociation constant can be written from this chemical equation.

$K_{\text{a}}=\frac{[{\text{H}}^{+}][{\text{OBr}}^{-}]}{[\text{HOBr}]}$

The concentrations of each of the ions at equilibrium can be obtained from the ICE table. Where ICE represents the Initial, Change and Equilibrium concentrations of the weak acid.

$\begin{array}{l}\text{ HOBr}(\text{a}q)\underset{}{\rightleftharpoons }{\text{H}}^{+}(\text{a}q)+{\text{OBr}}^{-}(\text{a}q)\\ {\text{ [HOBr] [H}}^{+}{\text{] [OBr}}^{-}]\\ \begin{array}{l}\text{i} 0.040 0 0\hfill \\ \hfill \\ \text{C} -\text{X} +\text{X} +\text{X}\hfill \\ \hfill \\ \text{E} 0.040-\text{X} +\text{X} +\text{X}\hfill \end{array}\end{array}$

The hydrogen ion concentration can be obtained from the given pH. The pH is defined as the negative logarithm of the hydrogen ion concentration.

$p\text{H}=-\log [{\text{H}}^{+}]$

The pH of the weak acid solution at equilibrium is 5.05. Thus, we can calculate the concentration of the hydrogen ion.

$\begin{array}{l}p\text{H}=-\log [{\text{H}}^{+}]\\ [{\mathrm{H}}^{+}]={10}^{-5.05}=8.9125\times {10}^{-6}\text{mol}{\text{L}}^{-1}\\ \text{X}=[{\text{H}}^{+}]=8.9125\times {10}^{-6}\text{mol}{\text{L}}^{-1}\end{array}$

We calculated the “x” which is the concentration of hydrogen ion. The concentration of the anion is also “x”. Thus,

$\begin{array}{c}\text{X}=[{\text{OBr}}^{-}]\\ =8.9125\times {10}^{-6}\text{mol}{\text{L}}^{-1}\end{array}$

Now, we need to calculate the concentration of $[\text{HOBr}]$ = 0.040 − x. The value of x is very negligible hence, it can be ignored, and put just 0.040. Thus, the concentration of hypobromous acid is.

$[\text{HOBr}]$ = 0.040.

The concentrations of the anion, hydrogen ion and hypobromous acid are used in the equation used for acid dissociation constant.

$\begin{array}{c}{K}_{\text{a}}=\frac{[{\text{H}}^{+}][{\text{OBr}}^{-}]}{[\text{HOBr}]}\\ =\frac{[8.9125\times {10}^{-6}][8.9125\times {10}^{-6}]}{0.040}\\ K_{\text{a}}=1.9858\times {10}^{-9}\end{array}$

Thus, the acid dissociation constant of hypobromous acid is $K_{\text{a}}=1.9858\times {10}^{-9}$.

Conclusion

Weak acids do not dissociate completely. Each weak acid has a specific dissociation constant. Here, ICE table is made from the given chemical equation. Thus, the acid dissociation constant of hypobromous acid is $K_{\text{a}}=1.9858\times {10}^{-9}$.