Chapter 21, Problem 43QAP

Interpretation Introduction

Interpretation:

To calculate the $\text{pH}$ of bromine water.

Concept introduction:

$\text{pH}$ is −ve logarithm of molar hydrogen ion concentration.

$\text{pH}=-\mathrm{l}\text{og}\left[{\text{H}}^{\text{+}}\right]$

Answer to Problem 43QAP

$\text{pH}$ of bromine water is $2.973$.

Explanation of Solution

At $25°\text{C}$, reaction of bromine is:

${\text{Br}}_2\left(I\right)+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}^{+}\left(aq\right){\text{+Br}}^{-}\left(aq\right)+\text{HBrO}\left(aq\right)$

$\text{Initial}\text{concentration:}\text{C}000$

$\text{After}\text{change}\text{in}\text{concentration}:-xxxx$

$\text{At}\text{equilibrium}:\text{C}\left(1-x\right)\text{C}x\text{C}x\text{C}x$

For above reaction equilibrium constant K is:

$\begin{array}{c}\text{K}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{Br}}^{-}\right]\left[\text{HBrO}\right]}{\left[{\text{Br}}_{\text{2}}\right]}\\ =\frac{\left(\text{c}x\right)\left(\text{c}x\right)\left(\text{c}x\right)}{\text{c}\left(1-x\right)}\\ =\frac{{\left(\text{c}x\right)}^3}{\text{c}\left(1-x\right)}\end{array}$

$x<<\text{c}$ and neglect $\text{c}\left(1-x\right)$ and concentration of ${\text{H}}^{\text{+}}$ is:

$\text{K}={\left(\text{c}x\right)}^3$

Given,

The value of equilibrium constant $\text{K}=1.2\times {10}^{-9}$

$\begin{array}{c}\text{K}={\left(\text{c}x\right)}^3\\ 1.2\times {10}^{-9}={\left(\text{c}x\right)}^3\\ \text{c}x=1.063\times {10}^{-3}\end{array}$

Hence,

$\begin{array}{c}\left[{\text{H}}^{\text{+}}\right]=\text{c}x\\ =1.063\times {10}^{-3}\end{array}$

$\begin{array}{c}\text{pH}=-\mathrm{l}\text{og}\left[{\text{H}}^{\text{+}}\right]\\ =-\log \left(1.063\times {10}^{-3}\right)\\ =-\log \left(1.063\right)+(-\log ({10}^{-3}))\\ =-0.02653+3\\ =2.973\end{array}$

Thus, $\text{pH}$ of the solution is $2.973$

Conclusion

$\text{pH}$ of bromine water is $2.973$.