Chapter 19, Problem 114A

Interpretation Introduction

Interpretation: To calculate the pH of the solution.

Concept Introduction: pH of the solution is described as the negative log of hydrogen ion concentration. It can be expressed as follows:

  $\text{pH}=-\text{log}\left[{\text{H}}^{\text{+}}\right]$

Answer to Problem 114A

The pH of the solution is $5.3$ .

Explanation of Solution

The density of the solution is $1.0\text{ g/mL}$ .

The acid dissociation constant is $\text{3}\text{.5}\times {\text{10}}^{-8}$ .

The mass of $\text{NaClO}$ is 5.1%.

Let’s assume, the mass of the solute is $\text{5 g}$ and $\text{95 g}$ of solvent in $\text{100 g}$ solution.

The molality of the solution can be expressed as follows:

  $\begin{array}{c}\text{Molality}=\frac{5/74.5}{0.095}\\ \text{m}=0.706\text{ mol}\end{array}$

The molarity can be given as the multiplication of density and molality.

  $\begin{array}{c}\text{Molarity}=\text{Density×Molality}\\ {\text{=10}}^{-3}{\text{kgL}}^{-1}\times 0.706\text{ mol}\\ \text{=7}\text{.06}\times {\text{10}}^{-4}\text{ M}\end{array}$

The pH of the solution can be calculated as follows:

  $\begin{array}{c}\left[H^{+}\right].\sqrt{{\text{K}}_{\text{a}}\text{.c}}=\sqrt{3.5\times {10}^{-8}\times 7.06\times {10}^{-4}}\\ =\sqrt{25\times {10}^{-12}}\\ =5\times {10}^{-6}\end{array}$

  $\begin{array}{c}\text{pH=}-\log \left[{\text{H}}^{+}\right]\\ =-\log \left[5\times {10}^{-6}\right]\\ =5.3\end{array}$

Conclusion

The pH of the solution is $5.3$ .