Chapter 13, Problem 135QP

Interpretation Introduction

Interpretation:

All the given solutions are to be arranged in accordance with increasing hydroxide concentration present in the solutions.

Concept Introduction

The $\text{pH}$ for the given solution of hydrochloride is calculated in equation (1) where, $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ion in the given solution.

$\text{pH}=\log \left[{\text{H}}^{+}\right]$ ……(1)

The relation between hydroxyl concentration, $\left[{\text{OH}}^{-}\right]$ , and $\text{pOH}$ of the given solution.

$\text{pOH}=\log \left[{\text{OH}}^{-}\right]$ ……(2)

The relation between the ionic product of water, $K_{\text{w}}$ , at $298.15\text{K}$ , $\left[{\text{H}}^{+}\right]$ , and $\left[{\text{OH}}^{-}\right]$ . The ionic product of water at $298.15\text{K}$ is ${10}^{-14}$ .

$\begin{array}{c}{K}_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]\\ {10}^{-14}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]\end{array}$ ……(3)

Answer to Problem 135QP

Solution:

$0.5\text{M}\text{HCl}<0.1{\text{M HNO}}_3<{\text{HNO}}_3\left(\text{pH}=4.0\right)<\text{water}<\text{Buffer}\left(\text{pOH}=11.0\right)$

Explanation of Solution

Given Information: The given concentrations of the solutions are $0.1M{\text{HNO}}_3$ , $0.5M\text{ HCl}$ , a solution of ${\text{HNO}}_2$ having $\text{pH}$ value $4.0$ , a buffer solution having $\text{pOH}$ value $11.0$ , and water.

${\text{HNO}}_{\text{3}}$ acid is a strong electrolyte. Thus, the hydrogen ion concentration in $0.1\text{M}{\text{HNO}}_3$ is $0.1\text{M}$ . This value for hydrogen ion concentration is substituted in equation (3) to calculate the hydroxyl ion concentration.

$\begin{array}{l}{10}^{-14}=\left[0.1\right]\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]={10}^{-13}\text{M}\end{array}$

$\text{HCl}$ acid is a strong electrolyte. Thus, the hydrogen ion concentration in $0.1\text{M}\text{HCl}$ is $0.5\text{M}$ . This value for hydrogen ion concentration is substituted in equation (3) to calculate the hydroxyl ion concentration.

$\begin{array}{c}{10}^{-14}=\left[0.5\right]\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]=\frac{{10}^{-14}}{0.5}\\ =\text{2}\times {10}^{-14}\text{M}\end{array}$

In equation (1), $4$ is substituted for $\text{pH}$ of a solution of ${\text{HNO}}_{\text{2}}$ to calculate the hydrogen ion concentration.

$\begin{array}{l}4=\log \left[{\text{H}}^{+}\right]\\ \left[{\text{H}}^{+}\right]={10}^{-4}\text{M}\end{array}$

In equation (3), ${10}^{-4}\text{M}$ is substituted for $\left[{\text{H}}^{+}\right]$ of a solution of ${\text{HNO}}_{\text{2}}$ to calculate the hydroxyl ion concentration.

$\begin{array}{c}{10}^{-14}=\left[{10}^{-4}\right]\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]=\frac{{10}^{-14}}{{10}^{-4}}\text{M}\\ ={10}^{-10}\text{M}\end{array}$

The hydroxyl ion concentration in the given buffer solution is calculated by the substitution of $\text{pOH}$ in equation (2).

$\begin{array}{l}11=-\log \left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]={10}^{-11}\text{M}\end{array}$

The hydroxyl ion concentration of pure water at $298.15\text{K}$ is equal to ${10}^{-7}\text{M}$ .

Conclusion

Therefore, the order of increasing hydroxyl ion concentration is given as follows, $0.5\text{M}\text{HCl}<0.1{\text{M HNO}}_3<{\text{HNO}}_3\left(\text{pH}=4.0\right)<\text{water}<\text{Buffer}\left(\text{pOH}=11.0\right)$ .