Chapter 19, Problem 111A

a)

Interpretation Introduction

Interpretation : The pH of water for each temperature is to be calculated.

Concept Introduction : The negative logarithm of the concentration of hydrogen ions is the pH of a solution. The formula to calculate pH is given as

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

Answer to Problem 111A

The pH of water at various temperature is calculated as follows:

At temperature $0^{\circ }\text{C}$ , the pH is $7.47$ .

At temperature ${10}^{\circ }\text{C}$ , the pH is $7.27$ .

At temperature ${20}^{\circ }\text{C}$ , the pH is $7.08$ .

At temperature ${30}^{\circ }\text{C}$ , the pH is $6.92$ .

At temperature ${40}^{\circ }\text{C}$ , the pH is $6.77$ .

At temperature ${50}^{\circ }\text{C}$ , the pH is $6.63$ .

Explanation of Solution

Given information:

At temperature $0^{\circ }\text{C}$ , the ion product constant $\left({\text{K}}_{\text{w}}\right)$ is $\text{1}\text{.137}\times {10}^{-15}$ .

At temperature ${10}^{\circ }\text{C}$ , the ion product constant $\left({\text{K}}_{\text{w}}\right)$ is $2.917\times {10}^{-15}$ .

At temperature ${20}^{\circ }\text{C}$ , the ion product constant $\left({\text{K}}_{\text{w}}\right)$ is $6.807\times {10}^{-15}$ .

At temperature ${30}^{\circ }\text{C}$ , the ion product constant $\left({\text{K}}_{\text{w}}\right)$ is $\text{1}\text{.469}\times {10}^{-14}$ .

At temperature ${40}^{\circ }\text{C}$ , the ion product constant $\left({\text{K}}_{\text{w}}\right)$ is $2.917\times {10}^{-14}$ .

At temperature ${50}^{\circ }\text{C}$ , the ion product constant $\left({\text{K}}_{\text{w}}\right)$ is $5.470\times {10}^{-14}$ .

The ion product constant is the square of the concentration of hydrogen ions.

The formula to calculate pH is given as

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

The hydrogen ion at $0^{\circ }\text{C}$ is given as

  $\begin{array}{c}{\text{K}}_{\text{w}}=\text{1}\text{.137}\times {10}^{-15}\\ {\left[{\text{H}}^{+}\right]}^2=\text{1}\text{.137}\times {10}^{-15}\\ \left[{\text{H}}^{+}\right]=\sqrt{\text{1}\text{.137}\times {10}^{-15}}\\ =3.37\times {10}^{-8}\end{array}$

To calculate the pH at $0^{\circ }\text{C}$, substitute the values in above formula:

  $\begin{array}{c}\text{pH=}-{\text{log[H}}^{\text{+}}\text{]}\\ \text{=}-\log \left(3.37\times {10}^{-8}\right)\\ \text{pH}=7.47\end{array}$

The hydrogen ion at ${10}^{\circ }\text{C}$ is given as

  $\begin{array}{c}{\text{K}}_{\text{w}}=2.917\times {10}^{-15}\\ {\left[{\text{H}}^{+}\right]}^2=2.917\times {10}^{-15}\\ \left[{\text{H}}^{+}\right]=\sqrt{2.917\times {10}^{-15}}\\ =5.4\times {10}^{-8}\end{array}$

To calculate the pH at ${10}^{\circ }\text{C}$, substitute the values in above formula:

  $\begin{array}{c}\text{pH=}-{\text{log[H}}^{\text{+}}\text{]}\\ \text{=}-\log \left(5.4\times {10}^{-8}\right)\\ \text{pH}=7.27\end{array}$

The hydrogen ion at ${20}^{\circ }\text{C}$ is given as

  $\begin{array}{c}{\text{K}}_{\text{w}}=6.807\times {10}^{-15}\\ {\left[{\text{H}}^{+}\right]}^2=6.807\times {10}^{-15}\\ \left[{\text{H}}^{+}\right]=\sqrt{6.807\times {10}^{-15}}\\ =8.25\times {10}^{-8}\end{array}$

To calculate the pH at ${20}^{\circ }\text{C}$, substitute the values in above formula:

  $\begin{array}{c}\text{pH=}-{\text{log[H}}^{\text{+}}\text{]}\\ \text{=}-\log \left(8.25\times {10}^{-8}\right)\\ \text{pH}=7.08\end{array}$

The hydrogen ion at ${30}^{\circ }\text{C}$ is given as

  $\begin{array}{c}{\text{K}}_{\text{w}}=\text{1}\text{.469}\times {10}^{-14}\\ {\left[{\text{H}}^{+}\right]}^2=\text{1}\text{.469}\times {10}^{-14}\\ \left[{\text{H}}^{+}\right]=\sqrt{\text{1}\text{.469}\times {10}^{-14}}\\ =12.12\times {10}^{-8}\end{array}$

To calculate the pH at ${30}^{\circ }\text{C}$, substitute the values in above formula:

  $\begin{array}{c}\text{pH=}-{\text{log[H}}^{\text{+}}\text{]}\\ \text{=}-\log \left(12.12\times {10}^{-8}\right)\\ \text{pH}=6.92\end{array}$

The hydrogen ion at ${40}^{\circ }\text{C}$ is given as

  $\begin{array}{c}{\text{K}}_{\text{w}}=2.917\times {10}^{-14}\\ {\left[{\text{H}}^{+}\right]}^2=2.917\times {10}^{-14}\\ \left[{\text{H}}^{+}\right]=\sqrt{2.917\times {10}^{-14}}\\ =17.08\times {10}^{-8}\end{array}$

To calculate the pH at ${40}^{\circ }\text{C}$, substitute the values in above formula:

  $\begin{array}{c}\text{pH=}-{\text{log[H}}^{\text{+}}\text{]}\\ \text{=}-\log \left(17.08\times {10}^{-8}\right)\\ \text{pH}=6.77\end{array}$

The hydrogen ion at ${50}^{\circ }\text{C}$ is given as

  $\begin{array}{c}{\text{K}}_{\text{w}}=5.470\times {10}^{-14}\\ {\left[{\text{H}}^{+}\right]}^2=5.470\times {10}^{-14}\\ \left[{\text{H}}^{+}\right]=\sqrt{5.470\times {10}^{-14}}\\ =23.39\times {10}^{-8}\end{array}$

To calculate the pH at ${50}^{\circ }\text{C}$, substitute the values in above formula:

  $\begin{array}{c}\text{pH=}-{\text{log[H}}^{\text{+}}\text{]}\\ \text{=}-\log \left(23.39\times {10}^{-8}\right)\\ \text{pH}=6.63\end{array}$

b)

Interpretation Introduction

Interpretation : The graph between pH versus temperature is to be plotted and pH is to be estimated.

Concept Introduction : The negative logarithm of the concentration of hydrogen ions is the pH of a solution. The values from 0 to 14 are the range of a pH scale.

Answer to Problem 111A

The pH of water at $5^{\circ }\text{C}$ is $7.4$ .

Explanation of Solution

From the calculated pH, the graph is plotted between pH and temperature.

  

From the graph, the pH of water at $5^{\circ }\text{C}$ is $7.4$ .

c)

Interpretation Introduction

Interpretation : The temperature at a pH of 6.85 is to be evaluated.

Concept Introduction : The negative logarithm of the concentration of hydrogen ions is the pH of a solution. The values from 0 to 14 are the range of a pH scale.

Answer to Problem 111A

The temperature at which pH of water approximately $6.85$ is ${35}^{\circ }\text{C}$ .

Explanation of Solution

From the calculated pH, the graph is plotted between pH and temperature.

  

From the graph, the temperature at which pH of water approximately $6.85$ is ${35}^{\circ }\text{C}$ .