Chapter 17, Problem 20CR

(a)

Interpretation Introduction

Interpretation:

pH and pOH of 0.00141 M ${\text{HNO}}_{\text{3}}$ is to be calculated.

Concept Introduction:

To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of ${\text{[H}}^{+}\text{]}$ in aqueous solution is small, by using the p scale in the form of $\text{pH}$ scale, it is better way to represent acidity of solution.

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

And, at $25°\text{C}$ , the actual concentrations of the products are:

  ${\text{[H}}^{+}{\text{][OH}}^{-}\text{]=1}\times {\text{10}}^{-14}$

The mathematical product of both ${\text{[H}}^{\text{+}}\text{]}$ and ${\text{[OH}}^{-}\text{]}$ concentrations are always constant known as ion product constant ( $K_w$ )

  ${\text{[H}}^{+}{\text{][OH}}^{-}\text{]=1}\times {\text{10}}^{-14}=K_w$

The relation between $\text{pH}$ and $\text{pOH}$ can be derived from $K_w$ .

  $\text{pH+pOH=14}$

Answer to Problem 20CR

pH of 0.00141 M ${\text{HNO}}_{\text{3}}$ is 2.85 and pOH is 11.15.

Explanation of Solution

Concentration of H+ ion is 0.00141 M HNO₃ in given solution. From this information, pH can be calculated.

  $\begin{array}{l}{\text{[H}}^{\text{+}}\text{]=0}\text{.00141}\text{M}\\ {\text{pH=-log[H}}^{\text{+}}\text{]=-log(0}\text{.00141)}\\ \text{pH=-(-2}\text{.85)=2}\text{.85}\end{array}$

From this pOH can be calculated as follows:

$\begin{array}{l}\text{pH+pOH=14}\\ \text{pOH=14-pH}\\ \text{pOH=14-2}\text{.85=11}\text{.15}\end{array}$

Hence, pH and pOH of 0.00141 M ${\text{HNO}}_{\text{3}}$ are 2.85 and 11.15 respectively.

(b)

Interpretation Introduction

Interpretation:

pH and pOH of $\text{2}\text{.13}\times {\text{10}}^{-3}M\text{ NaOH}$ is to be calculated.

Concept Introduction:

To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of ${\text{[H}}^{+}\text{]}$ in aqueous solution is small, by using the p scale in the form of $\text{pH}$ scale, it is better way to represent acidity of solution.

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

And, at $25°\text{C}$ , the actual concentrations of the products are:

  ${\text{[H}}^{+}{\text{][OH}}^{-}\text{]=1}\times {\text{10}}^{-14}$

The mathematical product of both ${\text{[H}}^{\text{+}}\text{]}$ and ${\text{[OH}}^{-}\text{]}$ concentrations are always constant known as ion product constant ( $K_w$ )

  ${\text{[H}}^{+}{\text{][OH}}^{-}\text{]=1}\times {\text{10}}^{-14}=K_w$

The relation between $\text{pH}$ and $\text{pOH}$ can be derived from $K_w$ .

  $\text{pH+pOH=14}$

Answer to Problem 20CR

pH of $\text{2}\text{.13}\times {\text{10}}^{-3}M\text{ NaOH}$ is 11.33 and pOH is 2.67.

Explanation of Solution

Concentration of OH- ion is 2.13x10-3 M NaOH in given solution. From this information, pOH can be calculated.

  $\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]=2}{\text{.13×10}}^{\text{-3}}\text{M}\\ {\text{pOH=-log[OH}}^{\text{-}}\text{]=-log(2}{\text{.13×10}}^{\text{-3}}\text{)}\\ \text{pOH=-(-2}\text{.67)=2}\text{.67}\end{array}$

From this pH can be calculated as follows.

$\begin{array}{l}\text{pH+pOH=14}\\ \text{pH=14-pOH}\\ \text{pOH=14-2}\text{.67=11}\text{.33}\end{array}$

Hence, pH and pOH $\text{2}\text{.13}\times {\text{10}}^{-3}M\text{ NaOH}$ are 2.67 and 11.33 respectively.

(c)

Interpretation Introduction

Interpretation:

pH and pOH of 0.00515 M $\text{HCl}$ is to be calculated.

Concept Introduction:

To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of ${\text{[H}}^{+}\text{]}$ in aqueous solution is small, by using the p scale in the form of $\text{pH}$ scale, it is better way to represent acidity of solution.

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

And, at $25°\text{C}$ , the actual concentrations of the products are:

  ${\text{[H}}^{+}{\text{][OH}}^{-}\text{]=1}\times {\text{10}}^{-14}$

The mathematical product of both ${\text{[H}}^{\text{+}}\text{]}$ and ${\text{[OH}}^{-}\text{]}$ concentrations are always constant known as ion product constant ( $K_w$ )

  ${\text{[H}}^{+}{\text{][OH}}^{-}\text{]=1}\times {\text{10}}^{-14}=K_w$

The relation between $\text{pH}$ and $\text{pOH}$ can be derived from $K_w$ .

  $\text{pH+pOH=14}$

Answer to Problem 20CR

pH of 0.00515 M $\text{HCl}$ is 2.29 and pOH is 11.71.

Explanation of Solution

Concentration of H+ ion is 0.00515 M $\text{HCl}$ in given solution. From this information, pH can be calculated.

  $\begin{array}{l}{\text{[H}}^{\text{+}}\text{]=0}\text{.00515}\text{M}\\ {\text{pH=-log[H}}^{\text{+}}\text{]=-log(0}\text{.00515)}\\ \text{pH=-(-2}\text{.29)=2}\text{.29}\end{array}$

From this pOH can be calculated as follows.

$\begin{array}{l}\text{pH+pOH=14}\\ \text{pOH=14-pH}\\ \text{pOH=14-2}\text{.29=11}\text{.71}\end{array}$

Hence, pH and pOH of 0.00515 M $\text{HCl}$ are 2.29 and 11.71 respectively.

(d)

Interpretation Introduction

Interpretation:

pH and pOH of $\text{5}\text{.65}\times {\text{10}}^{-5}{\text{M Ca(OH)}}_2$ is to be calculated.

Concept Introduction:

To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of ${\text{[H}}^{+}\text{]}$ in aqueous solution is small, by using the p scale in the form of $\text{pH}$ scale, it is better way to represent acidity of solution.

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

And, at $25°\text{C}$ , the actual concentrations of the products are:

  ${\text{[H}}^{+}{\text{][OH}}^{-}\text{]=1}\times {\text{10}}^{-14}$

The mathematical product of both ${\text{[H}}^{\text{+}}\text{]}$ and ${\text{[OH}}^{-}\text{]}$ concentrations are always constant known as ion product constant ( $K_w$ )

  ${\text{[H}}^{+}{\text{][OH}}^{-}\text{]=1}\times {\text{10}}^{-14}=K_w$

The relation between $\text{pH}$ and $\text{pOH}$ can be derived from $K_w$ .

$\text{pH+pOH=14}$

Answer to Problem 20CR

pH of $\text{5}\text{.65}\times {\text{10}}^{-5}{\text{M Ca(OH)}}_2$ is 10.05 and pOH is 3.95.

Explanation of Solution

Concentration of OH- ion is $\text{5}\text{.65}\times {\text{10}}^{-5}{\text{M Ca(OH)}}_2$ in given solution. Here in calcium hydroxide there are two OH- groups therefore concentration of OH- will be obtained by multiplying given concentration by two. From this information pOH can be calculated.

  $\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]=2×5}{\text{.65×10}}^{\text{-5}}\text{M}\\ {\text{[OH}}^{\text{-}}\text{]=11}{\text{.3×10}}^{\text{-5}}\text{M}\\ {\text{pOH=-log[OH}}^{\text{-}}\text{]=-log(11}{\text{.3×10}}^{\text{-5}})\\ \text{pOH=-(-3}\text{.95)=3}\text{.95}\end{array}$

From this pH can be calculated as follows.

$\begin{array}{l}\text{pH+pOH=14}\\ \text{pH=14-pOH}\\ \text{pOH=14-3}\text{.95=10}\text{.05}\end{array}$

Hence, pH and pOH of $\text{5}\text{.65}\times {\text{10}}^{-5}{\text{M Ca(OH)}}_2$ are 10.05 and 3.95 respectively.