Chapter 16, Problem 29PE

(a)

Interpretation Introduction

Interpretation:

For a solution of $0.250M\text{ HBr}$ value of $\text{pH}$ and $\text{pOH}$ has to be calculated.

Concept Introduction:

The concentration of ${\text{H}}^{+}$ decides the acidity in solution and concentration of ${\text{OH}}^{-}$ decides the basicity in solution.

$\text{pH}$ is defined as concentration of hydrogen ion. It also explains about the acidity of solution.

$\text{pOH}$ is defined as the concentration of hydroxide ion. It also explains about the basicity of solution.

Explanation of Solution

The complete ionization of $\text{HBr}$ is as follows:

  $\text{HBr}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)$

The $0.250M\text{ HBr}$ completely ionizes to give $0.250M{\text{ H}}^{+}$ and $0.250M{\text{ Br}}^{-}$. Hence, $\left[{\text{H}}^{+}\right]$ is $0.250M$.

$\text{pH}$ is defined as negative logarithm of concentration of ${\text{H}}^{+}$. It is calculated as follows:

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

Substitute $0.250M$ for ${\text{H}}^{+}$ in equation (1).

  $\begin{array}{c}\text{pH}=-\log \left(0.250M\right)\\ =5.49\end{array}$

Hence, $\text{pH}$ in $0.250M\text{ HBr}$ is 5.49.

The expression that relates $\text{pH}$ and $\text{pOH}$ is as follows:

  $\text{pH}+\text{pOH}=\text{14}$        (2)

Rearrange equation (2) for $\text{pOH}$.

  $\text{pOH}=\text{14}-\text{pH}$        (3)

Substitute $0.903$ for $\text{pH}$ in equation (3).

  $\begin{array}{c}\text{pOH}=\text{14}-5.49\\ =8.51\end{array}$

Hence, $\text{pOH}$ in $0.250M\text{ HBr}$ is 8.51.

(b)

Interpretation Introduction

Interpretation:

For a solution of $0.333M\text{ KOH}$, $\text{pH}$ and $\text{pOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The complete ionization of $\text{KOH}$ is as follows:

  $\text{KOH}\left(aq\right)\rightleftharpoons {\text{K}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

The $0.333M\text{ KOH}$ completely ionizes to give $0.333M{\text{ K}}^{+}$ and $0.333M{\text{ OH}}^{-}$. Hence, $\left[{\text{OH}}^{-}\right]$ is $0.333M$.

$\text{pOH}$ is defined as negative logarithm of concentration of ${\text{OH}}^{-}$. It is calculated as follows:

  $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$        (4)

Substitute $0.333$ for $\left[{\text{OH}}^{-}\right]$ in equation (4).

  $\begin{array}{c}\text{pOH}=-\log \left[0.333\right]\\ =0.48\end{array}$

Hence, $\text{pOH}$ in $0.333M\text{ KOH}$ is 0.48.

The expression that relates $\text{pH}$ and $\text{pOH}$ is as follows:

  $\text{pH}+\text{pOH}=\text{14}$        (2)

Rearrange equation (2) for $\text{pH}$.

  $\text{pH}=\text{14}-\text{pOH}$        (6)

Substitute $0.48$ for $\text{pH}$ in equation (6).

  $\begin{array}{c}\text{pH}=\text{14}-0.48\\ =13.52\end{array}$

Hence, $\text{pH}$ in $0.333M\text{ KOH}$ is 13.52.

(c)

Interpretation Introduction

Interpretation:

For a solution of $0.895M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$, $\text{pH}$ and $\text{pOH}$ has to be calculated.

Concept Introduction:

Equilibrium constant for ionization of weak acid is called ionization constant $K_a$. It is ratio of concentration of products to concentration of reactants.

Explanation of Solution

The ionization of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ is as follows:

  ${\text{HC}}_2{\text{H}}_3{\text{O}}_2\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{C}}_2{\text{H}}_3{\text{O}}_2^{-}\left(aq\right)$

The expression for $K_a$ for given reaction is as follows:

  $K_a=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_2{\text{H}}_3{\text{O}}_2^{-}\right]}{\left[{\text{HC}}_2{\text{H}}_3{\text{O}}_2\right]}$        (7)

Through chemical equation it is evident that one ${\text{C}}_2{\text{H}}_3{\text{O}}_2^{-}$ is formed for every ${\text{H}}^{\text{+}}$. Thus concentration of ${\text{H}}^{\text{+}}$ is equal to concentration of ${\text{C}}_2{\text{H}}_3{\text{O}}_2^{-}$. Thus consider $\text{x}$ for ${\text{H}}^{\text{+}}$ and ${\text{C}}_2{\text{H}}_3{\text{O}}_2^{-}$.

Rearrange equation (7) for $\left[{\text{H}}^{\text{+}}\right]$.

  $\left[{\text{H}}^{\text{+}}\right]=\frac{K_a\left[{\text{HC}}_2{\text{H}}_3{\text{O}}_2\right]}{\left[{\text{C}}_2{\text{H}}_3{\text{O}}_2^{-}\right]}$        (8)

Substitute $\text{x}$ for $\left[{\text{C}}_2{\text{H}}_3{\text{O}}_2^{-}\right]$, $\text{x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $1.8\times {10}^{-5}$ for $K_a$ and $0.895$ for $\left[{\text{HC}}_2{\text{H}}_3{\text{O}}_2\right]$ in equation (8).

  $\begin{array}{c}\text{x}=\frac{\left(1.8\times {10}^{-5}\right)\left(0.895\right)}{\text{x}}\\ {\text{x}}^2=1.61\times {10}^{-5}\end{array}$

Solve this equation for $\text{x}$.

  $\text{x}=4.01\times {10}^{-3}$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in $0.895M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ is $4.01\times {10}^{-3}M$.

$\text{pH}$ is defined as negative logarithm of concentration of ${\text{H}}^{+}$. It is calculated as follows:

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

Substitute $4.01\times {10}^{-3}$ for ${\text{H}}^{+}$ in equation (1).

  $\begin{array}{c}\text{pH}=-\log \left[4.01\times {10}^{-3}\right]\\ =-\log 4.01+3\text{log10}\\ \text{=}-0.603+3\\ =2.40\end{array}$

Hence, $\text{pH}$ in $0.895M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ is 2.40.

The expression that relates $\text{pH}$ and $\text{pOH}$ is as follows:

  $\text{pH}+\text{pOH}=\text{14}$        (2)

Rearrange equation (2) for $\text{pOH}$.

  $\text{pOH}=\text{14}-\text{pH}$        (3)

Substitute $2.40$ for $\text{pH}$ in equation (3).

  $\begin{array}{c}\text{pOH}=\text{14}-2.40\\ =11.6\end{array}$

Hence, $\text{pOH}$ in $0.895M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ is 11.16.