Chapter 16, Problem 31PE

(a)

Interpretation Introduction

Interpretation:

Hydroxide concentration when concentration of ${\text{H}}^{+}$ is $1.0\times {10}^{-2}$ 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

$\text{pH}$ 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 $1.0\times {10}^{-2}$ for $\left[{\text{H}}^{\text{+}}\right]$ in equation (1).

  $\begin{array}{c}\text{pH}=-\log \left[1.0\times {10}^{-2}\right]\\ =-\log 1+2\log 10\\ =2\end{array}$

Hence, $\text{pH}$ for given $\left[{\text{H}}^{\text{+}}\right]$ is 2.

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

  $\begin{array}{c}\text{pOH}=\text{14}-2\\ =12\end{array}$

Hence, $\text{pOH}$ for given $\left[{\text{H}}^{\text{+}}\right]$ is 12.

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

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

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

  $\left[{\text{OH}}^{-}\right]={10}^{-\text{pOH}}$        (5)

Substitute 12 for $\text{pOH}$ in equation (5).

  $\left[{\text{OH}}^{-}\right]={10}^{-\text{12}}$

Hence, $\left[{\text{OH}}^{-}\right]$ for given $\left[{\text{H}}^{\text{+}}\right]$ is ${10}^{-\text{12}}$.

(b)

Interpretation Introduction

Interpretation:

Hydroxide concentration when concentration of ${\text{H}}^{+}$ is $3.2\times {10}^{-7}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

$\text{pH}$ 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 $3.2\times {10}^{-7}$ for $\left[{\text{H}}^{\text{+}}\right]$ in equation (1).

  $\begin{array}{c}\text{pH}=-\log \left[3.2\times {10}^{-7}\right]\\ =-\log 3.2+7\log 10\\ =-0.505+7\\ =6.50\end{array}$

Hence, $\text{pH}$ for given $\left[{\text{H}}^{\text{+}}\right]$ is 6.50.

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

  $\begin{array}{c}\text{pOH}=\text{14}-6.50\\ =7.5\end{array}$

Hence, $\text{pOH}$ for given $\left[{\text{H}}^{\text{+}}\right]$ is 7.5.

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

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

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

  $\left[{\text{OH}}^{-}\right]={10}^{-\text{pOH}}$        (5)

Substitute 7.5 for $\text{pOH}$ in equation (5).

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]={10}^{-7.5}\\ =3.16\times {10}^{-8}\end{array}$

Hence, $\left[{\text{OH}}^{-}\right]$ for given $\left[{\text{H}}^{\text{+}}\right]$ is $3.16\times {10}^{-8}$.

(c)

Interpretation Introduction

Interpretation:

For a solution of $1.25M\text{ KOH}$, hydroxide concentration 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 $1.25M\text{ KOH}$ completely ionizes to give $1.25M{\text{ K}}^{+}$ and $1.25M{\text{ OH}}^{-}$. Hence, $\left[{\text{OH}}^{-}\right]$ is $1.25M$.

(d)

Interpretation Introduction

Interpretation:

For a solution of $0.75M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ , hydroxide concentration has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The ionization of weak acid ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ 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 equilibrium constant for ionization of weak acid is called ionization constant $K_a$ and is expressed as follows:

  $K_a=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_2{\text{H}}_3{\text{O}}_2^{-}\right]}{\left[{\text{HC}}_2{\text{H}}_3{\text{O}}_2\right]}$        (6)

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 ${\text{C}}_2{\text{H}}_3{\text{O}}_2^{-}$. Value of $\text{x}$ is small and therefore, concentration of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ is $0.75M$.

Rearrange equation (6) for $\left[{\text{H}}^{\text{+}}\right]$.

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

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 $0.75$ for $\left[{\text{HC}}_2{\text{H}}_3{\text{O}}_2\right]$ in equation (7).

  $\begin{array}{l}\text{x}=\frac{\left(1.8\times {10}^{-5}\right)\left(0.75\right)}{\text{x}}\\ {\text{x}}^2=1.35\times {10}^{-5}\end{array}$

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

  $\text{x}=3.67\times {10}^{-}{}^3$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in $0.75M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ is $3.67\times {10}^{-}{}^{3}M$.

$\text{pH}$ 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 $3.67\times {10}^{-}{}^3$ for ${\text{H}}^{+}$ in equation (1).

  $\begin{array}{c}\text{pH}=-\log \left[3.67\times {10}^{-}{}^3\right]\\ =-\log \text{3}\text{.67}+3\text{log10}\\ \text{=}-0.565+3\\ =2.44\end{array}$

Hence, $\text{pH}$ in $0.75M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ is 2.44.

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

  $\begin{array}{c}\text{pOH}=\text{14}-2.44\\ =11.56\end{array}$

Hence, $\text{pOH}$ in $0.75M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ is 11.56.

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

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

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

  $\left[{\text{OH}}^{-}\right]={10}^{-\text{pOH}}$        (5)

Substitute 11.56 for $\text{pOH}$ in equation (5).

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]={10}^{-11.56}\\ =2.8\times {10}^{-12}\end{array}$

Hence, $\left[{\text{OH}}^{-}\right]$ in $0.75M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ is $2.8\times {10}^{-12}M$.