Chapter 16, Problem 32PE

(a)

Interpretation Introduction

Interpretation:

Concentration of hydroxide ions for concentration of ${\text{H}}^{+}$ to be $4\times {10}^{-9}$ 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 $4\times {10}^{-9}$ for $\left[{\text{H}}^{\text{+}}\right]$ in equation (1).

  $\begin{array}{c}\text{pH}=-\log \left[4\times {10}^{-9}\right]\\ =-\log 4+9\log 10\\ =-0.602+9\\ =8.40\end{array}$

Hence, $\text{pH}$ for given $\left[{\text{H}}^{\text{+}}\right]$ is 8.40.

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

  $\begin{array}{c}\text{pOH}=\text{14}-8.40\\ =5.6\end{array}$

Hence, $\text{pOH}$ for given $\left[{\text{H}}^{\text{+}}\right]$ is 5.6.

$\text{pOH}$ 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 5.6 for $\text{pOH}$ in equation (5).

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]={10}^{-\text{5}\text{.6}}\\ =2.5\times {10}^{-6}\end{array}$

Hence, $\left[{\text{OH}}^{-}\right]$ for given $\left[{\text{H}}^{\text{+}}\right]$ is $2.5\times {10}^{-6}$.

(b)

Interpretation Introduction

Interpretation:

Concentration of hydroxide ions for concentration of ${\text{H}}^{+}$ to be $1.2\times {10}^{-5}$ 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 $1.2\times {10}^{-5}$ for $\left[{\text{H}}^{\text{+}}\right]$ in equation (1).

  $\begin{array}{c}\text{pH}=-\log \left[1.2\times {10}^{-5}\right]\\ =-\log 1.2+5\log 10\\ =-0.079+5\\ =4.92\end{array}$

Hence, $\text{pH}$ for given $\left[{\text{H}}^{\text{+}}\right]$ is $4.92$.

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

  $\begin{array}{c}\text{pOH}=\text{14}-4.92\\ =9.08\end{array}$

Hence, $\text{pOH}$ for given $\left[{\text{H}}^{\text{+}}\right]$ is 9.08.

$\text{pOH}$ 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 9.08for $\text{pOH}$ in equation (5).

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]={10}^{-9.08}\\ =8.31\times {10}^{-10}\end{array}$

Hence, $\left[{\text{OH}}^{-}\right]$ for given $\left[{\text{H}}^{\text{+}}\right]$ is $8.31\times {10}^{-10}$.

(c)

Interpretation Introduction

Interpretation:

Concentration of hydroxide ions in solution of $1.25M\text{ HCN}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The ionization of weak acid $\text{HCN}$ is as follows:

  $\text{HCN}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{CN}}^{-}\left(aq\right)$

The equilibrium constant for ionization of weak acid is called ionization constant $K_a$ and is expressed as follows:

  $K_a=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{CN}}^{-}\right]}{\left[\text{HCN}\right]}$        (6)

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

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

  $\left[{\text{H}}^{\text{+}}\right]=\frac{K_a\cdot \left[\text{HCN}\right]}{\left[{\text{CN}}^{-}\right]}$        (7)

Substitute $\text{x}$ for $\left[{\text{CN}}^{-}\right]$, $\text{x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $4\times {10}^{-10}$ for $K_a$ and $1.25$ for $\left[\text{HCN}\right]$ in equation (7).

  $\begin{array}{l}\text{x}=\frac{\left(4\times {10}^{-10}\right)\left(1.25\right)}{\text{x}}\\ {\text{x}}^2=5\times {10}^{-10}\end{array}$

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

  $\text{x}=2.24\times {10}^{-}{}^5$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in $1.25M\text{ HCN}$ is $2.24\times {10}^{-}{}^{5}M$.

$\text{pH}$ 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 $2.24\times {10}^{-}{}^5$ for ${\text{H}}^{+}$ in equation (1).

  $\begin{array}{c}\text{pH}=-\log \left[2.24\times {10}^{-}{}^5\right]\\ =-\log \text{2}\text{.24}+5\text{log10}\\ \text{=}-0.35+5\\ =4.65\end{array}$

Hence, $\text{pH}$ in $1.25M\text{ HCN}$ is 4.65.

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 4.65for $\text{pH}$ in equation (3).

  $\begin{array}{c}\text{pOH}=\text{14}-4.65\\ =9.35\end{array}$

Hence, $\text{pOH}$ in $1.25M\text{ HCN}$ is 9.35.

$\text{pOH}$ 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 9.35 for $\text{pOH}$ in equation (5).

  $\begin{array}{c}\left[{\text{OH}}^{-}\right]={10}^{-9.35}\\ =4.5\times {10}^{-10}\end{array}$

Hence, $\left[{\text{OH}}^{-}\right]$ in $1.25M\text{ HCN}$ is $4.5\times {10}^{-10}$.

(d)

Interpretation Introduction

Interpretation:

Concentration of hydroxide ions in solution of $0.333M\text{ NaOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

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

  $\text{NaOH}\left(aq\right)\rightleftharpoons {\text{Na}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

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