Chapter 15, Problem 15.91QE

(a)

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ solution of $0.010M{\text{ CH}}_3\text{COONa}$ has to becalculated.

Concept Introduction:

The ionization of a hypothetical weak base is given as follows:

  $\text{B}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{BH}}^{\text{+}}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)$

The expression of $K_{\text{b}}$ is given as follows:

  $K_{\text{b}}=\frac{\left[{\text{BH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{B}\right]}$

Here,

$\left[\text{B}\right]$ denotes the concentration of hypothetical weak base.

$\left[{\text{OH}}^{-}\right]$ denotes the concentration of ${\text{OH}}^{-}$ ions.

$\left[{\text{BH}}^{+}\right]$ denotes the concentration of ${\text{BH}}^{+}$ ions.

$K_{\text{b}}$ denotes ionization constant of weak base.

Answer to Problem 15.91QE

The value of $\text{pH}$ of $0.010M{\text{ CH}}_3\text{COONa}$  is 8.37.

Explanation of Solution

The relation among $K_{\text{a}}$, ${\text{K}}_{\text{b}}$ and $K_{\text{w}}$ is given as follows:

  $K_{\text{a}}{K}_{\text{b}}=K_{\text{w}}$        (1)

Refer table 15.6 for $K_{\text{a}}$ values of acids.

Substitute $1.8\times {10}^{-5}$ for $K_{\text{a}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (1).

  $\left(1.8\times {10}^{-5}\right)K_{\text{b}}=1\times {10}^{-14}$

Rearrange to obtain the value of $K_{\text{b}}$.

  $\begin{array}{c}{K}_{\text{b}}=\frac{1\times {10}^{-14}}{1.8\times {10}^{-5}}\\ =5.555\times {10}^{-10}\end{array}$

The ionization of ${\text{CH}}_3\text{COONa}$ salt occurs as follows:

  ${\text{Na}}^{+}\left(\text{aq}\right)+{\text{CH}}_3{\text{COO}}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{aq}\right)\rightleftharpoons {\text{CH}}_3\text{COOH}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)+{\text{Na}}^{+}\left(\text{aq}\right)$

The expression of $K_{\text{b}}$ is given as follows:

  $K_{\text{b}}=\frac{\left[{\text{CH}}_3\text{COOH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_3{\text{COO}}^{-}\right]}$        (2)

Here,

$\left[{\text{CH}}_3\text{COOH}\right]$ denotes the concentration of ${\text{CH}}_3\text{COOH}$.

$\left[{\text{OH}}^{-}\right]$ denotes the concentration of ${\text{OH}}^{-}$ ions.

$\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ denotes the concentration of ${\text{CH}}_3{\text{COO}}^{-}$ .

$K_{\text{b}}$ denotes equilibrium constant for this ionization.

The ICE table is given as follows:

  $\begin{array}{cccccccc}& {\text{CH}}_3{\text{COO}}^{-}& +& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & {\text{OH}}^{-}& +& {\text{CH}}_3\text{COOH}\\ \text{Initial, }M& 0.010& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.010-x\right)& & & & x& & x\end{array}$

Substitute $0.010-x$ for $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$, $x$ for $\left[{\text{CH}}_3\text{COOH}\right]$, and $5.050\times {10}^{-10}$ for $K_{\text{b}}$ in equation (2).

  $5.555\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.010-x\right)}$

Assume $x<<0.010M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^2=\left(5.555\times {10}^{-10}\right)\left(0.010\right)\\ x=\sqrt{0.0555\times {10}^{-10}}\\ =2.3569\times {10}^{-6}\end{array}$

The formula to calculate $\text{pOH}$ is given as follows:

  $\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$        (3)

Substitute $2.3569\times {10}^{-6}M$ for $\left[{\text{OH}}^{-}\right]$ in equation (3).

  $\begin{array}{c}\text{pOH}=-\text{log}\left(2.3569\times {10}^{-6}\right)\\ =5.6276\end{array}$

The formula to calculate $\text{pH}$ from $\text{pOH}$ is given as follows:

  $\text{pH}=14-\text{pOH}$        (4)

Substitute 2.6782 for $\text{pOH}$ in equation (4).

  $\begin{array}{c}\text{pH}=\text{14}-5.6276\\ =8.37\end{array}$

(b)

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ solution $0.125M$ of ammonium nitrate has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.91QE

The value of $\text{pH}$ the solution $0.125M$ of ammonium nitrate is 5.07.

Explanation of Solution

The ionization of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{2}}$ salt occurs as follows:

  ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{2}}\left(\text{aq}\right)\rightleftharpoons {\text{NH}}_{\text{4}}{}^{+}\left(\text{aq}\right)+{\text{NO}}_{\text{2}}{}^{-}\left(\text{aq}\right)$

${\text{NH}}_{\text{4}}{}^{+}$ is the conjugate acid of strong base ammonia and reacts with water as follows:

  ${\text{NH}}_4^{+}+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_3{\text{O}}^{+}+{\text{NH}}_3$

The expression of $K_{\text{a}}$ is given as follows:

  $K_{\text{a}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{NH}}_3\right]}{\left[{\text{NH}}_4^{+}\right]}$

Here,

$\left[{\text{NH}}_3\right]$ denotes the concentration of ${\text{NH}}_3$.

$\left[{\text{H}}_3{\text{O}}^{+}\right]$ denotes the concentration of ${\text{H}}_3{\text{O}}^{+}$ ions.

$\left[{\text{NH}}_4^{+}\right]$ denotes the concentration of ${\text{NH}}_4^{+}$.

$K_{\text{a}}$ denotes ionization constant of ${\text{NH}}_4^{+}$ ion.

Substitute $1.8\times {10}^{-5}$ for $K_{\text{b}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (1).

  $K_{\text{a}}\left(1.8\times {10}^{-5}\right)=1\times {10}^{-14}$

Rearrange to obtain the value of $K_{\text{a}}$.

  $\begin{array}{c}{K}_{\text{a}}=\frac{1\times {10}^{-14}}{1.8\times {10}^{-5}}\\ =5.6\times {10}^{-10}\end{array}$

The ICE table is given as follows:

  $\begin{array}{cccccccc}& {\text{NH}}_{\text{4}}{}^{+}& +& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & {\text{H}}_3{\text{O}}^{+}& +& {\text{NH}}_3\\ \text{Initial, }M& 0.125& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.125-x\right)& & & & x& & x\end{array}$

Substitute $0.125-x$ for $\left[{\text{NH}}_4^{+}\right]$, $x$ for $\left[{\text{NH}}_3\right]$, $x$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$ and $5.6\times {10}^{-10}$ for $K_{\text{a}}$ in equation (1).

  $5.6\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.125-x\right)}$

Assume $x<<0.125M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^2=\left(5.6\times {10}^{-10}\right)\left(0.125\right)\\ x=\sqrt{0.7\times {10}^{-10}}\\ =8.3666\times {10}^{-6}\end{array}$

The formula to calculate $\text{pH}$ is given as follows:

  $\text{pH}=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$        (5)

Substitute $8.3666\times {10}^{-6}M$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in equation (5).

  $\begin{array}{c}\text{pH}=-\text{log}\left(8.3666\times {10}^{-6}\right)\\ =5.077\end{array}$

(c)

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ of the solution $0.400M$ of potassium chlorite has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.91QE

The value of $\text{pH}$ of the solution $0.400M$ of potassium chlorite is 7.78.

Explanation of Solution

The ionization of ${\text{KClO}}_2$ salt occurs as follows:

  ${\text{K}}^{+}\left(\text{aq}\right)+{\text{ClO}}_2^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{HClO}}_2\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)+{\text{K}}^{+}\left(\text{aq}\right)$

The expression of $K_{\text{b}}$ is given as follows:

  $K_{\text{b}}=\frac{\left[{\text{HClO}}_2\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{ClO}}_2^{-}\right]}$        (7)

Here,

$\left[{\text{HClO}}_2\right]$ denotes the concentration of ${\text{HClO}}_2$.

$\left[{\text{OH}}^{-}\right]$ denotes the concentration of ${\text{OH}}^{-}$ ions.

$\left[{\text{ClO}}_2^{-}\right]$ denotes the concentration of ${\text{ClO}}_2^{-}$ .

$K_{\text{b}}$ denotes equilibrium constant for this ionization.

The relation among $K_{\text{a}}$, ${\text{K}}_{\text{b}}$ and $K_{\text{w}}$ is given as follows:

  $K_{\text{a}}{K}_{\text{b}}=K_{\text{w}}$

Substitute $1.1\times {10}^{-2}$ for $K_{\text{a}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (1).

  $\left(1.1\times {10}^{-2}\right)K_{\text{b}}=1\times {10}^{-14}$

Rearrange to obtain the value of $K_{\text{b}}$.

  $\begin{array}{c}{K}_{\text{b}}=\frac{1\times {10}^{-14}}{1.1\times {10}^{-2}}\\ =0.9090\times {10}^{-12}\end{array}$

The ICE table is given as follows:

  $\begin{array}{cccccccc}& {\text{ClO}}_2^{-}& +& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & {\text{OH}}^{-}& +& {\text{HClO}}_2\\ \text{Initial, }M& 0.4& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.4-x\right)& & & & x& & x\end{array}$

Substitute $0.4-x$ for $\left[{\text{ClO}}_2^{-}\right]$, $x$ for $\left[{\text{HClO}}_2\right]$, $x$ for $\left[{\text{OH}}^{-}\right]$ and $0.9090\times {10}^{-12}$ for $K_{\text{b}}$ in equation (7).

  $0.9090\times {10}^{-12}=\frac{\left(x\right)\left(x\right)}{\left(0.4-x\right)}$

Assume $x<<0.4M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^2=\left(0.9090\times {10}^{-12}\right)\left(0.4\right)\\ x=\sqrt{0.3636\times {10}^{-12}}\\ =0.603022\times {10}^{-6}\end{array}$

Substitute $0.603022\times {10}^{-6}M$ for $\left[{\text{OH}}^{-}\right]$ in equation (3).

  $\begin{array}{c}\text{pOH}=-\text{log}\left(0.603022\times {10}^{-6}\right)\\ =6.2196\end{array}$

Substitute 6.2196 for $\text{pOH}$ in equation (4).

  $\begin{array}{c}\text{pH}=\text{14}-6.2196\\ =7.78\end{array}$