Chapter 15, Problem 15.113QE

(a)

Interpretation Introduction

Interpretation:

Solution of $0.45M\text{ NaCl}$ has to be classified as either strongly acidic, weakly acidic, neutral, weakly basic, and strongly basic.

Concept Introduction:

Salts formed from alkali metal cation and halogens as anions are neutral. This is because the cation acts as spectator ion while the anion is not strong enough to react with water as halides are conjugate bases of strong hydrohalic acids. Thus salts such as $\text{NaCl}$, $\text{KCl}$ are regarded as neutral.

Answer to Problem 15.113QE

$0.45M\text{ NaCl}$ is weakly basic and its $\text{pH}$ is estimated to be 7.

Explanation of Solution

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

  $\text{NaCl}\left(\text{aq}\right)\rightleftharpoons {\text{Na}}^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)$

${\text{Cl}}^{-}$ the conjugate base of strong acid hydrochloric acid hence it is weak and not able to accept proton from water. Similarly ${\text{Na}}^{+}$ is also not strong enough to undergo any reaction with water. ${\text{Na}}^{+}$ acts as a spectator ion and does not alter the $\text{pH}$. Therefore $\text{NaCl}$ salt is neutral as there is no reaction with water and thus its $\text{pH}$ is estimated to be 7.

(b)

Interpretation Introduction

Interpretation:

Solution of $0.18M\text{ Ba}{\left(\text{F}\right)}_2$ has to be classified as either strongly acidic, weakly acidic, neutral, weakly basic, and strongly basic.

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 ${\text{K}}_{\text{b}}$ 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 15.113QE

$0.18M\text{ Ba}{\left(\text{F}\right)}_2$ is weakly basic and its $\text{pH}$ is estimated to be 8.05.

Explanation of Solution

The ionization of $\text{Ba}{\left(\text{F}\right)}_2$ salt occurs as follows:

  ${\text{Ba}}^{2+}\left(\text{aq}\right)+2{\text{F}}^{-}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{aq}\right)\rightleftharpoons 2\text{HF}\left(\text{aq}\right)+2{\text{OH}}^{-}\left(\text{aq}\right)+{\text{Ba}}^{2+}\left(\text{aq}\right)$

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

  $K_{\text{b}}=\frac{{\left[\text{HF}\right]}^2{\left[{\text{OH}}^{-}\right]}^2}{{\left[{\text{F}}^{-}\right]}^2}$        (1)

Here,

$\left[{\text{F}}^{-}\right]$ denotes the concentration of fluoride ion.

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

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

$K_{\text{b}}$ denotes ionization constant of fluoride ion.

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

  $K_{\text{a}}{K}_{\text{b}}=K_{\text{w}}$        (2)

Substitute $6.3\times {10}^{-5}$ for $K_{\text{b}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (2).

  $\left(6.3\times {10}^{-5}\right){\text{K}}_{\text{b}}=1\times {10}^{-14}$

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

  $\begin{array}{c}{K}_{\text{b}}=\frac{1\times {10}^{-14}}{6.3\times {10}^{-5}}\\ =1.587\times {10}^{-10}\end{array}$

The ICE table is given as follows:

  $\begin{array}{cccccccc}& {\text{2F}}^{-}& +& {\text{2H}}_{\text{2}}\text{O}& \rightleftharpoons & \text{2HF}& +& {\text{2OH}}^{-}\\ \text{Initial, }M& 0.18& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.18-x\right)& & & & x& & x\end{array}$

Substitute $0.18-x$ for $\left[{\text{F}}^{-}\right]$, $x$ for $\left[\text{HF}\right]$, $x$ for $\left[{\text{OH}}^{-}\right]$ and $1.587\times {10}^{-10}$  for $K_{\text{b}}$ in equation (1).

  $1.587\times {10}^{-10}=\frac{\left(x^2\right)\left(x^2\right)}{{\left(0.18-x\right)}^2}$

Assume $x<<0.18M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^4=\left(1.587\times {10}^{-10}\right){\left(0.18\right)}^2\\ x={\left(5.1418\times {10}^{-12}\right)}^{1/4}\\ =1.5058\times {10}^{-3}\end{array}$

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

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

Substitute $1.5058\times {10}^{-3}M$ for $\left[{\text{OH}}^{-}\right]$ in equation (3).

  $\begin{array}{c}\text{pOH}=-\text{log}\left(1.5058\times {10}^{-3}\right)\\ =2.8222\end{array}$

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

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

Substitute 2.8222 for $\text{pOH}$ in equation (4).

  $\begin{array}{c}\text{pH}=\text{14}-2.8222\\ =11.1778\end{array}$

(c)

Interpretation Introduction

Interpretation:

The solution of $0.25M{\text{ KHSO}}_4$ has to be classified as either strongly acidic, weakly acidic, neutral, weakly basic, and strongly basic.

Concept Introduction:

Refer to part(b).

Answer to Problem 15.113QE

$0.25M{\text{ KHSO}}_4$ is weakly acidic and its $\text{pH}$ is estimated to be 5.6989.

Explanation of Solution

 Since ${\text{HSO}}_4{}^{-}$ is a stronger acid than water it donates its proton to water and therefore the ionization of ${\text{KHSO}}_4$ salt occurs as follows:

  ${\text{K}}^{+}\left(\text{aq}\right)+{\text{HSO}}_4{}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{aq}\right)\rightleftharpoons {\text{SO}}_4^{2-}\left(\text{aq}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{K}}^{+}\left(\text{aq}\right)$

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

  $K_{\text{a}}=\frac{\left[{\text{SO}}_4^{2-}\right]\left[{\text{H}}_3{\text{O}}^{+}\right]}{\left[{\text{HSO}}_4{}^{-}\right]}$        (5)

Here,

$\left[{\text{SO}}_4^{2-}\right]$ denotes the concentration of ${\text{HSO}}_4{}^{-}$ ion.

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

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

$K_{\text{a}}$ denotes ionization constant of ${\text{HSO}}_4{}^{-}$ ion.

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 $1.0\times {10}^{-2}$ for $K_{\text{b}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (2).

  $K_{\text{a}}\left(1.0\times {10}^{-2}\right)=1\times {10}^{-14}$

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

  $\begin{array}{c}{K}_{\text{a}}=\frac{1\times {10}^{-14}}{1.0\times {10}^{-2}}\\ =1\times {10}^{-12}\end{array}$

The ICE table is given as follows:

  $\begin{array}{cccccccc}& {\text{HSO}}_4{}^{-}& +& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & {\text{H}}_3{\text{O}}^{+}& +& {\text{SO}}_4^{2-}\\ \text{Initial, }M& 0.25& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.25-x\right)& & & & x& & x\end{array}$

Substitute $0.25-x$ for $\left[{\text{HSO}}_4{}^{-}\right]$, $x$ for $\left[{\text{SO}}_4^{2-}\right]$, $x$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$ and $1\times {10}^{-12}$ for $K_{\text{a}}$ in equation (5).

  $1\times {10}^{-12}=\frac{\left(x\right)\left(x\right)}{\left(0.25-x\right)}$

Assume $x<<0.25M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^2=\left(1\times {10}^{-12}\right)\left(0.25\right)\\ x=\sqrt{0.25\times {10}^{-12}}\\ =0.5\times {10}^{-6}\end{array}$

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

  $\text{pH}=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$        (6)

Substitute $0.5\times {10}^{-6}M$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in equation (6).

  $\begin{array}{c}\text{pH}=-\text{log}\left(0.5\times {10}^{-6}\right)\\ =5.6989\end{array}$

(d)

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ solution of $0.33M{\text{ NaNO}}_2$ has to be calculated.

Concept Introduction:

Refer to part (b).

Answer to Problem 15.113QE

$0.33M{\text{ NaNO}}_2$ is weakly basic and the value of $\text{pH}$ is estimated as 8.325.

Explanation of Solution

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

  ${\text{Na}}^{+}\left(\text{aq}\right)+{\text{NO}}_2^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{HNO}}_2\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)+{\text{Na}}^{+}\left(\text{aq}\right)$

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

  $K_{\text{b}}=\frac{\left[{\text{HNO}}_2\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NO}}_2^{-}\right]}$        (7)

Here,

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

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

$\left[{\text{NO}}_2^{-}\right]$ denotes the concentration of ${\text{NO}}_2^{-}$ .

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

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

  $K_{\text{a}}{K}_{\text{b}}=K_{\text{w}}$        (2)

Substitute $5.6\times {10}^{-4}$ for $K_{\text{a}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (2).

  $\left(5.6\times {10}^{-4}\right)K_{\text{b}}=1\times {10}^{-14}$

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

  $\begin{array}{c}{K}_{\text{b}}=\frac{1\times {10}^{-14}}{5.6\times {10}^{-4}}\\ =1.7857\times {10}^{-11}\end{array}$

The ICE table is given as follows:

  $\begin{array}{cccccccc}& {\text{NO}}_2^{-}& +& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & {\text{OH}}^{-}& +& {\text{HNO}}_2\\ \text{Initial, }M& 0.33& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.33-x\right)& & & & x& & x\end{array}$

Substitute $0.33-x$ for $\left[{\text{NO}}_2^{-}\right]$, $x$ for $\left[{\text{HNO}}_2\right]$, $x$ for $\left[{\text{OH}}^{-}\right]$ and $1.7857\times {10}^{-11}$ for $K_{\text{b}}$ in equation (7).

  $1.7857\times {10}^{-11}=\frac{\left(x\right)\left(x\right)}{\left(0.33-x\right)}$

Assume $x<<0.33M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^2=\left(1.7857\times {10}^{-11}\right)\left(0.33\right)\\ x=\sqrt{5.89281\times {10}^{-12}}\\ =2.4275\times {10}^{-6}\end{array}$

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

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

Substitute $2.4275\times {10}^{-6}M$ for $\left[{\text{OH}}^{-}\right]$ in equation (3).

  $\begin{array}{c}\text{pOH}=-\text{log}\left(2.4275\times {10}^{-6}\right)\\ =5.6148\end{array}$

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

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

Substitute 5.6858 for $\text{pOH}$ in equation (4).

  $\begin{array}{c}\text{pH}=\text{14}-5.6148\\ =8.3852\end{array}$