Chapter 14, Problem 54QRT

(a)

Interpretation Introduction

Interpretation:

The $pH$ of $0.10M$ aqueous solutions of $HN{O}_2$ given below has to be calculated.

Explanation of Solution

The equilibrium reaction is given below.

  $HN{O}_2\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons N{O}_2{}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$

The acid ionization constant can be written as given below.

  $K_a=\frac{\left[N{O}_2{}^{-}\right]\left[H_3{O}^{+}\right]}{\left[HN{O}_2\right]}$

A table can be set up as shown below.

  $\begin{array}{cccc}& HN{O}_2\left(aq\right)& H_3{O}^{+}\left(aq\right)& N{O}_2{}^{-}\left(aq\right)\\ Initialconc.\left(M\right)& 0.10& 1\times {10}^{-7}& 0\\ Changeinconc.\left(M\right)& -x& +x& +x\\ Equilibriumconc.\left(M\right)& 0.10-x& x& x\end{array}$

The concentration of $H_3{O}^{+}$ comes from its autoionization that is $1.0\times {10}^{-7}.$  It is assumed to be small compared to the change.

At equilibrium, $K_a=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}=7.4\times {10}^{-4}$

Then,

  $\begin{array}{l}{K}_a=\frac{\left(x\right)\left(x\right)}{\left(0.15-x\right)}=7.4\times {10}^{-4}\\ \\ \Rightarrow x^2=\left(7.4\times {10}^{-4}\right)\times \left(0.10-x\right)\\ \\ \Rightarrow x^2+7.4\times {10}^{-4}x-7.4\times {10}^{-5}=0\\ \\ \Rightarrow x=8.2\times {10}^{-3}=\left[H_3{O}^{+}\right]\end{array}$

The $pH$ of the solution can be calculated as shown below.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(8.2\times {10}^{-3}\right)=2.\text{09}\text{.}$

Therefore, the $pH$ of $0.10M$ aqueous solution of $HN{O}_2$ is $2.09.$

(b)

Interpretation Introduction

Interpretation:

The $pH$ of $0.10M$ aqueous solutions of $N{H}_{4}Cl$ given below has to be calculated.

Explanation of Solution

The equilibrium reaction is given below.

  $N{H}_4{}^{+}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons N{H}_3\left(aq\right)+H_3{O}^{+}\left(aq\right)$

The acid ionization constant can be written as given below.

  $K_a=\frac{\left[N{H}_3\right]\left[H_3{O}^{+}\right]}{\left[N{H}_4{}^{+}\right]}$

A table can be set up as shown below.

  $\begin{array}{cccc}& N{H}_4{}^{+}\left(aq\right)& H_3{O}^{+}\left(aq\right)& N{H}_3\left(aq\right)\\ Initialconc.\left(M\right)& 0.10& 1\times {10}^{-7}& 0\\ Changeinconc.\left(M\right)& -x& +x& +x\\ Equilibriumconc.\left(M\right)& 0.10-x& x& x\end{array}$

The concentration of $H_3{O}^{+}$ comes from its autoionization that is $1.0\times {10}^{-7}.$  It is assumed to be small compared to the change.

At equilibrium, $K_a=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}=5.6\times {10}^{-10}$

Assuming x is very small, it can be written $0.10-x\cong \mathrm{0.10.}$

Then,

  $\begin{array}{l}{K}_a=\frac{\left(x\right)\left(x\right)}{\left(0.15-x\right)}=5.6\times {10}^{-10}\\ \\ \Rightarrow x^2=\left(5.6\times {10}^{-10}\right)\times \left(0.10\right)\\ \\ \Rightarrow x=\sqrt{5.6\times {10}^{-11}}=7.5\times {10}^{-6}M=\left[H_3{O}^{+}\right]\end{array}$

The $pH$ of the solution can be calculated as shown below.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(7.5\times {10}^{-6}\right)=\mathrm{5.13.}$

Therefore, the $pH$ of $0.10M$ aqueous solution of $N{H}_{4}Cl$ is $5.13.$

(c)

Interpretation Introduction

Interpretation:

The $pH$ of $0.10M$ aqueous solutions of $NaF$ given below has to be calculated.

Explanation of Solution

The equilibrium reaction is given below.

  $F^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons HF\left(aq\right)+O{H}^{-}\left(aq\right)$

The base ionization constant can be written as given below.

  $K_b=\frac{\left[HF\right]\left[O{H}^{-}\right]}{\left[F^{-}\right]}$

A table can be set up as shown below.

  $\begin{array}{cccc}& F^{-}\left(aq\right)& HF\left(aq\right)& O{H}^{-}\left(aq\right)\\ Initialconc.\left(M\right)& 0.10& 0& 1\times {10}^{-7}\\ Changeinconc.\left(M\right)& -x& +x& +x\\ Equilibriumconc.\left(M\right)& 0.10-x& x& x\end{array}$

The concentration of $O{H}^{-}$ comes from its autoionization that is $1.0\times {10}^{-7}.$  It is assumed to be small compared to the change.

At equilibrium, $K_b=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}=1.5\times {10}^{-11}$

Assuming x is very small, it can be written $0.10-x\cong \mathrm{0.10.}$

Then,

  $\begin{array}{l}{K}_b=\frac{\left(x\right)\left(x\right)}{\left(0.15-x\right)}=1.5\times {10}^{-11}\\ \\ \Rightarrow x^2=\left(1.5\times {10}^{-11}\right)\times \left(0.10\right)\\ \\ \Rightarrow x=\sqrt{1.5\times {10}^{-12}}=1.2\times {10}^{-6}M=\left[O{H}^{-}\right]\end{array}$

The $pH$ of the solution can be calculated as shown below.

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(1.2\times {10}^{-6}\right)=\mathrm{5.91.}\\ \\ pH=14.00-pOH=14.00-5.91=8.09.\end{array}$

Therefore, the $pH$ of $0.10M$ aqueous solution of $HF$ is $8.09.$

(d)

Interpretation Introduction

Interpretation:

The $pH$ of $0.10M$ aqueous solutions of $Mg{\left(C{H}_{3}COO\right)}_2$ given below has to be calculated.

Explanation of Solution

The concentration of $C{H}_{3}CO{O}^{-}$ can be calculated as shown below.

  $\left[C{H}_{3}CO{O}^{-}\right]=0.10MMg{\left(C{H}_{3}COO\right)}_2\times \frac{2molC{H}_{3}CO{O}^{-}}{1molMg{\left(C{H}_{3}COO\right)}_2}=0.20M.$

The equilibrium reaction is given below.

  $C{H}_{3}CO{O}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons C{H}_{3}COOH\left(aq\right)+O{H}^{-}\left(aq\right)$

The base ionization constant can be written as given below.

  $K_b=\frac{\left[C{H}_{3}COOH\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}CO{O}^{-}\right]}$

A table can be set up as shown below.

  $\begin{array}{cccc}& C{H}_{3}CO{O}^{-}\left(aq\right)& C{H}_{3}COOH\left(aq\right)& O{H}^{-}\left(aq\right)\\ Initialconc.\left(M\right)& 0.20& 0& 1\times {10}^{-7}\\ Changeinconc.\left(M\right)& -x& +x& +x\\ Equilibriumconc.\left(M\right)& 0.20-x& x& x\end{array}$

The concentration of $O{H}^{-}$ comes from its autoionization that is $1.0\times {10}^{-7}.$  It is assumed to be small compared to the change.

At equilibrium, $K_b=\frac{\left(x\right)\left(x\right)}{\left(0.20-x\right)}=5.6\times {10}^{-10}$

Assuming x is very small, it can be written $0.20-x\cong \mathrm{0.20.}$

Then,

  $\begin{array}{l}{K}_b=\frac{\left(x\right)\left(x\right)}{\left(0.20-x\right)}=5.6\times {10}^{-10}\\ \\ \Rightarrow x^2=\left(5.6\times {10}^{-10}\right)\times \left(0.20\right)\\ \\ \Rightarrow x=\sqrt{\left(5.6\times {10}^{-10}\right)\times \left(0.20\right)}=1.1\times {10}^{-5}M=\left[O{H}^{-}\right]\end{array}$

The $pH$ of the solution can be calculated as shown below.

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(1.1\times {10}^{-5}\right)=\mathrm{4.98.}\\ \\ pH=14.00-pOH=14.00-4.98=9.02.\end{array}$

Therefore, the $pH$ of $0.10M$ aqueous solution of $Mg{\left(C{H}_{3}COO\right)}_2$ is $9.02.$

(e)

Interpretation Introduction

Interpretation:

The $pH$ of $0.10M$ aqueous solutions of $BaO$ given below has to be calculated.

Explanation of Solution

The reaction of $BaO$ with water is given below.

  $O^{2-}\left(aq\right)+H_{2}O\left(l\right)\to 2O{H}^{-}\left(aq\right)$

The concentration of $O{H}^{-}$ can be calculated as shown below.

  $\left[O{H}^{-}\right]=0.10M{O}^{2-}\times \frac{2molO{H}^{-}}{1mol{O}^{2-}}=0.20M.$

The $pH$ of the solution can be calculated as shown below.

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(0.20\right)=\mathrm{0.70.}\\ \\ pH=14.00-pOH=14.00-0.70=13.30.\end{array}$

Therefore, the $pH$ of $0.10M$ aqueous solution of $BaO$ is $13.30.$

(f)

Interpretation Introduction

Interpretation:

The $pH$ of $0.10M$ aqueous solutions of $KHS{O}_4$ given below has to be calculated.

Explanation of Solution

$HS{O}_4{}^{-}$ acts as both acid and base.

  $\begin{array}{l}HS{O}_4{}^{-}\left(aq\right)+H_{2}O\left(aq\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+S{O}_4{}^{2-}\left(aq\right)\\ \\ K_a=\frac{\left[H_3{O}^{+}\right]\left[S{O}_4{}^{2-}\right]}{\left[HS{O}_4{}^{-}\right]}=1.1\times {10}^{-2}\\ \\ HS{O}_4{}^{-}\left(aq\right)+H_{2}O\left(aq\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+H_{2}S{O}_4\left(aq\right)\\ \\ K_b=\frac{\left[H_{2}S{O}_4\right]\left[O{H}^{-}\right]}{\left[HS{O}_4{}^{-}\right]}=verysmall\end{array}$

$HS{O}_4{}^{-}$ is more acidic than basic.

A table can be set up as shown below.

  $\begin{array}{cccc}& HS{O}_4{}^{-}\left(aq\right)& H_3{O}^{+}\left(aq\right)& S{O}_4{}^{2-}\left(aq\right)\\ Initialconc.\left(M\right)& 0.10& 1\times {10}^{-7}& 0\\ Changeinconc.\left(M\right)& -x& +x& +x\\ Equilibriumconc.\left(M\right)& 0.10-x& x& x\end{array}$

The concentration of $H_3{O}^{+}$ comes from its autoionization that is $1.0\times {10}^{-7}.$  It is assumed to be small compared to the change.

At equilibrium, $K_a=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}=1.1\times {10}^{-2}$

Then,

  $\begin{array}{l}{x}^2=\left(1.1\times {10}^{-2}\right)\times \left(0.10-x\right)\\ \\ \Rightarrow x^2+1.1\times {10}^{-2}x-1.1\times {10}^{-3}=0\\ \\ \Rightarrow x=2.8\times {10}^{-2}M=\left[H_3{O}^{+}\right]\end{array}$

The $pH$ of the solution can be calculated as shown below.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(2.8\times {10}^{-2}\right)=1.55.$

Therefore, the $pH$ of $0.10M$ aqueous solution of $KHS{O}_4$ is $1.55.$

(g)

Interpretation Introduction

Interpretation:

The $pH$ of $0.10M$ aqueous solutions of $NaHC{O}_3$ given below has to be calculated.

Explanation of Solution

$HC{O}_3{}^{-}$ acts as both acid and base.

  $\begin{array}{l}HC{O}_3{}^{-}\left(aq\right)+H_{2}O\left(aq\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+C{O}_3{}^{2-}\left(aq\right)\\ \\ K_a=\frac{\left[H_3{O}^{+}\right]\left[C{O}_3{}^{2-}\right]}{\left[HC{O}_3{}^{-}\right]}=4.7\times {10}^{-11}\\ \\ HC{O}_3{}^{-}\left(aq\right)+H_{2}O\left(aq\right)\rightleftharpoons O{H}^{-}\left(aq\right)+H_{2}C{O}_3\left(aq\right)\\ \\ K_b=\frac{\left[H_{2}C{O}_3\right]\left[O{H}^{-}\right]}{\left[HC{O}_3{}^{-}\right]}=2.3\times {10}^{-8}\end{array}$

$HC{O}_3{}^{-}$ is more basic than acidic.

A table can be set up as shown below.

  $\begin{array}{cccc}& HC{O}_3{}^{-}\left(aq\right)& H_{2}C{O}_3\left(aq\right)& O{H}^{-}\left(aq\right)\\ Initialconc.\left(M\right)& 0.10& 0& 1\times {10}^{-7}\\ Changeinconc.\left(M\right)& -x& +x& +x\\ Equilibriumconc.\left(M\right)& 0.10-x& x& x\end{array}$

The concentration of $O{H}^{-}$ comes from its autoionization that is $1.0\times {10}^{-7}.$  It is assumed to be small compared to the change.

At equilibrium, $K_b=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}=2.3\times {10}^{-8}$

Then,

Assuming x is very small, it can be written $0.20-x\cong \mathrm{0.20.}$

  $\begin{array}{l}{K}_b=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}=2.3\times {10}^{-8}\\ \\ \Rightarrow x^2=\left(2.3\times {10}^{-8}\right)\times \left(0.10\right)\\ \\ \Rightarrow x=\sqrt{\left(2.3\times {10}^{-8}\right)\times \left(0.10\right)}=4.8\times {10}^{-5}M=\left[O{H}^{-}\right]\end{array}$

The $pH$ of the solution can be calculated as shown below.

  $\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]=-\log \left(4.8\times {10}^{-5}\right)=\mathrm{4.32.}\\ \\ pH=14.00-pOH=14.00-4.32=9.68.\end{array}$

Therefore, the $pH$ of $0.10M$ aqueous solution of $NaHC{O}_3$ is $9.68.$

(h)

Interpretation Introduction

Interpretation:

The $pH$ of $0.10M$ aqueous solutions $BaC{l}_2$ given below has to be calculated.

Explanation of Solution

$BaC{l}_2\to B{a}^{2+}+2C{l}^{-}$

Neither of these ions affect the $pH$ of a water solution, because $B{a}^{2+}$ is the cation of $Ba{\left(OH\right)}_2$ and $C{l}^{–}$ is a weak base.  So $pH$ is $7.00$ and no calculation is needed.