Chapter 15, Problem 15.114QE

(a)

Interpretation Introduction

Interpretation:

Solution of $0.30M{\text{ NH}}_4\text{Cl}$ has to be classified as either strongly acidic, weakly acidic, neutral, weakly basic, and strongly basic.

Concept Introduction:

The ionization of a hypothetical weak acid $\left(\text{HA}\right)$ is given as follows:

  $\text{HA}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{A}}^{-}\left(\text{aq}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)$

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

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

Here,

$K_{\text{w}}$ denotes the ionic product of water.

$K_{\text{a}}$ denotes the ionization constant of weak acid.

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

Answer to Problem 15.114QE

$0.30M{\text{ NH}}_4\text{Cl}$ is weakly acidic and its $\text{pH}$ is estimated to be 4.887.

Explanation of Solution

The ionization of ${\text{NH}}_4\text{Cl}$ salt occurs as follows:

  ${\text{NH}}_4\text{Cl}\left(\text{aq}\right)\rightleftharpoons {\text{NH}}_4^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)$

${\text{Cl}}^{-}$ the conjugate base of strong acid hydrochloric acid hence it is weak and not able to accept proton from water. It remains in the medium as spectator ion.

${\text{NH}}_{\text{4}}{}^{+}$ is the conjugate acid of strong base ammonia and is therefore weak 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:

  $K_{\text{a}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{NH}}_3\right]}{\left[{\text{NH}}_4^{+}\right]}$        (1)

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.

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

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

Refer to table 15.8 for the value of $K_{\text{b}}$ of ammonia.

Substitute $1.8\times {10}^{-5}$ for $K_{\text{b}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (2).

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

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:

  $\begin{array}{cccccccc}& {\text{NH}}_{\text{4}}{}^{+}& +& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & {\text{H}}_3{\text{O}}^{+}& +& {\text{NH}}_3\\ \text{Initial, }M& 0.30& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.30-x\right)& & & & x& & x\end{array}$

Substitute $0.30-x$ 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).

  $5.6\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.30-x\right)}$

Assume $x<<0.30M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^2=\left(5.6\times {10}^{-10}\right)\left(0.30\right)\\ x=\sqrt{1.68\times {10}^{-10}}\\ =1.29614\times {10}^{-5}\end{array}$

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

  $\text{pH}=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$        (3)

Substitute $1.29614\times {10}^{-5}M$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in equation (3).

  $\begin{array}{c}\text{pH}=-\text{log}\left(1.29614\times {10}^{-5}\right)\\ =4.887\end{array}$

(b)

Interpretation Introduction

Interpretation:

Solution of $0.15M{\text{ N}}_2{\text{H}}_5\text{Cl}$ has to be classified as either strongly acidic, weakly acidic, neutral, weakly basic, and strongly basic.

Concept Introduction:

Refer to part (a).

Answer to Problem 15.114QE

$0.45M{\text{ N}}_2{\text{H}}_5\text{Cl}$ is weakly acidic and its $\text{pH}$ is estimated to be 4.968.

Explanation of Solution

The ionization of ${\text{N}}_2{\text{H}}_5\text{Cl}$ salt occurs as follows:

  ${\text{N}}_2{\text{H}}_5\text{Cl}\left(\text{aq}\right)\rightleftharpoons {\text{N}}_2{\text{H}}_5{}^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)$

${\text{Cl}}^{-}$ the conjugate base of strong acid hydrochloric acid hence it is weak and not able to accept proton from water. It remains in the medium as spectator ion.

${\text{N}}_2{\text{H}}_5{}^{+}$ is the conjugate acid of hydrazine base and reacts with water as follows:

  ${\text{N}}_2{\text{H}}_5{}^{+}+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_3{\text{O}}^{+}+{\text{N}}_2{\text{H}}_4$

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

  $K_{\text{a}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{N}}_2{\text{H}}_5{}^{+}\right]}{\left[{\text{N}}_2{\text{H}}_5\right]}$        (4)

Here,

$\left[{\text{N}}_2{\text{H}}_4\right]$ denotes the concentration of ${\text{N}}_2{\text{H}}_4$.

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

$\left[{\text{N}}_2{\text{H}}_5{}^{+}\right]$ denotes the concentration of ${\text{N}}_2{\text{H}}_5{}^{+}$.

$K_{\text{a}}$ denotes ionization constant of ${\text{N}}_2{\text{H}}_6{}^{+}$ ion.

Refer to table 15.8 for the value of $K_{\text{b}}$ of hydrazine.

Substitute $1.3\times {10}^{-6}$ for $K_{\text{b}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (2).

  $K_{\text{a}}\left(1.3\times {10}^{-6}\right)=1\times {10}^{-14}$

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

  $\begin{array}{c}{K}_{\text{a}}=\frac{1\times {10}^{-14}}{1.3\times {10}^{-6}}\\ =7.692\times {10}^{-9}\end{array}$

The ICE table is given as follows:

  $\begin{array}{cccccccc}& {\text{N}}_2{\text{H}}_5{}^{+}& +& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & {\text{H}}_3{\text{O}}^{+}& +& {\text{N}}_2{\text{H}}_4\\ \text{Initial, }M& 0.15& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.15-x\right)& & & & x& & x\end{array}$

Substitute $0.15-x$ for $\left[{\text{NH}}_4^{+}\right]$, $x$ for $\left[{\text{NH}}_3\right]$, $x$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$ and $7.692\times {10}^{-9}$ for $K_{\text{a}}$ in equation (4).

  $7.692\times {10}^{-9}=\frac{\left(x\right)\left(x\right)}{\left(0.15-x\right)}$

Assume $x<<0.15M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^2=\left(7.692\times {10}^{-9}\right)\left(0.15\right)\\ x=\sqrt{1.1538\times {10}^{-10}}\\ =1.074\times {10}^{-5}\end{array}$

Substitute $1.074\times {10}^{-5}M$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in equation (3).

  $\begin{array}{c}\text{pH}=-\text{log}\left(1.074\times {10}^{-5}\right)\\ =4.968\end{array}$

(c)

Interpretation Introduction

Interpretation:

Solution of $0.50M{\text{ KNO}}_3$ has to be classified as either strongly acidic, weakly acidic, neutral, weakly basic, and strongly basic.

Concept Introduction:

Answer to Problem 15.114QE

$0.50M{\text{ KNO}}_3$ is neutral and The value of $\text{pH}$ is estimated as7.

Explanation of Solution

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

  ${\text{KNO}}_3\left(\text{aq}\right)\rightleftharpoons {\text{K}}^{+}\left(\text{aq}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)$

${\text{NO}}_3^{-}$ ion is conjugate base of strong acid nitric acid hence it remains as spectator ion. ${\text{K}}^{+}$ conjugate acid of strong base potassium hydroxide and hence it is also not strong enough to undergo any reaction with water. ${\text{K}}^{+}$ acts as a spectator ion and does not alter the $\text{pH}$.  Therefore $\text{KCl}$ salt is neutral as there is no reaction with water and thus its $\text{pH}$ is estimated to be 7.

(d)

Interpretation Introduction

Interpretation:

Solution of $0.50M\text{ HCOONa}$ 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 15.114QE

$0.50M\text{ HCOONa}$ is weakly basic and the value of $\text{pH}$ of   is 8.726.

Explanation of Solution

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

  ${\text{Na}}^{+}\left(\text{aq}\right)+{\text{HCOO}}^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{aq}\right)\rightleftharpoons \text{HCOOH}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)+{\text{Na}}^{+}\left(\text{aq}\right)$

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

  $K_{\text{b}}=\frac{\left[\text{HCOOH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{HCOO}}^{-}\right]}$        (5)

Here,

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

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

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

$K_{\text{b}}$ 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 $1.8\times {10}^{-4}$ for $K_{\text{a}}$ and $1\times {10}^{-14}$ for $K_{\text{w}}$ in equation (2).

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

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

  $\begin{array}{c}{K}_{\text{b}}=\frac{1\times {10}^{-14}}{1.8\times {10}^{-4}}\\ =0.555\times {10}^{-10}\end{array}$

The ICE table is given as follows:

  $\begin{array}{cccccccc}& {\text{HCOO}}^{-}& +& {\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & {\text{OH}}^{-}& +& \text{HCOOH}\\ \text{Initial, }M& 0.50& & & & 0& & 0\\ \text{Change, }M& -x& & & & +x& & +x\\ \text{Equilibrium, }M& \left(0.50-x\right)& & & & x& & x\end{array}$

Substitute $0.50-x$ for $\left[{\text{HCOO}}^{-}\right]$, $x$ for $\left[\text{HCOOH}\right]$, $x$ for $\left[{\text{OH}}^{-}\right]$ and $0.5556\times {10}^{-10}$ for $K_{\text{b}}$ in equation (5).

  $0.5556\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.50-x\right)}$

Assume $x<<0.50M$ and rearrange to obtain the value of $x$.

  $\begin{array}{c}{x}^2=\left(0.5556\times {10}^{-10}\right)\left(0.50\right)\\ x=\sqrt{0.2778\times {10}^{-10}}\\ =0.52706\times {10}^{-6}\end{array}$

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

  $\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$        (6)

Substitute $0.52706\times {10}^{-6}M$ for $\left[{\text{OH}}^{-}\right]$ in equation (6).

  $\begin{array}{c}\text{pOH}=-\text{log}\left(0.52706\times {10}^{-6}\right)\\ =5.274\end{array}$

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

  $\text{pH}=14-\text{pOH}$        (7)

Substitute 5.274 for $\text{pOH}$ in equation (7).

  $\begin{array}{c}\text{pH}=\text{14}-5.274\\ =8.726\end{array}$