Chapter 19, Problem 19.55P

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$$\text{65}\text{.5}\text{mL}$ of $\text{0}\text{.234}\text{M}{\text{NH}}_{\text{3}}$ at the equivalence point and the volume of $0.125MHCl$ needed to the reach the point’s in titrations has to be calculated. 

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

$\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Acid ionization constant${\text{K}}_{\text{a}}$: 

The equilibrium expression for the reaction ${\text{HA}}_{(aq)}\stackrel{}{\to }{\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$ is given below.

$K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where $K_a$ acid ionization is constant, $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion, $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion, $[\text{HA]}$ is concentration of the acid

Base ionization constant $K_b$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{B}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{}{\to }{\text{HB}}^{\text{+}}{}_{(aq)}+{\text{OH}}^{\text{-}}{}_{(aq)}$

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where $K_b$ is base ionization constant, $[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion, $[{\text{HB}}^{\text{+}}\text{]}$ is concentration of conjugate acid, $[\text{B]}$ is concentration of the base

Answer to Problem 19.55P

The pH of a solution made by mixing $\text{65}\text{.5}\text{mL}$ and $\text{0}\text{.234}\text{M}{\text{NH}}_{\text{3}}$ with $0.125MHCl$ is $5.71.$

Explanation of Solution

Given,

The balance equation is given below,

  $\text{HCl(aq) +}N{H}_3\text{(aq)}\underset{}{\to }N{H}_4^{+}\text{(aq) +}C{l}^{-}\text{(aq)}$

The chlorine ions on the product side are written as separate spices because they have no effect on $pH$ of the solution.

Calculate the required volume of $HCl$,

Volume $mL$ of $HCl$,

$\begin{array}{c}HCl=\left(\frac{0.234molN{H}_3}{L}\right)\left(\frac{{10}^{-3}L}{1mL}\right)\left(65.5mL\right)\left(\frac{1molHCl}{1molN{H}_3}\right)\left(\frac{L}{0.125molHCl}\right)\left(\frac{1mL}{{10}^{-3}L}\right)\\ \\ =122.616\\ \\ =123mLofHCl\end{array}$

Therefore, the required volume of $HCl=123mL.$

Determination the moles of $N{H}_3$present,

$\begin{array}{c}MolesofN{H}_3=\left(\frac{0.234molN{H}_3}{1mL}\right)\left(\frac{{10}^{-3}L}{1mL}\right)\left(65.5mL\right)\\ \\ =0.015327molN{H}_3\end{array}$

At the equivalent point, $\text{0}\text{.015327}\text{mol}{\text{NH}}_{\text{3}}$ will be added so, the moles acid = moles base.

The $HCl$ will react with an equal amount of the acid $0molN{H}_3$ will remain and $\text{0}\text{.015327}\text{mol}of{\text{NH}}_4^{+}$ will be formed.

ICE Table of equilibrium,

$\begin{array}{l}\text{HCl(aq) }\text{+}N{H}_3\text{(aq)}\underset{}{\rightleftharpoons }N{H}_4^{+}\text{(l) +}C{l}^{-}\text{(aq)}\\ \text{Initial }\text{(M)}\text{0}\text{.015327}\text{mol}\text{0}\text{.015327}\text{mol }-\text{0}\\ \text{Change}\left(\text{M}\right)-\text{0}\text{.015327}\text{mol}-\text{0}\text{.015327}\text{mol}-+\text{0}\text{.015327}\text{mol}\\ -----------------------------------------\\ \text{Final }00-+\text{0}\text{.015327}\text{mol}\end{array}$

Determination the liters of solution present at the equivalent point,

Volume

   $\begin{array}{l}=\left[\left(65.5+122.616\right)mL\right]\left({10}^{-3}L/1mL\right)\\ \\ =0.188116L\end{array}$

Concentration of $N{H}_4^{+}$ at equivalent point,

Molarity,

  $\begin{array}{l}=\left(0.015327molN{H}_4^{+}\right)/\left(0.188116L\right)\\ \\ =0.081476M\end{array}$

Next calculate $K_a$for $N{H}_4^{+}$

The equilibrium value for $K_{b}N{H}_3=1.76\times {10}^{-5}$

So,

  $K_b=K_W/K_a=\frac{\left(1.0\times {10}^{-14}\right)}{\left(1.76\times {10}^{-5}\right)}=5.6818\times {10}^{-10}$ 

Using a reaction table for the equilibrium reaction of $N{H}_4^{+}$

  $\begin{array}{l}N{H}_4^{+}\text{+}{H}_{2}O\underset{}{\rightleftharpoons }N{H}_3\text{ +}{H}_3{O}^{+}\\ \text{Initial }\text{(M)}:0.081476M-0\text{0}\\ \text{Change}\left(\text{M}\right):-x-+\text{x}+\text{x}\\ --------------------------------\\ \text{Equilibrium: }0.081476-x+x+x\end{array}$

Determination the hydroxide ion concentration from the $K_a$ and then determine the $pH$ from the $pOH$.

  $\begin{array}{l}{K}_a=5.6818\times {10}^{-10}=\frac{\left[H_3{O}^{+}\right]\left[N{H}_3\right]}{\left[N{H}_4^{+}\right]}=\frac{\left[x\right]\left[x\right]}{\left[0.081476-x\right]}=\frac{x^2}{\left[0.081476\right]}\\ \left[H_3{O}^{+}\right]=x=6.803898\times {10}^{-6}M\end{array}$

Calculation for $pH$ and $pOH$

  $\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ \\ =-\log \left(6.803898\times {10}^{-6}\right)\\ \\ pH=5.1672=5.71\end{array}$

(b)

Interpretation Introduction

Interpretation:

$21.8\text{mL}$ of $1.11MC{H}_{3}N{H}_2$ at the equivalence point and the volume of $0.125MHCl$ needed to the reach the point’s in titrations has to be calculated.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Acid ionization constant${\text{K}}_{\text{a}}$: 

The equilibrium expression for the reaction ${\text{HA}}_{(aq)}\stackrel{}{\to }{\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$ is given below.

${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{\text{[HA]}}$

Where $K_a$ acid ionization is constant, $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion, $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion, $[\text{HA]}$ is concentration of the acid

Base ionization constant ${\text{K}}_{\text{b}}$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{B}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{}{\to }{\text{HB}}^{\text{+}}{}_{(aq)}+{\text{OH}}^{\text{-}}{}_{(aq)}$

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where $K_b$ is base ionization constant, $[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion, $[{\text{HB}}^{\text{+}}\text{]}$ is concentration of conjugate acid, $[\text{B]}$ is concentration of the base

Answer to Problem 19.55P

The pH of a solution made by mixing $21.8\text{mL}$ and $1.11MC{H}_{3}N{H}_2$ with $0.125MHCl$ is $5.80.$

Explanation of Solution

The balance equation is given below,

  $HCl\left(aq\right)+C{H}_{3}N{H}_2\left(aq\right)\underset{}{\to }C{H}_{3}N{H}_3^{+}\text{(aq)}+C{l}^{-}\left(aq\right)$

The chloride ion on the product side are written as separate spices because they have no effect on $pH$ of the solution.

Next calculate the volume of $HCl$,

Volume of $HCl$ in $mL$,

$\begin{array}{c}HCl=\left(\frac{1.11molC{H}_{3}N{H}_2}{L}\right)\left(\frac{{10}^{-3}L}{1mL}\right)\left(21.8mL\right)\left(\frac{1molKOH}{1mol{H}_{2}C{O}_3}\right)\left(\frac{L}{0.125molKOH}\right)\left(\frac{1mL}{{10}^{-3}L}\right)\\ =193.584\\ =194mLofHCl\end{array}$

The solution requires an equal volume to reach second equivalence point $194mLofHCl.$

Determination the moles of $C{H}_{3}N{H}_2$present,

  $\begin{array}{c}MolesofC{H}_{3}N{H}_2=\left(\frac{1.11molC{H}_{3}N{H}_2}{L}\right)\left(\frac{{10}^{-3}L}{1mL}\right)\left(21.8mL\right)\\ \\ =0.024198molofC{H}_{3}N{H}_2\end{array}$

At the equivalence point, $\text{0}\text{.024198}\text{mol}\text{HCl}$ will be added the moles acid=moles of base.

The $HCl$ will react with an equal amount of the base $0molC{H}_{3}N{H}_2$ will remain and $0.024198molofC{H}_{3}N{H}_3^{+}$ will be formed.

ICE Table of equilibrium,

$\begin{array}{l}\text{HCl(aq) }\text{+}C{H}_{3}N{H}_2\text{(aq)}\underset{}{\rightleftharpoons }C{H}_{3}N{H}_3^{+}\text{(aq) +}C{l}^{-}\text{(aq)}\\ \text{Initial }\text{(M)}\text{0}\text{.024198}\text{mol}\text{0}\text{.024198}\text{mol }0-\\ \text{Change}\left(\text{M}\right)-\text{0}\text{.024198}\text{mol}-\text{0}\text{.024198}\text{mol}+\text{0}\text{.024198}\text{mol}-\\ --------------------------------------------\\ \text{Final }00+\text{0}\text{.024198}\text{mol}\text{-}\end{array}$

Determination the liters of solution present at the equivalent point,

Volume

   $\begin{array}{l}=\left[\left(21.8+193.584\right)mL\right]\left({10}^{-3}L/1mL\right)\\ \\ =0.0215384L.\end{array}$

Concentration of $HC{O}_3^{-}$ at equivalent point,

Molarity,

  $\begin{array}{l}=\frac{\left(0.022249molHC{O}_3^{-}\right)}{\left(0.055548L\right)}\\ \\ =0.0404875M.\end{array}$

Concentration of $C{H}_{3}N{H}_3^{+}$ at equivalent point,

Molarity,

  $\begin{array}{l}=\frac{\left(0.024198molC{H}_{3}N{H}_3^{+}\right)}{\left(0.215384L\right)}\\ \\ =0.1123482M.\end{array}$

Next calculate $K_a$for $C{H}_{3}N{H}_3^{+}$

The equilibrium value for $K_b\left(C{H}_{3}N{H}_3^{+}\right)=4.4\times {10}^{-4}$

  $K_b=K_W/K_a=\frac{\left(1.0\times {10}^{-14}\right)}{\left(4.5\times {10}^{-7}\right)}=2.222\times {10}^{-8}$ 

The equilibrium value for,

  $K_b=K_W/K_a=\frac{\left(1.0\times {10}^{-14}\right)}{\left(4.4\times {10}^{-4}\right)}=2.2727\times {10}^{-11}$

Following the reaction table for the equilibrium reaction of $C{H}_{3}N{H}_3^{+}$

$\begin{array}{l}C{H}_{3}N{H}_3^{+}\text{(aq) }\text{+}{H}_2\text{O(aq)}\underset{}{\rightleftharpoons }C{H}_{3}N{H}_2\text{(aq) +}{H}_3{\text{O}}^{+}\text{(aq)}\\ \text{Initial }\text{(M)}\text{0}\text{.1123482}\text{mol}-00\\ \text{Change}\left(\text{M}\right)-\text{x}-00\\ --------------------------------------------\\ \text{Final }\text{0}\text{.1123482-x}xxx\end{array}$

Determination the hydrogen ion concentration from the $K_a$ and then determine the $pH$ from the $pOH$.

Calculation for first equilibrium point:

  $\begin{array}{l}{K}_a=2.2727\times {10}^{-11}=\frac{\left[H_2{O}^{+}\right]\left[C{H}_{3}N{H}_2\right]}{\left[C{H}_{3}N{H}_3^{+}\right]}=\frac{\left[x\right]\left[x\right]}{\left[0.1123482-x\right]}=\frac{x^2}{\left[0.1123482\right]}\\ \\ x=\left[H_3{O}^{+}\right]=1.5979\times {10}^{-6}M\end{array}$

Calculation for $pH$

  $\begin{array}{c}pH=-\log \left[H{3}_{}{O}^{+}\right]\\ \\ =-\log \left(1.5979\times {10}^{-6}\right)\\ \\ =5.7964\\ \\ pH=5.80.\end{array}$