Chapter 19, Problem 19.61P

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$ of $\text{2}\text{.65}\text{mL}\text{of}\text{0}\text{.0750}\text{M}\text{pyridine}$  at the equivalence point and the volume of $0.447MHN{O}_3$ 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

Relationship between ${\text{K}}_{\text{a}}$and $K_b$

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

Answer to Problem 19.61P

The pH of a solution made by mixing $\text{2}\text{.65}\text{mL}\text{of}\text{0}\text{.0750}\text{M}\text{pyridine}$ with $0.447MHN{O}_3$ is $3.21.$

Explanation of Solution

Given,

The balance equation is given below,

  ${\text{HNO}}_3\text{(aq) +}{C}_5{\text{H}}_5\text{N(aq)}\underset{}{\to }{C}_5{\text{H}}_5{\text{NH}}^{+}\text{(aq) +}N{O}_3^{+}\text{(aq)}$

The nitrate 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 ${\text{HNO}}_3$,

Volume $mL$ of ${\text{HNO}}_3$,

$\begin{array}{c}{\text{HNO}}_3=\left(\frac{0.0750mol{C}_5{H}_{5}N}{L}\right)\left(2.65mL\right)\left(\frac{1mol{\text{HNO}}_3}{1mol{C}_5{H}_{5}N}\right)\left(\frac{L}{0.447mol{\text{HNO}}_3}\right)\left(\frac{1mL}{{10}^{-3}L}\right)\\ \\ =444.63087\\ \\ =445mLofHN{O}_3.\end{array}$

Therefore, the required volume of ${\text{HNO}}_{\text{3}}=445mL$

Determination the moles of $C_5{\text{H}}_5{\text{NH}}^{+}$produced,

$\begin{array}{c}Molesof{C}_5{\text{H}}_5{\text{NH}}^{+}=\left(\frac{0.0750mol{C}_5{\text{H}}_5\text{N}}{1mL}\right)\left(2.65mL\right)\left(\frac{1mol{C}_5{\text{H}}_5{\text{NH}}^{+}}{1mol{C}_5{\text{H}}_5\text{N}}\right)\\ \\ =0.19875molof{C}_5{\text{H}}_5{\text{NH}}^{+}\end{array}$

Determination the liters of solution present at the equivalent point,

Volume

   $\begin{array}{l}=\left[2.65L+\left(444.63087mL\right)\right]\left({10}^{-3}L/1mL\right)\\ \\ =3.09463L\end{array}$

Concentration of $C_5{\text{H}}_5{\text{NH}}^{+}$ at equivalent point,

Molarity,

  $\begin{array}{l}=\left(0.19875mol{C}_5{\text{H}}_5{\text{NH}}^{+}\right)/\left(3.09463L\right)\\ \\ =0.064224M\end{array}$

Calculate$K_a$for $C_5{\text{H}}_5{\text{NH}}^{+}$

The equilibrium value for $K_b\left(C_5{\text{H}}_5\text{N}\right)=1.7\times {10}^{-9}$

  $K_b=K_W/K_a=\frac{\left(1.0\times {10}^{-14}\right)}{\left(1.7\times {10}^{-9}\right)}=5.88253\times {10}^{-6}$ 

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

  $\begin{array}{l}{K}_a=5.88253\times {10}^{-6}=\frac{\left[H_3{O}^{+}\right]\left[C_5{H}_{5}N\right]}{\left[C_5{H}_{5}N{H}^{+}\right]}=\frac{\left[x\right]\left[x\right]}{\left[0.064224-x\right]}=\frac{x^2}{\left[0.064224\right]}\\ \\ \left[H_3{O}^{+}\right]=x=6.1464\times {10}^{-4}M\end{array}$

Calculation for $pH$

  $\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ \\ =-\log \left(6.1464\times {10}^{-4}\right)\\ \\ pH=3.211379=3.21\end{array}$

Therefore, the volume of $0.447MHN{O}_3$ $pH$ reachable point is $\underset{\_}{3.21}.$

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $0.188\text{L}\text{of}\text{0}\text{.250}\text{M}$ ethylenediamine at the equivalence point and the volume of $0.447MHN{O}_3$ 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

Relationship between ${\text{K}}_{\text{a}}$and $K_b$

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

Answer to Problem 19.61P

The pH of a solution made by mixing $0.188\text{L}\text{of}\text{0}\text{.250}\text{M}$ ethylenediamine with $0.447MHN{O}_3$ is $5.36.$

The second equivalent pH of a given buffer solution after the addition of $0.188\text{L}\text{of}\text{0}\text{.250}\text{M}$ ethylenediamine and $0.447MHN{O}_3$ is $3.89.$

Explanation of Solution

The balance equation is given below,

  $\begin{array}{l}{\text{HNO}}_3\text{(aq) +}{H}_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_2\text{(aq)}\underset{}{\to }{H}_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{+}\text{(aq) +}N{O}_3^{-}\text{(aq)}\\ \\ {\text{HNO}}_3\text{(aq) +}{H}_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{+}\text{(aq)}\underset{}{\to }{H}_3{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{2+}\text{(aq) +}N{O}_3^{-}\text{(aq)}\end{array}$

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

Calculate the volume of ${\text{HNO}}_3$ needed,

Volume of ${\text{HNO}}_3$ in $mL$,

$\begin{array}{c}=\left(\frac{0.250mol{H}_{2}NC{H}_{2}C{H}_{3}N{H}_2}{L}\right)\left(0.188L\right)\left(\frac{1mol{\text{HNO}}_3}{1mol{H}_{2}NC{H}_{2}C{H}_{3}N{H}_2}\right)\left(\frac{L}{0.447mol{\text{HNO}}_3}\right)\left(\frac{1mL}{{10}^{-3}L}\right)\\ \\ =105.1454\\ \\ =105mLofHCl.\end{array}$

The solution requires an equal volume to reach second equivalence point $105mLofHCl.$

Determination the moles of $H_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{+}$present,

$\begin{array}{c}Molesof{H}_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{+}=\left(\frac{0.250mol{H}_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_2}{L}\right)\left(0.188mL\right)\left(\frac{1mol{H}_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{+}}{1mol{H}_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_2}\right)\\ \\ =0.0470molof{H}_{2}NC{H}_{2}C{H}_{2}N{H}_3^{+}\end{array}$

An equal number of moles of $H_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{+}$ will be present at the second equivalent point.

Determination the liters of solution at the first equilibrium point:

Volume

   $\begin{array}{l}=\left[0.188L+\left(105.1454mL\right)\right]\left({10}^{-3}L/1mL\right)\\ \\ =0.293145L\end{array}$

Determination the liters of solution at the second equilibrium point:

Volume

   $\begin{array}{l}=\left[0.188L+2\left(105.1454mL\right)\right]\left({10}^{-3}L/1mL\right)\\ \\ =0.39829L\end{array}$

Concentration of $H_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{+}$ at equivalent point,

Molarity,

  $\begin{array}{l}=\frac{\left(0.0470mol{H}_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{+}\right)}{\left(0.293145L\right)}\\ \\ =0.16033M\end{array}$

Concentration of $H_3{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{2+}$ at equivalent point,

Molarity,

  $\begin{array}{l}=\frac{\left(0.0470mol{H}_3{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{2+}\right)}{\left(0.39829L\right)}\\ \\ =0.11800M\end{array}$

Calculate $K_a$for $H_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{+}$

The equilibrium value for $K_b\left(H_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_2\right)=8.5\times {10}^{-5}$

  $K_b=K_W/K_a=\frac{\left(1.0\times {10}^{-14}\right)}{\left(8.5\times {10}^{-5}\right)}=1.17674\times {10}^{-10}$ 

Next calculate $K_a$for $H_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{2+}$

The equilibrium value for $K_b\left(H_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{+}\right)=7.1\times {10}^{-8}$

  $K_b=K_W/K_a=\frac{\left(1.0\times {10}^{-14}\right)}{\left(7.1\times {10}^{-8}\right)}=1.40845\times {10}^{-7}$ 

Determination the hydrogen ion concentration from the $K_a$ and then determine the $pH$ for the first equilibrium point.

Calculation for first equilibrium point:

  $\begin{array}{l}{K}_a=1.17647\times {10}^{-10}=\frac{\left[H_3{O}^{+}\right]\left[H_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_2\right]}{\left[H_3{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{+}\right]}=\frac{\left[x\right]\left[x\right]}{\left[0.16033-x\right]}=\frac{x^2}{\left[0.16033\right]}\\ \\ x=\left[H_3{O}^{+}\right]=4.3430780\times {10}^{-6}M\end{array}$

Calculation for $pH$

  $\begin{array}{c}pH=-\log \left[H{3}_{}{O}^{+}\right]\\ \\ =-\log \left(4.3430780\times {10}^{-6}\right)\\ \\ =5.36220\\ \\ pH=5.36\end{array}$

Determination the hydrogen ion concentration from $K_{a^{\text{'}}}$ and then determine the $pH$ for the second equilibrium point.

Calculation for second equilibrium point:

  $\begin{array}{l}{K}_a=1.40845\times {10}^{-7}=\frac{\left[H_3{O}^{+}\right]\left[H_2{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{+}\right]}{\left[H_3{\text{NCH}}_2{\text{CH}}_2{\text{NH}}_3^{2+}\right]}=\frac{\left[x\right]\left[x\right]}{\left[0.11800-x\right]}=\frac{x^2}{\left[0.11800\right]}\\ \\ x=\left[H_3{O}^{+}\right]=1.2891745\times {10}^{-4}M\end{array}$

Calculation for $pH$

  $\begin{array}{c}pH=-\log \left[H{3}_{}{O}^{+}\right]\\ \\ =-\log \left(1.2891745\times {10}^{-4}M\right)\\ \\ =3.889688\\ \\ pH=3.89\end{array}$