Chapter 17, Problem 17.21QP

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ at the equivalence point of titration $\text{HCl}\text{Vs}{\text{NH}}_{\text{3}}$ has to be calculated.

Concept Information:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution.
•  $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

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

Answer to Problem 17.21QP

The $\text{pH}$ at the equivalence point of titration $\text{HCl}\text{Vs}{\text{NH}}_{\text{3}}$ is $5.28$

Explanation of Solution

Concentration of $\text{HCl}$ is $\text{0}\text{.10}\text{M}$

Concentration of ${\text{NH}}_{\text{3}}$ is $\text{0}\text{.10}\text{M}$

Find the concentration of sodium formate

$\begin{array}{l}{\text{HCl}}_{\text{(aq)}}{\text{ + NH}}_{\text{3}}{}_{\text{(aq)}}\to {\text{ NH}}_{\text{4}}{\text{Cl}}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.010 }\text{0}\text{0}\\ \text{Change in concentration (M):}\text{-0}\text{.10}\text{-0}\text{.10}\text{+0}\text{.10}\\ \text{Final}\text{concentration (M): }\text{0}\text{0}\text{0}\text{.10}\\ \\ \text{Both the concentrations of}\text{are same and to reach equivalence point, equal amount of }\\ \text{solution should be needed}\text{. Therefore, the volume of solution is twofolded}\text{.}\\ \\ {\text{The concentration of NH}}_{\text{4}}\text{Cl is}\\ \text{=}\frac{\text{0}\text{.10 mol}}{\text{2}\text{L}}\text{=}\text{0}\text{.050M}\end{array}$

The concentration of ammonium chloride can be calculated from the final concentration of ammonium chloride in equilibrium table.  The solution gets doubled and the strength of ammonium chloride is divided by two.

Calculate the $\text{pH}$ of the titration

$\begin{array}{l}{\text{NH}}_{\text{4}}{{}^{+}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{ NH}}_{\text{3}}{}_{\text{(aq)}}{\text{ + H}}_{\text{3}}{\text{O}}^{+}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.050 }\text{0}\text{0}\\ \text{Change in concentration (M):}\text{-x}+\text{x}\text{+x}\\ \text{Equilibrium}\text{concentration (M): }\text{0}\text{.050-x}\text{x}\text{x}\\ \\ {\text{K}}_{\text{b}}{\text{ value for NH}}_{\text{3}}\text{ is 1}\text{.8 ×}{\text{10}}^{\text{-5}}\\ {\text{K}}_a\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_b}\text{=}\frac{\text{1}\text{.0 ×}{\text{10}}^{\text{-14}}}{\text{1}\text{.8 ×}{\text{10}}^{\text{-5}}}\text{=}\text{5}\text{.6×}{\text{10}}^{\text{-10}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[NH}}_{\text{3}}{\text{][H}}_{\text{3}}{\text{O}}^{+}\text{]}}{{\text{[NH}}_{\text{4}}{}^{+}\text{]}}\\ \text{5}\text{.56}\text{×}{\text{10}}^{\text{-10}}\text{=}\frac{{\text{(x}}^{\text{2}}\text{)}}{\text{(0}\text{.050-x)}}\\ {\text{x}}^{\text{2}}=(\text{5}\text{.56}\text{×}{\text{10}}^{\text{-10}})\times \text{(0}\text{.050)}\\ {\text{x}}^{\text{2}}=2.78\times {10}^{-11}\\ \text{Taking}\text{squre}\text{root}\text{both}\text{side}\\ \text{x}=5.27\times {10}^{\text{-6}}\end{array}$

$\begin{array}{l}\text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ {\text{x = [H}}_{\text{3}}{\text{O}}^{+}\text{] = 5}\text{.27 ×}{\text{10}}^{\text{-6}}\text{M}\\ \text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{+}\text{]}\\ \text{=}\text{-log(5}\text{.27 ×}{\text{10}}^{\text{-6}}\text{)}\\ \text{pH}\text{=}\text{5}\text{.28}\end{array}$

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ can be calculated from the final concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in equilibrium table using base dissociation constant.  The $\text{pH}$ is determined by taking negative logarithm of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ concentration.

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ at the equivalence point of titration ${\text{CH}}_{\text{3}}\text{COOH}\text{Vs}\text{NaOH}$ has to be calculated.

Concept Information:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution.
•  $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

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

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 17.21QP

The $\text{pH}$ at the equivalence point of titration ${\text{CH}}_{\text{3}}\text{COOH}\text{Vs}\text{NaOH}$ is 8.22

Explanation of Solution

Concentration of ${\text{CH}}_{\text{3}}\text{COOH}$ is $\text{0}\text{.10}\text{M}$

Concentration of $\text{NaOH}$ is $\text{0}\text{.10}\text{M}$

Find the concentration of sodium formate

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}{\text{ + NaOH}}_{\text{(aq)}}\to {\text{ CH}}_{\text{3}}{\text{COO}}^{\text{-}}{\text{Na}}^{\text{+}}{}_{\text{(aq)}}+{\text{H}}_{\text{2}}\text{O}\\ \\ \text{Initial concentration(M): }\text{0}\text{.010 }\text{0}\text{0}\\ \text{Change in concentration (M):}\text{-0}\text{.10}\text{-0}\text{.10}\text{+0}\text{.10}\\ \text{Final}\text{concentration (M): }\text{0}\text{0}\text{0}\text{.10}\\ \\ \text{Both the concentrations of}\text{are same and to reach equivalence point, equal amount of }\\ \text{solution should be needed}\text{. Therefore, the volume of solution is twofolded}\text{.}\\ \\ {\text{The concentration of CH}}_{\text{3}}{\text{COO}}^{\text{-}}{\text{Na}}^{\text{+}}\text{is}\\ \text{=}\frac{\text{0}\text{.10 mol}}{\text{2}\text{L}}\text{=}\text{0}\text{.050M}\end{array}$

The concentration of sodium acetate can be calculated from the final concentration of sodium acetate in equilibrium table.  The solution gets doubled and the strength of sodium acetate is divided by two.

Calculate the $\text{pH}$ of the titration

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{\text{Na}}^{\text{+}}{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{ CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}{\text{ + NaOH}}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.050 }\text{0}\text{0}\\ \text{Change in concentration (M):}\text{-x}+\text{x}\text{+x}\\ \text{Equilibrium}\text{concentration (M): }\text{0}\text{.050-x}\text{x}\text{x}\\ \\ {\text{K}}_a\text{ value for NaOH is 13}\text{.8}\\ {\text{K}}_b\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_a}\text{=}\frac{\text{1}\text{.0 ×}{\text{10}}^{\text{-14}}}{\text{13}\text{.8}}\text{=}7.\text{24×}{\text{10}}^{\text{-16}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[CH}}_{\text{3}}\text{COOH][NaOH]}}{{\text{[CH}}_{\text{3}}\text{COONa]}}\\ 7.\text{24×}{\text{10}}^{\text{-16}}\text{=}\frac{{\text{(x}}^{\text{2}}\text{)}}{\text{(0}\text{.050-x)}}\\ {\text{x}}^{\text{2}}=(7.\text{24×}{\text{10}}^{\text{-16}})\times \text{(0}\text{.050)}\\ {\text{x}}^{\text{2}}=3.62\times {10}^{-17}\\ \text{Taking}\text{squre}\text{root}\text{both}\text{side}\\ \text{x}=6.01\times {10}^{\text{-9}}\end{array}$

$\begin{array}{l}\text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ \text{x = [NaOH] = }6.01\times {10}^{\text{-9}}\text{M}\\ \text{pH}\text{=}\text{-log[NaOH]}\\ \text{=}\text{-log(}6.01\times {10}^{\text{-9}}\text{)}\\ \text{pH}\text{=}8.22\end{array}$

Therefore given solution pH  value is 8.22