Chapter 19, Problem 19.53P

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$$\text{42}\text{.2}\text{mL}$ of $\text{0}\text{.0520}\text{M}{\text{CH}}_{\text{3}}\text{COOH}$ at the equivalence point and the volume of $0.0372MNaOH$ 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.

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

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

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 19.53P

The pH of a given buffer solution after the addition of $\text{42}\text{.2 mL}$$\text{0}\text{.0520}\text{M}{\text{CH}}_{\text{3}}\text{COOH}$ and $0.0372MNaOH$ is $8.540.$

Explanation of Solution

Given,

The balance chemical equation is given below,

  $\text{NaOH(aq) +}C{H}_3\text{COOH(aq)}\underset{}{\to }N{a}^{+}\text{(aq) +}C{H}_{3}CO{O}^{-}\text{(aq)}\text{+}{\text{H}}_{2}O\left(l\right)$

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

Calculate the volume of $NaOH$ needed.

Volume $mL$ of $NaOH$,

$\begin{array}{c}=\left(\frac{0.0520molC{H}_{3}COOH}{L}\right)\left(\frac{{10}^{-3}L}{1mL}\right)\left(42.2mL\right)\left(\frac{1molNaOH}{1molC{H}_{3}COOH}\right)\left(\frac{L}{0.0327molNaOH}\right)\left(\frac{1mL}{{10}^{-3}L}\right)\\ \\ =58.989247\\ \\ =59.0mLofNaOH.\end{array}$

Hence, the required volume of $NaOH=59.0mL$

Determination the moles of initially$C{H}_{3}COOH$present,

$\begin{array}{c}MolesofC{H}_{3}COOH=\left(\frac{0.0250molC{H}_{3}COOH}{L}\right)\left(\frac{{10}^{-3}L}{1mL}\right)\left(42.2mL\right)\\ \\ =0.0021944molN{H}_3\end{array}$

At the equivalent point, $\text{0}\text{.0021944}\text{mol}\text{NaOH}$ will be added so, the moles acid = moles base.

The $NaOH$ will react with an equal amount of the acid $0molC{H}_{3}COOH$ will remain and $\text{0}\text{.0021944}\text{moles}ofC{H}_{3}CO{O}^{-}$ will be formed.

ICE Table of equilibrium,

$\begin{array}{l}C{H}_3\text{COOH(aq) }\text{+}NaOH\text{(aq)}\underset{}{\rightleftharpoons }{H}_2\text{O(l) +}C{H}_{3}CO{O}^{-}\text{(aq)}\text{+}N{a}^{+}\left(aq\right)\\ \text{Initial }\text{(M)}\text{0}\text{.0021944}\text{mol}\text{0}\text{.0021944}\text{mol }-\text{0}-\\ \text{Change}\left(\text{M}\right)-\text{0}\text{.0021944}\text{mol}-\text{0}\text{.0021944}\text{mol}-+\text{0}\text{.0021944}\text{mol}-\\ -------------------------------------------------\\ \text{Final }00-+\text{0}\text{.0021944}\text{mol}-\end{array}$

Determination the liters of solution present at the equivalent point,

Volume

   $\begin{array}{l}=\left[\left(42.2+58.989247\right)mL\right]\left({10}^{-3}L/1mL\right)\\ \\ =0.101189247L.\end{array}$

Concentration of $C{H}_{3}CO{O}^{-}$ at equivalent point,

Molarity,

  $\begin{array}{l}=\left(0.0021944molC{H}_{3}CO{O}^{-}\right)/\left(0.10118947L\right)\\ \\ =0.0216861M.\end{array}$

Next calculate $K_b$for $C{H}_{3}CO{O}^{-}$

The equilibrium value for $K_{a}C{H}_{3}COOH=1.8\times {10}^{-5}$

So,

  $K_b=K_W/K_a=\frac{\left(1.0\times {10}^{-14}\right)}{\left(1.8\times {10}^{-5}\right)}=5.556\times {10}^{-10}$ 

Using a reaction table for the equilibrium reaction of $C{H}_{3}CO{O}^{-}$

  $\begin{array}{l}C{H}_{3}CO{O}^{-}\text{+}{H}_{2}O\underset{}{\rightleftharpoons }C{H}_{3}COOH\text{ +}O{H}^{-}\\ \text{Initial }\text{(M)}:0.0216861M-0\text{0}\\ \text{Change}\left(\text{M}\right):-x-+\text{x}+\text{x}\\ ----------------------------------\\ \text{Equilibrium: }0.0216861-x-xx\end{array}$

Determination the hydroxide ion $O{H}^{-}$ concentration from the $K_b$ and then determine the $pH$ from $pOH$.

  $\begin{array}{l}{K}_b=5.556\times {10}^{-10}=\frac{\left[C{H}_{3}COOH\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}CO{O}^{-}\right]}=\frac{\left[x\right]\left[x\right]}{\left[0.0216861-x\right]}=\frac{x^2}{\left[0.0216861\right]}\\ \left[O{H}^{-}\right]=x=3.471138\times {10}^{-6}M\end{array}$

Calculation for $pH$ and $pOH$

  $\begin{array}{c}pH=-\log \left[HO{H}^{-}\right]\\ \\ =-\log \left(3.471138\times {10}^{-6}\right)\\ \\ =5.459528\\ \\ pH=14.00-pOH\\ =14.00-5.459528\\ =8.54047\\ \\ pH=8.540.\end{array}$

(b)

Interpretation Introduction

Interpretation:

$\text{pH}$ of $28.9\text{mL}$ and $0.0850M{H}_{2}S{O}_3$ at the equivalence point and the volume of $0.0372MNaOH$ 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 19.53P

The pH of a given buffer solution after the addition of $\text{28}\text{.9 mL}$$0.0850M{H}_{2}S{O}_3$ and $0.0372MNaOH$ is $4.387.$

The second equivalent pH of a given buffer solution after the addition of $\text{28}\text{.9 mL}$$0.0850M{H}_{2}S{O}_3$ and $0.0372MNaOH$ is $9.685.$

Explanation of Solution

The titration of a diprotic acid such as $H_{2}S{O}_3$ two $O{H}^{-}$ ions is required to react with the two $H^{+}$ ions of each acid molecule. Because the large difference in $K_a$ values each mole of $H^{+}$ is titrated separately.

So, $H_{2}S{O}_3$ molecule lose one $H^{+}$ before any $HS{O}_3^{-}$ ions,

The balance chemical equation is given below,

  $NaOH\left(aq\right)+H_{2}S{O}_3\left(aq\right)\underset{}{\to }N{a}^{+}\text{(aq)}+HS{O}_3^{-}\left(aq\right)+H_{2}O\left(l\right)$

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

The volume of $NaOH$ needed to reach the first equivalent point, 

$Volume\left(mL\right)ofNaOH=$

$\begin{array}{c}=\left(\frac{0.0850mol{H}_{2}S{O}_3}{L}\right)\left(\frac{{10}^{-3}L}{1mL}\right)\left(28.9mL\right)\left(\frac{1moNaOH}{1mol{H}_{2}S{O}_3}\right)\left(\frac{L}{0.0372molNaOH}\right)\left(\frac{1mL}{{10}^{-3}L}\right)\\ \\ =66.034946\\ \\ =66.0mLofNaOH\end{array}$

The solution requires an equal volume to reach second equivalence point $66.0mLofNaOH.$

Determination the moles of $HS{O}_3^{-}$present at the fist equivalent point,

$\begin{array}{c}MolesofHS{O}_3^{-}=\left(\frac{0.0850mol{H}_{2}S{O}_3}{L}\right)\left(\frac{{10}^{-3}L}{1mL}\right)\left(28.9mL\right)\left(\frac{1molHS{O}_3^{2-}}{1mol{H}_{2}S{O}_3}\right)\\ \\ =0.0024565molofHS{O}_3^{-}\end{array}$

Determination the liters of solution present at the first equivalent point,

Volume

   $\begin{array}{l}=\left[\left(28.9+66.034946\right)mL\right]\left({10}^{-3}L/1mL\right)\\ \\ =0.094934946L.\end{array}$

Concentration of $HS{O}_3^{-}$ at the first equivalent point,

Molarity,

  $\begin{array}{l}=\frac{\left(0.0024565molHS{O}_3^{-}\right)}{\left(0.094934946L\right)}\\ \\ =0.0258756M.\end{array}$

The $HS{O}_3^{-}$ is a amphoteric substance.

To calculate $pH$ at the first equivalent point, determine whether $HS{O}_3^{-}$ a stronger acid or stronger base is.

Calculate $K_a$for $HS{O}_3^{-}$

$K_a$ for $HS{O}_3^{-}$ is $6.5\times {10}^{-8}$ 

The equilibrium value for $K_{a_1}\left(H_{2}S{O}_3\right)=1.4\times {10}^{-2}$

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

The equilibrium value for,

  $K_b=K_W/K_{a_1}=\frac{\left(1.0\times {10}^{-14}\right)}{\left(1.4\times {10}^{-2}\right)}=7.142857\times {10}^{-13}$

Calculate for first equivalent point:

Following the reaction table for the equilibrium reaction of $HS{O}_3^{-}$

$\begin{array}{l}HS{O}_3^{-}\text{(aq) }\text{+}{H}_2\text{O(aq)}\underset{}{\rightleftharpoons }{H}_3{\text{O}}^{+}\text{(aq) +}S{O}_3^{2-}\text{(aq)}\\ \text{Initial }\text{(M)}:\text{0}\text{.0258756}\text{mol}-00\\ \text{Change}\left(\text{M}\right):-\text{x}-+x+x\\ ---------------------------------------\\ \text{Final }:\text{0}\text{.0258756}\text{-x}-xx\end{array}$

Determination the hydrogen ion concentration from $K_a\text{'}$ and then determine the $pH$ from the hydronium ion concentration.

  $K_a=6.5\times {10}^{-8}=\frac{\left[H_3{O}^{+}\right]\left[S{O}_3^{2-}\right]}{\left[HS{O}_3^{2-}\right]}=\frac{\left[x\right]\left[x\right]}{\left[0.0258756-x\right]}=\frac{x^2}{\left[0.0258756\right]}$

Here, x is very small compared with $0.0258756M$

$x=\left[H_3{O}^{+}\right]=4.101114\times {10}^{-5}M$

Calculation for $pH$ at the first equivalent point,

  $\begin{array}{c}pH=-\log \left[H{3}_{}{O}^{+}\right]\\ \\ =-\log \left(4.10111\times {10}^{-5}\right)\\ \\ =4.3870982\\ \\ pH=4.387\end{array}$

Calculation for second equivalent point

The balanced equation for neutralization of $HS{O}_3^{-}$ at the second equivalent point is.

   $NaOH\left(aq\right)+HS{O}_3^{-}\left(aq\right)\underset{}{\to }N{a}^{+}\text{(aq)}+S{O}_3^{2-}\left(aq\right)+H_{2}O\left(l\right)$

A total of $66.034946mL$ were required to reach the first equivalence point. It will require an equal volume of $NaOH$ to reach the second equivalent point.   

Determination the liters of solution present at the second equivalent point,

Volume

   $\begin{array}{l}=\left[\left(28.9+66.034946+66.034946\right)mL\right]\left({10}^{-3}L/1mL\right)\\ =0.1609698992L\end{array}$

$0.0024565molHS{O}_3^{-}$ were present at the first equivalent point. The equal number of moles $S{O}_3^{2-}$ will be present at the second equivalent point.

Determination the concentration of $S{O}_3^{2-}$ at the second equivalent point,

Molarity,

  $\begin{array}{l}=\frac{\left(0.0024565molS{O}_3^{2-}\right)}{\left(0.16096989L\right)}\\ =0.0152606M\end{array}$

$S{O}_3^{2-}$ ions does not have hydrogen ion to donate, so it acts as accept a hydrogen. Because it acts as a base, so we calculate for $K_b$

Calculate $K_b$for $S{O}_3^{2-}$

$K_a$ for $HS{O}_3^{-}$ is $6.5\times {10}^{-8}$ 

  $K_b=K_W/K_a=\frac{\left(1.0\times {10}^{-14}\right)}{\left(6.5\times {10}^{-8}\right)}=1.5384\times {10}^{-7}$ 

Calculate for second equivalent point:

Following the reaction table for the equilibrium reaction of $S{O}_3^{2-}$

$\begin{array}{l}S{O}_3^{2-}\text{(aq) }\text{+}{H}_2\text{O(aq)}\underset{}{\rightleftharpoons }HS{O}_3^{-}\text{(aq) +}O{H}^{-}\text{(aq)}\\ \text{Initial }\text{(M)}:\text{0}\text{.0152606}\text{mol}-00\\ \text{Change}\left(\text{M}\right):-\text{x}-+x+x\\ ---------------------------------------\\ \text{Final }:\text{0}\text{.0152606}M\text{-x}-xx\end{array}$

Determination the hydrogen ion concentration from the $K_b\text{'}$ and then determine the $pH$ from $pOH$.

  $K_b=1.53846\times {10}^{-7}=\frac{\left[HS{O}_3^{-}\right]\left[O{H}^{-}\right]}{\left[S{O}_3^{-}\right]}=\frac{\left[x\right]\left[x\right]}{\left[0.0152606-x\right]}=\frac{x^2}{0.0152606}$

Here, x is very small compared with $0.0152606M$

  $x=\left[H_3{O}^{+}\right]=4.84539\times {10}^{-5}M$

Calculation for $pHandpOH$ at the second equivalent point,

  $\begin{array}{c}pOH=-\log \left[O{H}^{-}\right]\\ =-\log \left(4.84539\times {10}^{-5}\right)\\ \\ pOH=4.31467\\ \\ pH=14.00-pOH\\ \\ pH=14.00-4.31467\\ \\ =9.68533=9.685.\end{array}$