Chapter 19, Problem 19.51P

(a)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M butanoic acid, ($K_a\text{=}1.54\times {10}^{-5}$ ) with $\text{0}\text{.1000 M NaOH}$ solution after the addition of $\text{0 mL}$ titrant 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 19.51P

The pH of a given buffer solution after the addition $\text{0}\text{.00 mL}$ titrant and addition of $20.00mL$ and 0.1000 M butanoic acid with $\text{0}\text{.1000 M NaOH}$ is $2.91.$

Explanation of Solution

Given,

At $0mL$ of base added, the concentration of ${\left[H_{3}O\right]}^{+}$ is dependent on the dissociation of Butanoic acid.

ICE Table of equilibrium,

$\begin{array}{l}\text{H-But(aq) }\text{+}{H}_2\text{O(aq)}\underset{}{\rightleftharpoons }{H}_3{\text{O}}^{+}\text{(l) +}Bu{t}^{-}\text{(aq)}\\ \text{Initial }\text{(M)}\text{0}\text{.100}M-0\text{0}\\ \text{Change}\left(\text{M}\right)-\text{x}-+x+x\\ -----------------------------------------\\ \text{Equilibrium }\text{0}\text{.100}-x-xx\end{array}$

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

  $\begin{array}{l}{K}_a=1.54\times {10}^{-5}=\frac{\left[H_3{O}^{+}\right]\left[Bu{t}^{-}\right]}{\left[H-But\right]}=\frac{\left[x\right]\left[x\right]}{\left[0.100-x\right]}=\frac{x^2}{\left[0.1000\right]}\\ \left[H_3{O}^{+}\right]=x=1.2409674\times {10}^{-3}M\end{array}$

Calculation for $pH$ and $pOH$

  $\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ =-\log \left(1.2409674\times {10}^{-3}\right)\\ pH=2.9062=2.91\end{array}$

(b)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M butanoic acid, ($K_a\text{=}1.54\times {10}^{-5}$ ) with $\text{0}\text{.1000 M NaOH}$ solution after the addition of $\text{10}\text{.00 mL}$ titrant 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.

Answer to Problem 19.51P

The pH of a given buffer solution after the $\text{10}\text{.00 mL}$ titrant and addition of $20.00mL$ and 0.1000 M butanoic acid with $\text{0}\text{.1000 M NaOH}$ is $4.81.$

Explanation of Solution

Given,

The initial number of moles of $H-But$,

$\begin{array}{l}=\left(M\right)\left(V\right)=\left(0.1000molH-But/L\right)\left({10}^{-3}L/1mL\right)\left(20.00mL\right)\\ \\ =2.00\times 10^{-3}molH-But\end{array}$

Identification the moles of $NaOH$ added

Moles added $NaOH$

$\begin{array}{l}=\left(0.1000molNaOH/L\right)\left({10}^{-3}L/1mL\right)\left(10.00mL\right)\\ \\ =1.000\times {10}^{-3}molNaOH\end{array}$

The $NaOH$ will react with an equal amount of the aid, and $1.00\times {10}^{-3}mol\text{H-But}$ will remain. An equal number of moles of $Bu{t}^{-}$ will form.

ICE Table of equilibrium,

$\begin{array}{l}\text{H-But(aq) }\text{+}NaOH\text{(aq)}\underset{}{\rightleftharpoons }{H}_2\text{O(l) +}Bu{t}^{-}\text{(aq)}\text{+}{\text{Na}}^{+}\left(aq\right)\\ \text{Initial }\text{(M)}2.000\times {10}^{-3}mol1.000\times {10}^{-3}mol-0-\\ \text{Change}\left(\text{M}\right)-1.000\times {10}^{-3}mol-1.000\times {10}^{-3}mol-+1.000\times {10}^{-3}mol-\\ ---------------------------------------------------\\ \text{Equilibrium }1.000\times {10}^{-3}mol0-1.000\times {10}^{-3}mol-\end{array}$

Determination the liters of solution present at the equivalent point,

Volume

   $\begin{array}{l}=\left[\left(20.00+10.00\right)mL\right]\left({10}^{-3}L/1mL\right)\\ \\ =0.03000L\end{array}$

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

Molarity of the excess $\text{H-But}$

  $\begin{array}{l}=\left(1.000\times {10}^{-3}mol\text{H-But}\right)/\left(0.03000L\right)\\ \\ =0.03333M\end{array}$

Molarity of the excess ${\text{But}}^{-}$formed is,

  $\begin{array}{l}=\left(1.000\times {10}^{-3}mol{\text{But}}^{-}\right)/\left(0.03000L\right)\\ \\ =0.03333M\end{array}$

Using a reaction table for the equilibrium reaction of $\text{H-But}$

  $\begin{array}{l}\text{H-But}\text{+}{H}_{2}O\underset{}{\rightleftharpoons }{H}_3{O}^{+}\text{ +}{\text{But}}^{-}\\ \text{Initial }\text{(M)}:0.03333M-00.03333M\\ \text{Change}\left(\text{M}\right):-x-+\text{x}+\text{x}\\ -------------------------------------\\ \text{Equilibrium: }0.03333M-x+x0.03333M+x\end{array}$

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

  $\begin{array}{l}{K}_a=1.54\times {10}^{-5}=\frac{\left[H_3{O}^{+}\right]\left[Bu{t}^{-}\right]}{\left[H-But\right]}=\frac{x\left(0.0333+x\right)}{\left(0.0333-x\right)}=\frac{x\left(0.03333\right)}{0.03333}\\ \\ \left[H_3{O}^{+}\right]=x=1.54\times {10}^{-5}M\end{array}$

Calculation for $pH$ and $pOH$

  $\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ \\ =-\log \left(1.54\times {10}^{-5}\right)\\ \\ pH=4.812479=4.81\end{array}$

(c)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M butanoic acid, ($K_a\text{=}1.54\times {10}^{-5}$ ) with $\text{0}\text{.1000 M NaOH}$ solution after the addition of $\text{15}\text{.00 mL}$ titrant 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 19.51P

The pH of a given buffer solution after the addition $\text{15}\text{.00 mL}$ titrant and addition of $20.00mL$ and 0.1000 M butanoic acid with $\text{0}\text{.1000 M NaOH}$ is $5.29.$

Explanation of Solution

Determination the moles of $NaOH$ added

Moles added $NaOH$

  $\begin{array}{l}=\left(0.1000molNaOH/L\right)\left({10}^{-3}L/1mL\right)\left(15.00mL\right)\\ =1.500\times {10}^{-3}molNaOH\end{array}$

The $NaOH$ will react with an equal amount of the aid, and $5.00\times {10}^{-4}mol\text{H-But}$ will remain. An equal number of moles of $Bu{t}^{-}$ will form.

ICE Table of equilibrium,

$\begin{array}{l}\text{H-But(aq) }\text{+}NaOH\text{(aq)}\underset{}{\rightleftharpoons }{H}_2\text{O(l) +}Bu{t}^{-}\text{(aq)}\text{+}{\text{Na}}^{+}\left(aq\right)\\ \text{Initial }\text{(M)}2.000\times {10}^{-3}mol1.500\times {10}^{-3}mol-0-\\ \text{Change}\left(\text{M}\right)-1.500\times {10}^{-3}mol-1.500\times {10}^{-3}mol-+1.500\times {10}^{-3}mol-\\ -------------------------------------------------\\ \text{Equilibrium }5.000\times {10}^{-4}mol0-1.500\times {10}^{-3}mol-\end{array}$

Determination the liters of solution present at the equivalent point,

Volume

   $\begin{array}{l}=\left[\left(20.00+15.00\right)mL\right]\left({10}^{-3}L/1mL\right)\\ \\ =0.03500L\end{array}$

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

Molarity of the excess $\text{H-But}$

  $\begin{array}{l}=\left(5.000\times {10}^{-4}mol\text{H-But}\right)/\left(0.03500L\right)\\ \\ =0.0142857M\end{array}$

Molarity of the excess ${\text{But}}^{-}$formed is,

  $\begin{array}{l}=\left(1.500\times {10}^{-3}mol{\text{But}}^{-}\right)/\left(0.03500L\right)\\ \\ =0.042857M\end{array}$

Using a reaction table for the equilibrium reaction of $\text{H-But}$

  $\begin{array}{l}\text{H-But}\text{+}{H}_{2}O\underset{}{\rightleftharpoons }{H}_3{O}^{+}\text{ +}{\text{But}}^{-}\\ \text{Initial }\text{(M)}:0.042857M-00.042857M\\ \text{Change}\left(\text{M}\right):-x-+\text{x}+\text{x}\\ -------------------------------------\\ \text{Equilibrium: }0.042857M-x+x0.042857M+x\end{array}$

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

  $\begin{array}{l}{K}_a=1.54\times {10}^{-5}=\frac{\left[H_3{O}^{+}\right]\left[Bu{t}^{-}\right]}{\left[H-But\right]}=\frac{x\left(0.042857+x\right)}{\left(0.042857-x\right)}=\frac{x\left(0.042857\right)}{0.042857}\\ \\ \left[H_3{O}^{+}\right]=x=5.1333\times {10}^{-6}M\end{array}$

Calculation for $pH$ and $pOH$

  $\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ \\ =-\log \left(5.1333\times {10}^{-6}\right)\\ \\ pH=5.2896=5.29\end{array}$

(d)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M butanoic acid, ($K_a\text{=}1.54\times {10}^{-5}$ ) with $\text{0}\text{.1000 M NaOH}$ solution after the addition of $\text{19}\text{.00 mL}$ titrant 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 19.51P

The pH of a given buffer solution after the addition $\text{19}\text{.00 mL}$ titrant and addition of $20.00mL$ and 0.1000 M butanoic acid with $\text{0}\text{.1000 M NaOH}$ is $6.09.$

Explanation of Solution

Determination the moles of $NaOH$ added

Moles added $NaOH$

  $\begin{array}{l}=\left(0.1000molNaOH/L\right)\left({10}^{-3}L/1mL\right)\left(19.00mL\right)\\ =1.900\times {10}^{-3}molNaOH\end{array}$

The $NaOH$ will react with an equal amount of the aid, and $1.00\times {10}^{-4}mol\text{H-But}$ will remain and $1.900\times {10}^{-3}$ moles of $Bu{t}^{-}$ will form.

ICE Table of equilibrium,

$\begin{array}{l}\text{H-But(aq) }\text{+}NaOH\text{(aq)}\underset{}{\rightleftharpoons }{H}_2\text{O(l) +}Bu{t}^{-}\text{(aq)}\text{+}{\text{Na}}^{+}\left(aq\right)\\ \text{Initial }\text{(M)}2.000\times {10}^{-3}mol1.900\times {10}^{-3}mol-0-\\ \text{Change}\left(\text{M}\right)1.900\times {10}^{-3}mol-1.900\times {10}^{-3}mol-+1.900\times {10}^{-3}mol-\\ -------------------------------------------------\\ \text{Equilibrium }1.000\times {10}^{-4}mol0-1.900\times {10}^{-3}mol-\end{array}$

Determination the liters of solution present at the equivalent point,

Volume

   $\begin{array}{l}=\left[\left(20.00+19.00\right)mL\right]\left({10}^{-3}L/1mL\right)\\ =0.03900L\end{array}$

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

Molarity of the excess $\text{H-But}$

  $\begin{array}{l}=\left(1.00\times {10}^{-4}mol\text{H-But}\right)/\left(0.03900L\right)\\ =0.0025641M\end{array}$

Molarity of the excess ${\text{But}}^{-}$formed is,

  $\begin{array}{l}=\left(1.900\times {10}^{-3}mol{\text{But}}^{-}\right)/\left(0.03900L\right)\\ =0.0487179M\end{array}$

Using a reaction table for the equilibrium reaction of $\text{H-But}$

  $\begin{array}{l}\text{H-But}\text{+}{H}_{2}O\underset{}{\rightleftharpoons }{H}_3{O}^{+}\text{ +}{\text{But}}^{-}\\ \text{Initial }\text{(M)}:0.0025641M-00.0487179M\\ \text{Change}\left(\text{M}\right):-x-+\text{x}+\text{x}\\ -------------------------------------\\ \text{Equilibrium: }0.0025641M-x+x0.0487179M+x\end{array}$

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

  $\begin{array}{l}{K}_a=1.54\times {10}^{-5}=\frac{\left[H_3{O}^{+}\right]\left[Bu{t}^{-}\right]}{\left[H-But\right]}=\frac{x\left(0.0487179+x\right)}{\left(0.0025641-x\right)}=\frac{x\left(0.0487179\right)}{0.0025641}\\ \\ \left[H_3{O}^{+}\right]=x=8.1052632\times {10}^{-7}M\end{array}$

Calculation for $pH$ and $pOH$

  $\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ \\ =-\log \left(8.1052632\times {10}^{-7}\right)\\ \\ pH=6.09123=6.09\end{array}$

(e)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M butanoic acid, ($K_a\text{=}1.54\times {10}^{-5}$ ) with $\text{0}\text{.1000 M NaOH}$ solution after the addition of $\text{19}\text{.95 mL}$ titrant 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

Answer to Problem 19.51P

The pH of a given buffer solution after the addition $\text{19}\text{.95 mL}$ titrant and addition of $20.00mL$ and 0.1000 M butanoic acid with $\text{0}\text{.1000 M NaOH}$ is $6.09.$

Explanation of Solution

Determination the moles of $NaOH$ added

Moles added $NaOH$

  $\begin{array}{l}=\left(0.1000molNaOH/L\right)\left({10}^{-3}L/1mL\right)\left(19.95mL\right)\\ =1.995\times {10}^{-3}molNaOH\end{array}$

The $NaOH$ will react with an equal amount of the aid, and $5.00\times {10}^{-4}mol\text{H-But}$ will remain and $1.995\times {10}^{-3}$ moles of $Bu{t}^{-}$ will form.

ICE Table of equilibrium,

$\begin{array}{l}\text{H-But(aq) }\text{+}NaOH\text{(aq)}\underset{}{\rightleftharpoons }{H}_2\text{O(l) +}Bu{t}^{-}\text{(aq)}\text{+}{\text{Na}}^{+}\left(aq\right)\\ \text{Initial }\text{(M)}2.000\times {10}^{-3}mol1.995\times {10}^{-3}mol-0-\\ \text{Change}\left(\text{M}\right)1.995\times {10}^{-3}mol-1.995\times {10}^{-3}mol-+1.995\times {10}^{-3}mol-\\ -------------------------------------------------\\ \text{Equilibrium }5.000\times {10}^{-6}mol0-1.995\times {10}^{-3}mol-\end{array}$

Determination the liters of solution present at the equivalent point,

Volume

   $\begin{array}{l}=\left[\left(20.00+19.95\right)mL\right]\left({10}^{-3}L/1mL\right)\\ \\ =0.03995L\end{array}$

Hence, volume of given solution is $0.03995L.$

Molarity of the excess $\text{H-But}$

  $\begin{array}{l}=\left(5.00\times {10}^{-6}mol\text{H-But}\right)/\left(0.03995L\right)\\ \\ =0.000125156M\end{array}$

Molarity of the excess ${\text{But}}^{-}$formed is,

  $\begin{array}{l}=\left(1.995\times {10}^{-3}mol{\text{But}}^{-}\right)/\left(0.03995L\right)\\ \\ =0.0499374M\end{array}$

Using a reaction table for the equilibrium reaction of $\text{H-But}$

  $\begin{array}{l}\text{H-But}\text{+}{H}_{2}O\underset{}{\rightleftharpoons }{H}_3{O}^{+}\text{ +}{\text{But}}^{-}\\ \text{Initial }\text{(M)}:0.000125156M-00.0499374M\\ \text{Change}\left(\text{M}\right):-x-+\text{x}+\text{x}\\ -------------------------------------\\ \text{Equilibrium: }0.000125156-x+x0.0499374+x\end{array}$

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

  $\begin{array}{l}{K}_a=1.54\times {10}^{-5}=\frac{\left[H_3{O}^{+}\right]\left[Bu{t}^{-}\right]}{\left[H-But\right]}=\frac{x\left(0.0499374+x\right)}{\left(0.000125156-x\right)}=\frac{x\left(0.0499374\right)}{0.000125156}\\ \\ \left[H_3{O}^{+}\right]=x=3.859637\times {10}^{-8}M\end{array}$

Calculation for $pH$ and $pOH$

  $\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ \\ =-\log \left(3.859637\times {10}^{-8}\right)\\ \\ pH=7.41354=7.41\end{array}$

(f)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M butanoic acid, ($K_a\text{=}1.54\times {10}^{-5}$ ) with $\text{0}\text{.1000 M NaOH}$ solution after the addition of $\text{20}\text{.00 mL}$ titrant 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

Answer to Problem 19.51P

The pH of a given buffer solution after the addition $\text{20}\text{.00 mL}$ titrant and addition of $20.00mL$ and 0.1000 M butanoic acid with $\text{0}\text{.1000 M NaOH}$ is $5.244.$

Explanation of Solution

Determination the moles of $NaOH$ added

Moles added $NaOH$

  $\begin{array}{l}=\left(0.1000molNaOH/L\right)\left({10}^{-3}L/1mL\right)\left(20.00mL\right)\\ =2.000\times {10}^{-3}molNaOH\end{array}$

The $NaOH$ will react with an equal amount of the aid, and $0mol\text{H-But}$ will remain and $2.000\times {10}^{-3}$ moles of $Bu{t}^{-}$ will form.

ICE Table of equilibrium,

$\begin{array}{l}\text{H-But(aq) }\text{+}NaOH\text{(aq)}\underset{}{\rightleftharpoons }{H}_2\text{O(l) +}Bu{t}^{-}\text{(aq)}\text{+}{\text{Na}}^{+}\left(aq\right)\\ \text{Initial }\text{(M)}2.000\times {10}^{-3}mol2.000\times {10}^{-3}mol-0-\\ \text{Change}\left(\text{M}\right)-2.000\times {10}^{-3}mol-2.000\times {10}^{-3}mol-+2.000\times {10}^{-3}mol-\\ -------------------------------------------------\\ \text{Equilibrium }00-2.000\times {10}^{-3}mol-\end{array}$

The $K_b$ of $Bu{t}^{-}$ is now important

The volume of the solution at this point is,

Volume

   $\begin{array}{l}=\left[\left(20.00+20.00\right)mL\right]\left({10}^{-3}L/1mL\right)\\ \\ =0.04000L\end{array}$

Hence, volume of given solution is $0.04000L.$

Molarity of the excess $\text{H-But}$

  $\begin{array}{l}=\left(5.00\times {10}^{-6}mol\text{H-But}\right)/\left(0.03995L\right)\\ \\ =0.000125156M\end{array}$

Molarity of the excess ${\text{But}}^{-}$formed is,

  $\begin{array}{l}=\left(2.000\times {10}^{-3}mol{\text{But}}^{-}\right)/\left(0.04000L\right)\\ \\ =0.05000M\end{array}$

So,

  $K_b=K_W/K_a=\frac{\left(1.0\times {10}^{-14}\right)}{\left(1.54\times {10}^{-5}\right)}=6.49351\times {10}^{-10}$ 

Using a reaction table for the equilibrium reaction of $\text{H-But}$

  $\begin{array}{l}\text{H-But}\text{+}{H}_{2}O\underset{}{\rightleftharpoons }{H}_3{O}^{+}\text{ +}{\text{But}}^{-}\\ \text{Initial }\text{(M)}:0.05000M-00\\ \text{Change}\left(\text{M}\right):-x-+\text{x}+\text{x}\\ ---------------------------------\\ \text{Equilibrium: }0.05000-x+xx\end{array}$

Determination the butoxide ion concentration from the $K_b$ and then determine the $pH$ from the $pOH$.

  $\begin{array}{l}{K}_a=6.49351\times {10}^{-10}=\frac{\left[\text{H-But}\right]\left[O{H}^{-}\right]}{\left[Bu{t}^{-}\right]}=\frac{\left[x\right]\left[x\right]}{\left(0.05000-x\right)}=\frac{x^2}{0.05000}\\ \\ \left[O{H}^{-}\right]=x=5.6980304\times {10}^{-6}M\end{array}$

Calculation for $pOH$

  $\begin{array}{c}pOH=-\log \left[O{H}^{-}\right]\\ \\ =-\log \left(5.6980304\times {10}^{-6}\right)\\ \\ pOH=5.244275238\end{array}$

Hence, the $pOH=5.244$

Calculation for $pH$

  $\begin{array}{c}pH=14.00-pOH\\ \\ =14.00-5.244275238\\ \\ =8.7557248\\ \\ =8.76.\end{array}$

(g)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M butanoic acid, ($K_a\text{=}1.54\times {10}^{-5}$ ) with $\text{0}\text{.1000 M NaOH}$ solution after the addition of $\text{20}\text{.05 mL}$ titrant 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

Answer to Problem 19.51P

The pH of a given buffer solution after the addition $\text{20}\text{.05 mL}$ titrant and addition of $20.00mL$ and 0.1000 M butanoic acid with $\text{0}\text{.1000 M NaOH}$ is $10.10.$

Explanation of Solution

Determination the moles of $NaOH$ added

Moles added $NaOH$

  $\begin{array}{l}=\left(0.1000molNaOH/L\right)\left({10}^{-3}L/1mL\right)\left(20.05mL\right)\\ =2.005\times {10}^{-3}molNaOH\end{array}$

The $NaOH$ will react with an equal amount of the aid, and $0mol\text{H-But}$ will remain and $5.00\times {10}^{-6}$ of $NaOH$ will excess.

There will be in excess, there will be $2.000\times {10}^{-3}$ mol of $Bu{t}^{-}$ produced, but this weak base will not affect the $pH$ compared to the excess strong base $NaOH$.  

ICE Table of equilibrium,

$\begin{array}{l}\text{H-But(aq) }\text{+}NaOH\text{(aq)}\underset{}{\rightleftharpoons }{H}_2\text{O(l) +}Bu{t}^{-}\text{(aq)}\text{+}{\text{Na}}^{+}\left(aq\right)\\ \text{Initial }\text{(M)}2.000\times {10}^{-3}mol2.005\times {10}^{-3}mol-0-\\ \text{Change}\left(\text{M}\right)-2.000\times {10}^{-3}mol-2.000\times {10}^{-3}mol-+2.000\times {10}^{-3}mol-\\ -------------------------------------------------\\ \text{Equilibrium }05.000\times {10}^{-6}mol-2.000\times {10}^{-3}mol-\end{array}$

The volume of the solution at this point is

Volume

   $\begin{array}{l}=\left[\left(20.00+20.05\right)mL\right]\left({10}^{-3}L/1mL\right)\\ \\ =0.04005L\end{array}$

Hence, volume of given solution is $0.04005L.$

Molarity of the excess $\text{H-But}$

  $\begin{array}{l}=\left(5.00\times {10}^{-6}mol\text{H-But}\right)/\left(0.04005L\right)\\ \\ =1.2484\times {10}^{-4}M\end{array}$

Calculation for $pOH$

  $\begin{array}{c}pOH=-\log \left[O{H}^{-}\right]\\ \\ =-\log \left(1.2484\times {10}^{-4}\right)\\ \\ pOH=3.9036\end{array}$

Hence, the $pOH=3.90$

Calculation for $pH$

  $\begin{array}{c}pH=14.00-pOH\\ \\ =14.00-3.9036\\ \\ =10.0964\\ \\ =10.10.\end{array}$

(h)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M butanoic acid, ($K_a\text{=}1.54\times {10}^{-5}$ ) with $\text{0}\text{.1000 M NaOH}$ solution after the addition of $\text{25}\text{.00 mL}$ titrant 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

Answer to Problem 19.51P

The pH of a given buffer solution after the addition $\text{25}\text{.00 mL}$ titrant and addition of $20.00mL$ and 0.1000 M butanoic acid with $\text{0}\text{.1000 M NaOH}$ is $12.05.$

Explanation of Solution

Determination the moles of $NaOH$ added

Moles added $NaOH$

  $\begin{array}{l}=\left(0.1000molNaOH/L\right)\left({10}^{-3}L/1mL\right)\left(25.00mL\right)\\ \\ =2.500\times {10}^{-3}molNaOH\end{array}$

The $NaOH$ will react with an equal amount of the aid, and $0mol\text{H-But}$ will remain and $5.00\times {10}^{-6}$ of $NaOH$ will excess.

There will be in excess, there will be $2.000\times {10}^{-3}$ mol of $Bu{t}^{-}$ produced, but this weak base will not affect the $pH$ compared to the excess strong base $NaOH$.  

ICE Table of equilibrium,

$\begin{array}{l}\text{H-But(aq) }\text{+}NaOH\text{(aq)}\underset{}{\rightleftharpoons }{H}_2\text{O(l) +}Bu{t}^{-}\text{(aq)}\text{+}{\text{Na}}^{+}\left(aq\right)\\ \text{Initial }\text{(M)}2.000\times {10}^{-3}mol2.500\times {10}^{-3}mol-0-\\ \text{Change}\left(\text{M}\right)-2.000\times {10}^{-3}mol-2.000\times {10}^{-3}mol-+2.000\times {10}^{-3}mol-\\ -------------------------------------------------\\ \text{Equilibrium }05.000\times {10}^{-4}mol-2.000\times {10}^{-3}mol-\end{array}$

The volume of the solution at this point is

Volume

   $\begin{array}{l}=\left[\left(20.00+25.00\right)mL\right]\left({10}^{-3}L/1mL\right)\\ \\ =0.04500L.\end{array}$

Molarity of the excess $\text{H-But}$

  $\begin{array}{l}=\left(5.00\times {10}^{-6}mol\text{H-But}\right)/\left(0.04500L\right)\\ \\ =1.1111\times {10}^{-2}M.\end{array}$

Calculation for $pOH$

  $\begin{array}{c}pOH=-\log \left[O{H}^{-}\right]\\ \\ =-\log \left(1.1111\times {10}^{-2}\right)\\ \\ pOH=\mathrm{1.9542.}\end{array}$

Hence, the $pOH=1.95$

Calculation for $pH$

  $\begin{array}{c}pH=14.00-pOH\\ \\ =14.00-1.9542\\ \\ =12.0458\\ \\ =12.05.\end{array}$