Chapter 5, Problem 5.42P

(a)

Interpretation Introduction

Interpretation:

The $pH$ of $0.10MC{H}_{3}COOH(aq)$ has to be determined. 

Concept Introduction:

For the titration of a weak acid with a strong base, the $pH$ before the equivalence point depends on the excess concentration of acid and the $pH$ after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so $\text{pH}$ is $\text{7}\text{.0}$.  The equivalence point occurs when total volume of base has been added. 

$pH=-\log \left[H_3{O}^{+}\right]$

$pH=-\log \left[O{H}^{-}\right]$

$pH+pOH=14$

Calculation of Moles:

1. $No.ofmoles\left(n\right)=\frac{W}{M.W}$

Where,

$\begin{array}{l}W =Weightofagivencompound\\ \\ M.W=Molecularweightofthesamecompound\end{array}$

2. $No.ofmoles\left(n\right)=\left(Molarity\right)\left( Volume\right)$

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

$Molarity=\frac{\text{moles of solute}}{\text{volume of solution in L}}$

Answer to Problem 5.42P

The $pH$ of $0.10MC{H}_{3}COOH(aq)$ has been determined to be $2.8723$.

Explanation of Solution

Given Data:

$\begin{array}{l}{K}_{a}forC{H}_{3}COOH=1.8\times {10}^{-5}\\ \\ \left[C{H}_{3}COOH(aq)\right]=0.10M\end{array}$

The equilibrium can be written as

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{ConcentrationC{H}_{3}COOH(aq)+H_{2}O(l)\rightleftharpoons C{H}_{3}CO{O}^{-}(aq)+H_3{O}^{+}(aq)}\\ Initial0.10M00\\ Change-x+x+x\end{array}}\\ Final\left(0.10-x\right)xx\end{array}$

The equilibrium constant can be represented for the above equation,

$\begin{array}{l}{K}_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C{H}_{3}COOH\right]}\\ \\ =\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}\end{array}$

The value of $x$ is very small, so can be neglected in the denominator.  Then

$\begin{array}{l}{K}_a=\frac{x^2}{0.10}\\ \\ x=\sqrt{\left(K_a\right)\left(0.10\right)}=1.34164\times {10}^{-3}.\\ \\ \left[H_3{O}^{+}\right]=1.34164\times {10}^{-3}\end{array}$

$\begin{array}{l}pH=-\log \left[H_3{O}^{+}\right]\\ \\ =-\log \left(1.34164\times {10}^{-3}\right)\\ \\ =\mathrm{2.8723.}\end{array}$

Therefore, the $pH$ of $0.10MC{H}_{3}COOH(aq)$ has been calculated as $2.8723$.

(b)

Interpretation Introduction

Interpretation:

The $pH$ of the solution after the addition of $10.0c{m}^3$ of $0.10MNaOH(aq)$ has to be  determined. 

Concept Introduction:

For the titration of a weak acid with a strong base, the $pH$ before the equivalence point depends on the excess concentration of acid and the $pH$ after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so $\text{pH}$ is $\text{7}\text{.0}$.  The equivalence point occurs when total volume of base has been added. 

$pH=-\log \left[H_3{O}^{+}\right]$

$pH=-\log \left[O{H}^{-}\right]$

$pH+pOH=14$

Calculation of Moles:

1. $No.ofmoles\left(n\right)=\frac{W}{M.W}$

Where,

$\begin{array}{l}W =Weightofagivencompound\\ \\ M.W=Molecularweightofthesamecompound\end{array}$

2. $No.ofmoles\left(n\right)=\left(Molarity\right)\left( Volume\right)$

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

$Molarity=\frac{\text{moles of solute}}{\text{volume of solution in L}}$

Answer to Problem 5.42P

The $pH$ of the solution after the addition of $10.0c{m}^3$ of $0.10MNaOH(aq)$ has been  determined to be $4.56$.

Explanation of Solution

Given Data:

$\begin{array}{l}Volumeof C{H}_{3}COOH =\text{ 25}.00ml=25.00\times {10}^{-3}L\\ \\ Concentrationof C{H}_{3}COOH =0.10M\\ \\ ConcentrationofNaOH=0.10M\\ \\ VolumeofNaOHadded=10.00ml=10.00\times {10}^{-3}L\\ \\ Thetotalvolumeofthesolutionatthispo\mathrm{int}=25.00\times {10}^{-3}L+10.00\times {10}^{-3}L\\ \\ =35.00\times {10}^{-3}L\end{array}$

$\begin{array}{l}TheinitialnumberofmolesofC{H}_{3}COOH=Molarity\times Volume\\ \\ =\left(0.10M\right)\left(25.00\times {10}^{-3}L\right)\\ \\ =2.5\times {10}^{-3}.\end{array}$

$\begin{array}{l}MolesofaddedNaOH=Molarity\times Volume\\ \\ =\left(0.10M\right)\left(10.00\times {10}^{-3}L\right)\\ \\ =1\times {10}^{-3}.\end{array}$

The reaction occurring in the titration is the neutralization of $O{H}^{-}$ (from $NaOH$) by $H_3{O}^{+}$ (from $C{H}_{3}COOH$):

$\begin{array}{l}NaOH\left(aq\right)+C{H}_{3}COOH\left(aq\right)\to H_{2}O\left(l\right)+C{H}_{3}COONa\left(aq\right)\\ \\ H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)\end{array}$

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{No.ofmolesC{H}_{3}COOH\left(aq\right)+NaOH\left(aq\right)\to H_{2}O\left(l\right)+C{H}_{3}COONa\left(aq\right)}\\ Initial2.5\times {10}^{-3}1.0\times {10}^{-3}-0\\ Change-1.0\times {10}^{-3}-1.0\times {10}^{-3}-+1.0\times {10}^{-3}\end{array}}\\ Final1.5\times {10}^{-3}0-1.0\times {10}^{-3}\end{array}$

$\begin{array}{l}{K}_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C{H}_{3}COOH\right]}\\ \\ =\frac{\left(\frac{1.0\times {10}^{-3}moles}{35\times {10}^{-3}L}\right)\left(x\right)}{\left(\frac{1.5\times {10}^{-3}moles}{35\times {10}^{-3}L}\right)}\\ \\ x=1.8\times {10}^{-5}\times 1.5=2.7\times {10}^{-5}.\\ \\ \left[H_3{O}^{+}\right]=2.7\times {10}^{-5}\end{array}$

$\begin{array}{l}pH=-\log \left[H_3{O}^{+}\right]\\ \\ =-\log \left(2.7\times {10}^{-5}\right)\\ \\ =\mathrm{4.56.}\end{array}$

Therefore, the $pH$ of the solution has been calculated to be $4.56$. 

(c)

Interpretation Introduction

Interpretation:

The volume of $0.10MNaOH(aq)$ is required to reach halfway to the stoichiometric point has to be determined. 

Concept Introduction:

For the titration of a weak acid with a strong base, the $pH$ before the equivalence point depends on the excess concentration of acid and the $pH$ after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so $\text{pH}$ is $\text{7}\text{.0}$.  The equivalence point occurs when total volume of base has been added. 

Answer to Problem 5.42P

The volume of $0.10MNaOH(aq)$ is required to reach halfway to the stoichiometric point has been determined to be $12.5mL$.

Explanation of Solution

Given Data:

$\begin{array}{l}VolumeofC{H}_{3}COOH=25.0c{m}^3=25.0\times {10}^{-3}L\\ \\ \left[C{H}_{3}COOH(aq)\right]=0.10M\\ \\ \left[NaOH(aq)\right]=0.10M\end{array}$

The reaction occurring in the titration is the neutralization of $O{H}^{-}$ (from $NaOH$) by $H_3{O}^{+}$ (from $C{H}_{3}COOH$):

$\begin{array}{l}NaOH\left(aq\right)+C{H}_{3}COOH\left(aq\right)\to H_{2}O\left(l\right)+C{H}_{3}COONa\left(aq\right)\\ \\ H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)\end{array}$

Shows that $1$ equivalent of acid react with $1$ equivalent of base.  For achieving equivalence point, we need the equivalent quantities of acid and base.  So according to the question $25.0mL$ of $0.10MNaOH(aq)$ is required to neutralize completely $25.0mL$ of $0.10MC{H}_{3}COOH(aq)$.  The volume of $0.10MNaOH(aq)$ is required to reach halfway to the stoichiometric point is $12.5mL$.

(d)

Interpretation Introduction

Interpretation:

The $pH$ at the halfway point has to be determined. 

Concept Introduction:

For the titration of a weak acid with a strong base, the $pH$ before the equivalence point depends on the excess concentration of acid and the $pH$ after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so $\text{pH}$ is $\text{7}\text{.0}$.  The equivalence point occurs when total volume of base has been added. 

$pH=-\log \left[H_3{O}^{+}\right]$

Answer to Problem 5.42P

The $pH$  at the halfway point has been determined to be $3.5$.

Explanation of Solution

At the equivalence point, there is not an excess of either acid or base so $\text{pH}$ is $\text{7}\text{.0}$.  The $pH$  at the halfway point will be $3.5$.

(e)

Interpretation Introduction

Interpretation:

The volume of $0.10MNaOH(aq)$ required to reach the stoichiometric point has to be determined. 

Concept Introduction:

For the titration of a weak acid with a strong base, the $pH$ before the equivalence point depends on the excess concentration of acid and the $pH$ after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so $\text{pH}$ is $\text{7}\text{.0}$.  The equivalence point occurs when total volume of base has been added. 

Answer to Problem 5.42P

The volume of $0.10MNaOH(aq)$ required to reach the stoichiometric point has been determined to be $25.0mL$.

Explanation of Solution

Given Data:

$\begin{array}{l}VolumeofC{H}_{3}COOH=25.0c{m}^3=25.0\times {10}^{-3}L\\ \\ \left[C{H}_{3}COOH(aq)\right]=0.10M\\ \\ \left[NaOH(aq)\right]=0.10M\end{array}$

The reaction occurring in the titration is the neutralization of $O{H}^{-}$ (from $NaOH$) by $H_3{O}^{+}$ (from $C{H}_{3}COOH$):

$\begin{array}{l}NaOH\left(aq\right)+C{H}_{3}COOH\left(aq\right)\to H_{2}O\left(l\right)+C{H}_{3}COONa\left(aq\right)\\ \\ H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)\end{array}$

Shows that $1$ equivalent of acid react with $1$ equivalent of base.  For achieving equivalence point, we need the equivalent quantities of acid and base.  So according to the question $25.0mL$ of $0.10MNaOH(aq)$ is required to neutralize completely $25.0mL$ of $0.10MC{H}_{3}COOH(aq)$.

(e)

Interpretation Introduction

Interpretation:

The $pH$ at the stoichiometric point has to be calculated. 

Concept Introduction:

For the titration of a weak acid with a strong base, the $pH$ before the equivalence point depends on the excess concentration of acid and the $pH$ after the equivalence point depends on the excess concentration of base.  At the equivalence point, there is not an excess of either acid or base so $\text{pH}$ is $\text{7}\text{.0}$.  The equivalence point occurs when total volume of base has been added. 

$pH=-\log \left[H_3{O}^{+}\right]$

Answer to Problem 5.42P

The $pH$ at the stoichiometric point has  been calculated to be $7.0$.

Explanation of Solution

Given Data:

$\begin{array}{l}VolumeofC{H}_{3}COOH=25.0c{m}^3=25.0\times {10}^{-3}L\\ \\ \left[C{H}_{3}COOH(aq)\right]=0.10M\\ \\ \left[NaOH(aq)\right]=0.10M\end{array}$

The reaction occurring in the titration is the neutralization of $O{H}^{-}$ (from $NaOH$) by $H_3{O}^{+}$ (from $C{H}_{3}COOH$):

$\begin{array}{l}NaOH\left(aq\right)+C{H}_{3}COOH\left(aq\right)\to H_{2}O\left(l\right)+C{H}_{3}COONa\left(aq\right)\\ \\ H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)\end{array}$

Shows that $1$ equivalent of acid react with $1$ equivalent of base.  For achieving equivalence point, we need the equivalent quantities of acid and base.  So according to the question $25.0mL$ of $0.10MNaOH(aq)$ is required to neutralize completely $25.0mL$ of $0.10MC{H}_{3}COOH(aq)$.  At the equivalence point, there is not an excess of either acid or base so $\text{pH}$ is $7.0$.