Chapter 19, Problem 19.55P

(a)

Interpretation Introduction

Interpretation:

The value of $pH$ during the titration of $\text{30}\text{.00 ml}$ of $\text{0}\text{.1000M}$$KOH$ with $\text{0}\text{.00 ml}$ of $\text{0}\text{.1000M}$$HBr$ solution has to be calculated.

Concept Introduction:

For the titration of a strong base with a strong acid, the $pH$ before the equivalence point depends on the excess concentration of base and the $pH$ after the equivalence point depends on the excess concentration of acid.  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 19.55P

The $pH$ during the titration of $\text{30}\text{.00 ml}$ of $\text{0}\text{.1000M}$ $KOH$ with $\text{0}\text{.00 ml}$ of $\text{0}\text{.1000M}$ $HBr$ solution has calculated as $13.00$.

Explanation of Solution

Given Data:

$\begin{array}{l}Volumeof KOH =\text{ 3}0.00ml=30.00\times {10}^{-3}L\\ \\ ConcentrationofKOH =0.1000M\\ \\ ConcentrationofHBr=0.1000M\\ \\ VolumeofHBradded=0.00ml=0.00\times {10}^{-3}L\\ \\ Thetotalvolumeofthesolutionatthispo\mathrm{int}=30.00\times {10}^{-3}L+0.00\times {10}^{-3}L\\ \\ =30.00\times {10}^{-3}L\end{array}$

The reaction occurring in the titration is the neutralization of $O{H}^{-}$ (from $KOH$) by $H_3{O}^{+}$ (from $HBr$):

$\begin{array}{l}HBr\left(aq\right)+KOH\left(aq\right)\to H_{2}O\left(l\right)+KBr\left(aq\right)\\ \\ H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)\end{array}$

$\begin{array}{l}TheinitialnumberofmolesofKOH=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(30.00\times {10}^{-3}L\right)\\ \\ =3.000\times {10}^{-3}.\end{array}$

$\begin{array}{l}MolesofaddedHBr=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(0.00\times {10}^{-3}L\right)\\ \\ =0.\end{array}$

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{No.ofmolesKOH\left(aq\right)+HBr\left(aq\right)\to H_{2}O\left(l\right)+KBr\left(aq\right)}\\ Initial3\times {10}^{-3}0-0\\ Change00-0\end{array}}\\ Final3\times {10}^{-3}0-0\end{array}$

$\begin{array}{l}Excess\left[O{H}^{-}\right]=\frac{3.000\times {10}^{-3}moles}{30.00\times {10}^{-3}L}\\ \\ =0.1000M.\end{array}$

$\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]\\ \\ =-\log \left(0.1000\right)\\ \\ =\mathrm{1.0000.}\end{array}$

$\begin{array}{l}pH=14-pOH\\ \\ =14-1.0000=\mathrm{13.00.}\end{array}$

Therefore, the $pH$ of the solution has been calculated to be $\text{13}\text{.00}$.

(b)

Interpretation Introduction

Interpretation:

The value of $pH$ during the titration of $\text{30}\text{.00 ml}$ of $KOH$ with $\text{0}\text{.1000M}$ $\text{15}\text{.00}\text{ml}$ of $\text{0}\text{.1000M}$ $HBr$ solution has to be calculated.

Concept Introduction:

For the titration of a strong base with a strong acid, the $pH$ before the equivalence point depends on the excess concentration of base and the $pH$ after the equivalence point depends on the excess concentration of acid.  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,

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

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 19.55P

The $pH$ during the titration of $\text{30}\text{.00 ml}$ of $KOH$ with $\text{0}\text{.1000M}$ $\text{15}\text{.00}\text{ml}$ of $\text{0}\text{.1000M}$ $HBr$ solution has calculated as $12.52.$

Explanation of Solution

Given Data:

$\begin{array}{l}Volumeof KOH =\text{ 3}0.00ml=30.00\times {10}^{-3}L\\ \\ ConcentrationofKOH =0.1000M\\ \\ ConcentrationofHBr=0.1000M\\ \\ VolumeofHBradded=15.00ml=15.00\times {10}^{-3}L\\ \\ Thetotalvolumeofthesolutionatthispo\mathrm{int}=30.00\times {10}^{-3}L+15.00\times {10}^{-3}L\\ \\ =45.00\times {10}^{-3}L\end{array}$

The reaction occurring in the titration is the neutralization of $O{H}^{-}$ (from $KOH$) by $H_3{O}^{+}$ (from $HBr$):

$\begin{array}{l}HBr\left(aq\right)+KOH\left(aq\right)\to H_{2}O\left(l\right)+KBr\left(aq\right)\\ \\ H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)\end{array}$

$\begin{array}{l}TheinitialnumberofmolesofKOH=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(30.00\times {10}^{-3}L\right)\\ \\ =3.000\times {10}^{-3}.\end{array}$

$\begin{array}{l}MolesofaddedHBr=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(15.00\times {10}^{-3}L\right)\\ \\ =1.5\times {10}^{-3}.\end{array}$

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{No.ofmolesKOH\left(aq\right)+HBr\left(aq\right)\to H_{2}O\left(l\right)+KBr\left(aq\right)}\\ Initial3\times {10}^{-3}1.5\times {10}^{-3}-0\\ Change-1.5\times {10}^{-3}-1.5\times {10}^{-3}-+1.5\times {10}^{-3}\end{array}}\\ Final1.5\times {10}^{-3}0-1.5\times {10}^{-3}\end{array}$

$\begin{array}{l}Excess\left[O{H}^{-}\right]=\frac{1.500\times {10}^{-3}moles}{45.00\times {10}^{-3}L}\\ \\ =0.03333M.\\ \\ pOH=-\log \left[O{H}^{-}\right]\\ \\ =-\log \left(0.03333\right)\\ \\ =\mathrm{1.4722.}\end{array}$

$\begin{array}{l}pH=14-pOH\\ \\ =14-1.4722=\mathrm{12.52.}\end{array}$

Therefore, the $pH$ value of the solution has been calculated to be $12.52$.

(c)

Interpretation Introduction

Interpretation:

The value of $pH$ during the titration of $\text{40}\text{.00 ml}$ of $KOH$ $\text{0}\text{.1000M}$ with $\text{29}\text{.00}\text{ml}$ of $\text{0}\text{.1000M}$ $HBr$ solution has to be calculated.

Concept Introduction:

For the titration of a strong base with a strong acid, the $pH$ before the equivalence point depends on the excess concentration of base and the $pH$ after the equivalence point depends on the excess concentration of acid.  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,

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

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 19.55P

The $pH$ during the titration of $\text{40}\text{.00 ml}$ of $KOH$ $\text{0}\text{.1000M}$ with $\text{29}\text{.00}\text{ml}$ of $\text{0}\text{.1000M}$ $HBr$ solution has calculated as $11.23.$

Explanation of Solution

Given Data:

$\begin{array}{l}VolumeofKOH =\text{ 3}0.00ml=30.00\times {10}^{-3}L\\ \\ ConcentrationofKOH =0.1000M\\ \\ ConcentrationofHBr=0.1000M\\ \\ VolumeofHBradded=29.00ml=29.00\times {10}^{-3}L\\ \\ Thetotalvolumeofthesolutionatthispo\mathrm{int}=30.00\times {10}^{-3}L+29.00\times {10}^{-3}L\\ \\ =59.00\times {10}^{-3}L\end{array}$

The reaction occurring in the titration is the neutralization of $O{H}^{-}$ (from $KOH$) by $H_3{O}^{+}$ (from $HBr$):

$\begin{array}{l}HBr\left(aq\right)+KOH\left(aq\right)\to H_{2}O\left(l\right)+KBr\left(aq\right)\\ \\ H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)\end{array}$

$\begin{array}{l}TheinitialnumberofmolesofKOH=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(30.00\times {10}^{-3}L\right)\\ \\ =3.000\times {10}^{-3}.\end{array}$

$\begin{array}{l}MolesofaddedHBr=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(29.00\times {10}^{-3}L\right)\\ \\ =2.9\times {10}^{-3}.\end{array}$

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{No.ofmolesKOH\left(aq\right)+HBr\left(aq\right)\to H_{2}O\left(l\right)+KBr\left(aq\right)}\\ Initial3\times {10}^{-3}2.9\times {10}^{-3}-0\\ Change-2.9\times {10}^{-3}-2.9\times {10}^{-3}-+2.9\times {10}^{-3}\end{array}}\\ Final0.1\times {10}^{-3}0-2.9\times {10}^{-3}\end{array}$

$\begin{array}{l}Excess\left[O{H}^{-}\right]=\frac{0.100\times {10}^{-3}moles}{59.00\times {10}^{-3}L}\\ \\ =0.0016949M.\\ \\ pOH=-\log \left[O{H}^{-}\right]\\ \\ =-\log \left(0.0016949\right)\\ \\ =\mathrm{2.7708559.}\\ \\ pH=14-pOH=14-2.7708559\\ \\ =11.2291441=\mathrm{11.23.}\end{array}$

Therefore, the $pH$ of the solution has been calculated to be $11.23$.

(d)

Interpretation Introduction

Interpretation:

The value of $pH$ during the titration of $\text{30}\text{.00 ml}$ of $KOH$ $\text{0}\text{.1000M}$ with $\text{29}\text{.90}\text{ml}$ of $\text{0}\text{.1000M}$ $HBr$ solution has to be calculated. 

Concept Introduction:

For the titration of a strong base with a strong acid, the $pH$ before the equivalence point depends on the excess concentration of base and the $pH$ after the equivalence point depends on the excess concentration of acid.  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,

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

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 19.55P

The value of $pH$ during the titration of $\text{30}\text{.00 ml}$ of $KOH$ $\text{0}\text{.1000M}$ with $\text{29}\text{.90}\text{ml}$ of $\text{0}\text{.1000M}$ $HBr$ solution has calculated as $10.2$.

Explanation of Solution

Given Data:

$\begin{array}{l}Volumeof KOH =\text{ 3}0.00ml=30.00\times {10}^{-3}L\\ \\ ConcentrationofKOH =0.1000M\\ \\ ConcentrationofHBr=0.1000M\\ \\ VolumeofHBradded=29.90ml=29.90\times {10}^{-3}L\\ \\ Thetotalvolumeofthesolutionatthispo\mathrm{int}=30.00\times {10}^{-3}L+29.90\times {10}^{-3}L\\ \\ =59.90\times {10}^{-3}L\end{array}$

The reaction occurring in the titration is the neutralization of $H_3{O}^{+}$ (from $KOH$) by $O{H}^{-}$ (from $HBr$):

$\begin{array}{l}HBr\left(aq\right)+KOH\left(aq\right)\to H_{2}O\left(l\right)+KBr\left(aq\right)\\ \\ H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)\end{array}$

$\begin{array}{l}TheinitialnumberofmolesofKOH=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(30.00\times {10}^{-3}L\right)\\ \\ =3.000\times {10}^{-3}.\end{array}$

$\begin{array}{l}MolesofaddedHBr=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(29.90\times {10}^{-3}L\right)\\ \\ =2.99\times {10}^{-3}.\end{array}$

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{No.ofmolesHBr\left(aq\right)+KOH\left(aq\right)\to H_{2}O\left(l\right)+KBr\left(aq\right)}\\ Initial3\times {10}^{-3}2.99\times {10}^{-3}-0\\ Change-2.99\times {10}^{-3}-2.99\times {10}^{-3}-+2.99\times {10}^{-3}\end{array}}\\ Final0.01\times {10}^{-3}0-2.99\times {10}^{-3}\end{array}$

$\begin{array}{l}Excess\left[O{H}^{-}\right]=\frac{0.001\times {10}^{-3}moles}{59.90\times {10}^{-3}L}\\ \\ =0.000166945M.\\ \\ pOH=-\log \left[O{H}^{-}\right]\\ \\ =-\log \left(0.000166945\right)\\ \\ =\mathrm{3.7774266.}\\ \\ pH=14-pOH\\ \\ =14-3.7774266\\ \\ =10.2225734=\mathrm{10.2.}\end{array}$

Therefore, the $pH$ of the solution has been calculated to be $10.2$.

(e)

Interpretation Introduction

Interpretation:

The value of $pH$ during the titration of $\text{30}\text{.00 ml}$ of $KOH$ $\text{0}\text{.1000M}$ with $\text{30}\text{.00}\text{ml}$ of $\text{0}\text{.1000M}$ $HBr$ solution has to be calculated. 

Concept Introduction:

For the titration of a strong base with a strong acid, the $pH$ before the equivalence point depends on the excess concentration of base and the $pH$ after the equivalence point depends on the excess concentration of acid.  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,

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

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 19.55P

The $pH$ during the titration of $\text{30}\text{.00 ml}$ of $KOH$ $\text{0}\text{.1000M}$ with $\text{30}\text{.00}\text{ml}$ of $\text{0}\text{.1000M}$ $HBr$ solution has calculated as $7.00.$

Explanation of Solution

Given Data:

$\begin{array}{l}Volumeof KOH =\text{ 3}0.00ml=30.00\times {10}^{-3}L\\ \\ ConcentrationofKOH =0.1000M\\ \\ ConcentrationofHBr=0.1000M\\ \\ VolumeofHBradded=30.00ml=30.00\times {10}^{-3}L\\ \\ Thetotalvolumeofthesolutionatthispo\mathrm{int}=30.00\times {10}^{-3}L+30.00\times {10}^{-3}L\\ \\ =60.00\times {10}^{-3}L\end{array}$

The reaction occurring in the titration is the neutralization of $O{H}^{-}$ (from $KOH$) by $H_3{O}^{+}$ (from $HBr$):

$\begin{array}{l}HBr\left(aq\right)+KOH\left(aq\right)\to H_{2}O\left(l\right)+KBr\left(aq\right)\\ \\ H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)\end{array}$

$\begin{array}{l}TheinitialnumberofmolesofKOH=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(30.00\times {10}^{-3}L\right)\\ \\ =3.000\times {10}^{-3}.\end{array}$

$\begin{array}{l}MolesofaddedHBr=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(30.00\times {10}^{-3}L\right)\\ \\ =3.00\times {10}^{-3}.\end{array}$

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{No.ofmolesKOH\left(aq\right)+HBr\left(aq\right)\to H_{2}O\left(l\right)+KBr\left(aq\right)}\\ Initial3\times {10}^{-3}3.00\times {10}^{-3}-0\\ Change-3.00\times {10}^{-3}-3.00\times {10}^{-3}-+3.00\times {10}^{-3}\end{array}}\\ Final00-3.00\times {10}^{-3}\end{array}$

The $HBr$ will react with an equal amount of $KOH$ and $\text{0}\text{.0 mol}$ $KOH$ will remain.  This is the equivalence point of a strong base –strong acid titration, thus, $pH$ is $\text{7}\text{.0}$.  Only the neutral salt $KBr$ is in solution at the equivalence point. 

(f)

Interpretation Introduction

Interpretation:

The value of $pH$ during the titration of $\text{30}\text{.00 ml}$ of $KOH$ $\text{0}\text{.1000M}$ with $\text{30}\text{.10}\text{ml}$ of $\text{0}\text{.1000M}$ $HBr$ solution has to be calculated. 

Concept Introduction:

For the titration of a strong base with a strong acid, the $pH$ before the equivalence point depends on the excess concentration of base and the $pH$ after the equivalence point depends on the excess concentration of acid.  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,

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

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 19.55P

The $pH$ during the titration of $\text{30}\text{.00 ml}$ of $KOH$ $\text{0}\text{.1000M}$ with $\text{30}\text{.10}\text{ml}$ of $\text{0}\text{.1000M}$ $HBr$ solution has calculated as $3.8$

Explanation of Solution

Given Data:

$\begin{array}{l}Volumeof KOH =\text{ 3}0.00ml=30.00\times {10}^{-3}L\\ \\ ConcentrationofKOH =0.1000M\\ \\ ConcentrationofHBr=0.1000M\\ \\ VolumeofHBradded=30.10ml=30.10\times {10}^{-3}L\\ \\ Thetotalvolumeofthesolutionatthispo\mathrm{int}=30.00\times {10}^{-3}L+30.10\times {10}^{-3}L\\ \\ =60.10\times {10}^{-3}L\end{array}$

The reaction occurring in the titration is the neutralization of $O{H}^{-}$ (from $KOH$) by $H_3{O}^{+}$ (from $HBr$):

$\begin{array}{l}HBr\left(aq\right)+KOH\left(aq\right)\to H_{2}O\left(l\right)+KBr\left(aq\right)\\ \\ H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)\end{array}$

$\begin{array}{l}TheinitialnumberofmolesofKOH=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(30.00\times {10}^{-3}L\right)\\ \\ =3.000\times {10}^{-3}.\end{array}$

$\begin{array}{l}MolesofaddedHBr=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(30.10\times {10}^{-3}L\right)\\ \\ =3.01\times {10}^{-3}.\end{array}$

The $KOH$ will react with an equal amount of the acid. 

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{No.ofmolesKOH\left(aq\right)+HBr\left(aq\right)\to H_{2}O\left(l\right)+NaCl\left(aq\right)}\\ Initial3\times {10}^{-3}3.01\times {10}^{-3}-0\\ Change-3.00\times {10}^{-3}-3.00\times {10}^{-3}-+3.00\times {10}^{-3}\end{array}}\\ Final00.01\times {10}^{-3}-3.01\times {10}^{-3}\end{array}$

The amount of $HBr$ is now in excess.

$\begin{array}{l}Excess\left[H_3{O}^{+}\right]=\frac{0.01\times {10}^{-3}moles}{60.10\times {10}^{-3}L}\\ \\ =0.000166389M.\\ \\ pH=-\log \left[H_3{O}^{+}\right]\\ \\ =-\log \left(0.000166389\right)\\ \\ =3.778875=\mathrm{3.8.}\end{array}$

Therefore, $pH$ of the solution has been calculated to be $3.8$.

(g)

Interpretation Introduction

Interpretation:

The value of $pH$ during the titration of $\text{30}\text{.00 ml}$ of $KOH$ $\text{0}\text{.1000M}$ with $\text{40}\text{.00 ml}$ of $\text{0}\text{.1000M}$ $HBr$ solution has to be calculated. 

Concept Introduction:

For the titration of a strong base with a strong acid, the $pH$ before the equivalence point depends on the excess concentration of base and the $pH$ after the equivalence point depends on the excess concentration of acid.  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,

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

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 19.55P

The $pH$ during the titration of $\text{30}\text{.00 ml}$ of $KOH$ $\text{0}\text{.1000M}$ with $\text{40}\text{.00 ml}$ of $\text{0}\text{.1000M}$ $HBr$ solution has calculated as $1.85$

Explanation of Solution

Given Data:

$\begin{array}{l}Volumeof KOH =\text{ 3}0.00ml=30.00\times {10}^{-3}L\\ \\ ConcentrationofKOH =0.1000M\\ \\ ConcentrationofHBr=0.1000M\\ \\ VolumeofHBradded=40.00ml=40.00\times {10}^{-3}L\\ \\ Thetotalvolumeofthesolutionatthispo\mathrm{int}=30.00\times {10}^{-3}L+40.00\times {10}^{-3}L\\ \\ =70.00\times {10}^{-3}L\end{array}$

The reaction occurring in the titration is the neutralization of $O{H}^{-}$ (from $KOH$) by $H_3{O}^{+}$ (from $HBr$):

$\begin{array}{l}HBr\left(aq\right)+KOH\left(aq\right)\to H_{2}O\left(l\right)+KBr\left(aq\right)\\ \\ H_3{O}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)\end{array}$

$\begin{array}{l}TheinitialnumberofmolesofKOH=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(30.00\times {10}^{-3}L\right)\\ \\ =3.000\times {10}^{-3}.\end{array}$

$\begin{array}{l}MolesofaddedHBr=Molarity\times Volume\\ \\ =\left(0.1000M\right)\left(40.00\times {10}^{-3}L\right)\\ \\ =4.00\times {10}^{-3}.\end{array}$

The $KOH$ will react with an equal amount of the acid. 

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{No.ofmolesKOH\left(aq\right)+HBr\left(aq\right)\to H_{2}O\left(l\right)+KBr\left(aq\right)}\\ Initial3\times {10}^{-3}4.00\times {10}^{-3}-0\\ Change-3.00\times {10}^{-3}-3.00\times {10}^{-3}-+3.00\times {10}^{-3}\end{array}}\\ Final01.00\times {10}^{-3}-3.00\times {10}^{-3}\end{array}$

The amount of $HBr$ is now in excess.

$\begin{array}{l}Excess\left[H_3{O}^{+}\right]=\frac{1.00\times {10}^{-3}moles}{70.00\times {10}^{-3}L}\\ \\ =0.0142875M.\\ \\ pH=-\log \left[H_3{O}^{+}\right]\\ \\ =-\log \left(0.0142875\right)\\ \\ =1.845098=\mathrm{1.85.}\end{array}$

Therefore, the $pH$ of the solution has been calculated to be $1.85$.