Chapter 19, Problem 19.50P

(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,

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

(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,

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

(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,

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

(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,

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

(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,

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

(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,

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

(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,

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