Chapter 15, Problem 15.76SP

(a)

Interpretation Introduction

Interpretation:

The number of millimoles of ${\text{HNO}}_{\text{3}}$ that are present at the start of the titration of $\text{60}\text{.0 mL of 0}{\text{.150 M HNO}}_{\text{3}}$ with $\text{0}\text{.450 M NaOH}$ should be calculated.

Concept introduction:

Mole:

The number of moles of solute that present in the solution is given by the multiplication of molarity of solution with volume of solution in litre

$\text{Mole}\text{=}\text{Molarity×litre}\text{(L)}$

pH:

$\text{pH}$ is a logarithmic expression to express a solution is acidic, basic or neutral.  The $\text{pH}$ scale has values between 1 and 14 and on which 7 is neutral, below 7 values are more acidic in nature and above 7 values are more basic in nature.

$\text{pH}\text{=}{\text{-log}}_{\text{10}}{\text{[H}}^{\text{+}}\text{]}$ Higher the amount of solute that dissolved into the solution than saturated solution by increasing temperature to give a supersaturated solution.

(b)

Interpretation Introduction

Interpretation:

The required millilitres of $\text{NaOH}$ to reach the end point of the titration of $\text{60}\text{.0 mL of 0}{\text{.150 M HNO}}_{\text{3}}$ with $\text{0}\text{.450 M NaOH}$ should be calculated.

Concept introduction:

Mole:

The number of moles of solute that present in the solution is given by the multiplication of molarity of solution with volume of solution in litre

$\text{Mole}\text{=}\text{Molarity×litre}\text{(L)}$

pH:

$\text{pH}$ is a logarithmic expression to express a solution is acidic, basic or neutral.  The $\text{pH}$ scale has values between 1 and 14 and on which 7 is neutral, below 7 values are more acidic in nature and above 7 values are more basic in nature.

$\text{pH}\text{=}{\text{-log}}_{\text{10}}{\text{[H}}^{\text{+}}\text{]}$ Higher the amount of solute that dissolved into the solution than saturated solution by increasing temperature to give a supersaturated solution.

(c)

Interpretation Introduction

Interpretation:

In the titration of $\text{60}\text{.0 mL of 0}{\text{.150 M HNO}}_{\text{3}}$ with $\text{0}\text{.450 M NaOH}$, at the end point $\text{pH}$ of the solution should be given.

Concept introduction:

Acid-base neutralization reaction:

The aqueous solution of acid react with aqueous solution of base to produce soluble salt with water is known as acid-base neutralization reaction because of where neutralization is occur.

pH:

$\text{pH}$ is a logarithmic expression to express a solution is acidic, basic or neutral.  The $\text{pH}$ scale has values between 1 and 14 and on which 7 is neutral, below 7 values are more acidic in nature and above 7 values are more basic in nature.

$\text{pH}\text{=}{\text{-log}}_{\text{10}}{\text{[H}}^{\text{+}}\text{]}$ Higher the amount of solute that dissolved into the solution than saturated solution by increasing temperature to give a supersaturated solution.

(d)

Interpretation Introduction

Interpretation:

The general shape of the $\text{pH}$ titration curve should be drawn.

Concept introduction:

$\text{pH}$Titration graph:

The graph drawn by $\text{pH}$ change of the solution respect with volume of acid or alkaline added to give a $\text{pH}$ titration graph.

The maximum change in $\text{pH}$ respect with volume of acid or alkaline is a end point of the titration.