Chapter 4, Problem 4.148P

(a)

Interpretation Introduction

Interpretation:

The concentration of the ${\text{HNO}}_3$ solution is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the amount of compound in moles when molarity of solution and volume of solution is given is as follows:

$\text{Moles of solute}=\left(\text{volume of solution}\right)\left(\text{molarity of solution}\right)$        (1)

An acid-base reaction is also known as neutralization reaction involves the reaction of acid and base. The water molecule is employed as the reactant as well as product unit in these reactions.

Neutralization reaction involves the reaction of acid and base in a stoichiometric amount that is equal mole of ${\text{H}}^{+}$ ions and ${\text{OH}}^{-}$ ions reacts to form the product.

(b)

Interpretation Introduction

Interpretation:

The moles of $\text{NaOH}$ that were in excess after the first addition is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the amount of compound in moles when molarity of solution and volume of solution is given is as follows:

$\text{Moles of solute}=\left(\text{volume of solution}\right)\left(\text{molarity of solution}\right)$        (1)

An acid-base reaction is also known as neutralization reaction involves the reaction of acid and base. The water molecule is employed as the reactant as well as product unit in these reactions.

Neutralization reaction involves the reaction of acid and base in a stoichiometric amount that is equal mole of ${\text{H}}^{+}$ ions and ${\text{OH}}^{-}$ ions reacts to form the product.