Chapter 4, Problem 81E

(a)

Interpretation Introduction

Interpretation: The volume of $0.115\text{M}{\text{HClO}}_4$ required to neutralize $\text{50}\text{mL}$ of $\text{0}\text{.0875}\text{M}$

$\text{NaOH}$ .

Concept introduction:
The volume of ${\text{HClO}}_{\text{4}}$ required for neutralization is calculated by the formula,

   $\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ V_1=\frac{M_2{V}_2}{M_1}\end{array}$

Here,

• $M_1$ is the molarity of ${\text{HClO}}_{\text{4}}$ .
• $M_2$ is the molarity of $\text{NaOH}$ .
• $V_1$ is the volume of ${\text{HClO}}_{\text{4}}$ .
• $V_2$ is the volume of $\text{NaOH}$ .

(b)

Interpretation Introduction

Interpretation: The volume of $\text{0}\text{.128}\text{M}$

$\text{HCl}$ required to neutralize $\text{2}\text{.87}\text{g}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ .

Concept introduction:

The number of moles is defined as the ratio of given mass of the compound in grams to the molar mass of the compound.
And,
The volume is calculated by the formula,

$\begin{array}{c}M=\frac{n\text{of}\text{HCl}}{V\left(\text{mL}\right)}\times 1000\text{mL}\\ V\left(\text{mL}\right)=\frac{n}{M}\times 1000\text{mL}\end{array}$

(c)

Interpretation Introduction

Interpretation: The molarity of ${\text{Cl}}^{-}$ solution.

Concept introduction:
The molarity is defined as the ratio of number of moles of solute to the volume of the solution in liters.
  $\text{Molarity}=\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}}$

(d)

Interpretation Introduction

Interpretation: The mass of $\text{KOH}$ present in the solution if $\text{45}\text{.3}\text{mL}$ of $\text{0}\text{.108}\text{M}$

$\text{HCl}$ solution is required to neutralize the $\text{KOH}$ solution.

Concept introduction:
The mass of $\text{KOH}$ is calculated by the formula,

   $\begin{array}{c}n=\frac{m}{M}\\ m=\left(n\times M\right)\end{array}$

Here,

• $n$ is the number of moles of $\text{KOH}$ .
• $m$ is the given mass of $\text{KOH}$ .
• $M$ is the molar mass of $\text{KOH}$ .