Chapter 15, Problem 28PE

(a)

Interpretation Introduction

Interpretation:

Molarity of $\text{31}\text{.91 mL NaOH}$ required for titration with $\text{37}\text{.70 mL}$ of $\text{0}\text{.126 }M\text{ HCl}$ has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

  $M_1{V}_1=M_2{V}_2$  

Here,

$M_1$ denotes molarity of  $\text{HCl}$ solution.

$V_1$ denotes the volume of $\text{HCl}$ solution.

$M_2$ denotes the molarity of $\text{NaOH}$ solution.

$V_2$ denotes the volume  of $\text{NaOH}$ solution.

(b)

Interpretation Introduction

Interpretation:

Molarity of $\text{24}\text{.02 mL NaOH}$ required for titration with $\text{48}\text{.04 mL}$ of $\text{0}\text{.482 }M\text{ HCl}$ has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

Molarity of $\text{39}\text{.39 mL NaOH}$ required for titration with $\text{13}\text{.13 mL}$ of $\text{1}\text{.425 }M\text{ HCl}$ has to be calculated.

Concept Introduction:

Refer to part (a).