Chapter 15, Problem 27PE

(a)

Interpretation Introduction

Interpretation:

Molarity of $\text{40}\text{.13 mL}$$\text{HCl}$ required for titration with $\text{37}\text{.70 mL}$ of $\text{0}\text{.728 }M\text{ NaOH}$ 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  initial solution before dilution.

$V_1$ denotes the volume before dilution.

$M_2$ denotes the molarity of final solution after dilution.

$V_2$ denotes the volume after dilution.

(b)

Interpretation Introduction

Interpretation:

Molarity of $\text{19}\text{.00 mL}$$\text{HCl}$ required for titration with $\text{33}\text{.66 mL}$ of $0.306M\text{ NaOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

Molarity of $\text{27}\text{.25 mL}$$\text{HCl}$ required for titration with $\text{18}\text{.00 mL}$ of $0.555M\text{ NaOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).