Chapter 20, Problem 98A

Interpretation Introduction

Interpretation: The millimeters of 4M $\text{KOH}$ to neutralize $45\text{mL}$ of 2.5M ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ are to be calculated.

Concept Introduction: The reaction of acid and base is a neutralization reaction. The formula that can be used for the calculation of the volume of acid or base is,

  $\frac{M_a\times V_a}{N_1}=\frac{M_b\times V_b}{N_2}$ …… (1)

Here, $M_a$ is the molarity of the acid

  $M_b$ molarity of base

  $N_1$ is the number of moles of acid

  $N_2$ is the number of moles of base

  $V_a$ is the volume of acid

  $V_b$ is the volume of base

Answer to Problem 98A

The volume of base required to neutralize ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $5.6\text{mL}$ .

Explanation of Solution

The balanced chemical equation of the given reaction is,

  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}+2\text{KOH}\to {\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}+2{\text{H}}_{\text{2}}\text{O}$

Given, Here, $M_a$ is the molarity of the acid is 2.5M.

  $M_b$ molarity of the base is 4M.

  $N_1$ is the number of moles of acid is 1.

  $N_2$ is the number of moles of the base is 2.

  $V_a$ is the volume of acid is $4.50\text{mL}$ .

  $V_b$ is the volume of base

Substitute 2.5M for $M_a$ , 4M for $M_b$ , 1 for $N_1$ , 2 for $N_2$ and $4.50\text{mL}$ for $V_a$ in equation (1) to calculate the volume of base, KOH .

  $\begin{array}{c}\frac{2.5\times 4.5}{1}=\frac{4\times V_b}{2}\\ V_b=5.6\text{mL}\end{array}$

So, the volume of base required is $5.6\text{mL}$ .