Chapter 15, Problem 48A

Interpretation Introduction

Interpretation: The volume of $\text{0}\text{.151}\text{N}\text{NaOH}$ required to neutralize $\text{24}\text{.2}\text{mL}$ of $\text{0}\text{.125}\text{N}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ has to be calculated. The volume of $\text{0}\text{.151}\text{M}\text{NaOH}$ required to neutralize $\text{24}\text{.1}\text{mL}$ of $\text{0}\text{.125}\text{M}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ has to be calculated.

Concept Introduction:

Molarity: Molarity is defined as the number of moles of solute in one liter of solution. Molarity is the preferred concentration unit for stoichiometry calculations. The formula is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Concentration of the solution is calculated using the formula,

$M_1{V}_1=M_2{V}_2$

Normality: Normality is defined as the number of equivalents of solute per liter of solution. The formula is,

$\text{Normality=}\frac{\text{Number}\text{of}\text{equivalents}}{\text{Liter}\text{of}\text{solution}}$

Normality of the solution is calculated using the formula,

  $N_{acid}{V}_{acid}=N_{base}{V}_{base}$

Answer to Problem 48A

The volume of $\text{0}\text{.151}\text{N}\text{NaOH}$ required to neutralize $\text{24}\text{.2}\text{mL}$ of $\text{0}\text{.125}\text{N}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{20}\text{.03}\text{mL}\text{.}$

The volume of $\text{0}\text{.151}\text{M}\text{NaOH}$ required to neutralize $\text{24}\text{.1}\text{mL}$ of $\text{0}\text{.125}\text{M}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{19}\text{.95}\text{mL}\text{.}$

Explanation of Solution

The chemical reaction is written as,

$2{\text{NaOH+H}}_{\text{2}}{\text{SO}}_4\rightleftharpoons {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{+2H}}_{\text{2}}\text{O}$

Given,

The normality of base $\left(N_{base}\right)$ is $\text{0}\text{.151}\text{N}\text{NaOH}\text{.}$

The normality of acid $\left(N_{acid}\right)$ is $\text{0}\text{.125}\text{N}{\text{H}}_2{\text{SO}}_4\text{.}$

The volume of acid $\left(V_{acid}\right)$ is $\text{24}\text{.2}\text{mL}{\text{H}}_2{\text{SO}}_4\text{.}$

Normality: Normality is defined as the number of equivalents of solute per liter of solution. The formula is,

$\text{Normality=}\frac{\text{Number}\text{of}\text{equivalents}}{\text{Liter}\text{of}\text{solution}}$

Normality of the solution is calculated using the formula,

  $N_{acid}{V}_{acid}=N_{base}{V}_{base}$

Here,

$N_{acid}=$ Normality of the acid

$N_{base}=$ Normality of the base

$V_{acid}=$ Volume of the acid

$V_{base}=$ Volume of the base

The volume of the solution is called using the formula,

$\begin{array}{l}{V}_{base}=\frac{N_{acid}\times V_{acid}}{N_{base}}\\ V_{base}=\frac{\left(\text{0}\text{.125}\text{N}\right)\left(\text{24}\text{.2}\text{mL}\right)}{\text{0}\text{.151}\text{N}}\\ V_{base}=\text{20}\text{.03}\text{mL}\end{array}$

The volume of $\text{0}\text{.151}\text{N}\text{NaOH}$ required to neutralize $\text{24}\text{.2}\text{mL}$ of $\text{0}\text{.125}\text{N}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{20}\text{.03}\text{mL}\text{.}$

Given,

The molarity of base $\left(M_1\right)$ is $\text{0}\text{.151}\text{M}\text{NaOH}\text{.}$

The molarity of acid $\left(M_2\right)$ is $\text{0}\text{.125}\text{M}{\text{H}}_2{\text{SO}}_4\text{.}$

The volume of acid $\left(V_2\right)$ is $\text{24}\text{.1}\text{mL}{\text{H}}_2{\text{SO}}_4\text{.}$

Molarity is defined as the number of moles of solute in one liter of solution. The formula is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Concentration of the solution is calculated using the formula,

$M_1{V}_1=M_2{V}_2$

Here,

$M_1\text{=}$ Initial molarity of the solution

$V_1=$ Initial volume of the solution

$M_2=$ Final molarity of the solution

$V_2=$ Final volume of the solution

The molarity of the solution is calculated using the formula,

$\begin{array}{l}{M}_1{V}_1=M_2{V}_2\\ V_2=\frac{\left(\text{0}\text{.125}\text{M}\right)\left(\text{24}\text{.1}\text{mL}\right)}{\left(\text{0}\text{.151}\text{M}\right)}\\ V_2=\text{19}\text{.95}\text{mL}\end{array}$

The volume of $\text{0}\text{.151}\text{M}\text{NaOH}$ required to neutralize $\text{24}\text{.1}\text{mL}$ of $\text{0}\text{.125}\text{M}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{19}\text{.95}\text{mL}\text{.}$

Conclusion

The volume of $\text{0}\text{.151}\text{N}\text{NaOH}$ required to neutralize $\text{24}\text{.2}\text{mL}$ of $\text{0}\text{.125}\text{N}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{20}\text{.03}\text{mL}\text{.}$

The volume of $\text{0}\text{.151}\text{M}\text{NaOH}$ required to neutralize $\text{24}\text{.1}\text{mL}$ of $\text{0}\text{.125}\text{M}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{19}\text{.95}\text{mL}\text{.}$