Chapter 18, Problem 93A

In an acid-base titration, 45.78 mL of a sulfuric acid solution is titrated to the end point by $74.30\text{mL}$ of $0.4388M$ sodium hydroxide solution. What is the molarity of the ${\text{H}}_2{\text{SO}}_4$ solution?

Interpretation Introduction

Interpretation:

The Molarity of given acid solution is to be calculated when it is neutralized by base.

Concept introduction:

Molarity is number of moles of solute present per liter of solution.

When acid and base are mixed, they neutralize each other. So their molar concentrations can be calculated by applying mole concept.

Answer to Problem 93A

Molarity of sulphuric acid is $0.3561M$

Explanation of Solution

Given data:

$\begin{array}{l}{V}_{NaOH}=74.30mL\\ MolarityofNaOH=0.4388\text{}M\\ V_{H_{2}S{O}_4}=45.78mL\end{array}$

$\begin{array}{c}{n}_{NaOH}=Molarity\times volumeinliters\\ =0.4388\times \frac{74.30}{1000}\\ =0.03260mol\end{array}$

When sulphuric acid reacts with sodium hydroxide reaction is given by:

$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$

Thus one mol acid reacts with two mol base.

This impies,

$\begin{array}{c}{n}_{H_{2}S{O}_4}=\frac{0.03260}{2}mol\\ =0.01630mol\end{array}$

$\begin{array}{c}Molarityof{H}_{2}S{O}_4=\frac{0.01630}{45.78/1000}\\ =0.3561M\end{array}$

Conclusion

$Molarity$ is $0.3561M$