Chapter 14.8, Problem 65PP

Interpretation Introduction

Interpretation: If $38.2 \text{mL}$ of a $0.163\text{MKOH}$ solution is required to titrate $25.0 \text{mL}$ of a ${\text{H}}_2{\text{SO}}_4$ solution, what is the molarity of the ${\text{H}}_2{\text{SO}}_4$ Solution?

${\text{H}}_2{\text{SO}}_4(aq)+2\text{KOH}(aq)\to 2{\text{H}}_2\text{O}(\text{l})+{\text{K}}_2{\text{SO}}_4(\text{aq})$

Concept introduction:

The general equation is as follows:

${\text{H}}_2{\text{SO}}_4(\text{aq})+2\text{KOH}(\text{aq})\to 2{\text{H}}_2\text{O}(\text{l})+{\text{K}}_2{\text{SO}}_4(\text{aq})$

The formula used:

$\frac{M_1{V}_1}{n_1}=\frac{M_2{V}_2}{n_2}$