Chapter 16, Problem 143E

Interpretation Introduction

Interpretation:

The mass of ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\cdot 2{\text{H}}_{\text{2}}\text{O}$ required to make $2.50\times {10}^2\text{mL}$ of $0.500\text{N}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ is to be calculated.

Concept introduction:

The normality of a solution is referred as the gram equivalent weight of a solute dissolved in one liter of the solution. The normality of solution is given by the expression as shown below.

$\text{N}=\frac{{\text{n}}_{\text{eq}}}{\text{V}}$

The number of equivalents of the acidic solute is the number of ${\text{H}}^{+}$ ions produced and the number of equivalents of the basic solute is the number of ${\text{OH}}^{-}$ ions produced.