Chapter 10.11, Problem 10.30P

Interpretation Introduction

Interpretation:

The volume of $\text{0}\text{.150M}\text{NaOH}$ needed to titrate $\text{50}\text{mL}\text{of}0.2M{H}_{2}S{O}_4$ should be determined in $\text{mL}$.

Concept Introduction:

Titration: It is a quantitative analysis method used to express the concentration of acid or base present in the solution.

For an acid-base titration ${\left(Eq\right)}_{acid}={\left(Eq\right)}_{base}$

That is,

$\begin{array}{l}{N}_{acid}\times V_{acid}=N_{base}\times V_{base}\\ where,\\ N=Normality\\ V=Volume\end{array}$

Normality: It generally expresses the concentration of acid or base as equivalents of acid or base present in one liter of the solution.

$\text{Normality}\left(\text{N}\right)\text{=}\frac{\text{Equivalents of acid/base}}{\text{Liters of solution}}$

Equivalent of Acid: Generally, 1 equivalent of ion is the number of ions which has charge of one mole. In case of acids positive charge is of interest therefore, one equivalent of acid consists of 1 mole of $H^{+}$ ions.

Equivalent of Base: In case of bases negative charge is of interest therefore, one equivalent of base consists of 1 mole of $O{H}^{-}$ ions.

$\text{Bronsted-Lowry Acids}$: If a species loses a proton then it is considered as $\text{Bronsted-Lowry acid}\text{.}$

$\text{Bronsted-Lowry Base}$: If a species receives one proton, then it is considered as $\text{Bronsted-Lowry base}\text{.}$