Chapter 10, Problem 10.93AP

Interpretation Introduction

Interpretation:

The equivalents in 15$\text{mL}$ of 0.12$\text{M}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ and the volume of 0.085$MHN{O}_3$ needed to reach end point when titrated against 15$\text{mL}$ of 0.12$\text{M}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ should be calculated.

Concept Introduction:

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.

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.

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