Titration is the process of measuring the volume of one reagent required to react with a measured volume or mass of another reagent. In titration the point of neutralization called the end point, is observed when an indicator placed in the solution being titrated changes color. In a titration calculation the identity of acid or base is known of unknown concentration.

The number of equivalents of a species in a reaction is the product of solution volume times normality. The number of equivalents of all species in a reaction is the same.

V$_{1}$N$_{1}$ = V$_{2}$N$_{2}$

## Titration Examples

Some of the solved problems based on titration formula are given below:

Question 1: What is the normality of a hydrochloric acid solution 25ml of which requires 37ml of 0.30N NaOH solution for neutralization?
Solution:
V$_{1}$ (NaOH) = 37ml
N$_{1}$ (NaOH) = 0.30N
V$_{2}$ (HCl) = 25ml
N$_{2}$ (HCl) = ?

Using V1N1 = V2N2

Solve for N$_{2}$

(37ml)(0.30N) = (25ml)(N2)

N$_{2}$ = $\frac{(37ml)(0.30N)}{25ml}$

N$_{2}$ = $\frac{11.1}{25}$ N

N$_{2}$ = 0.44N.

Question 2: A 25.0mL sample of an electroplating solution is analyzed for its sulfuric acid concentration. It takes 46.8mL of the 0.661N NaOH to neutralize the sample, Find the normality of the acid.
Solution:
V$_{1}$ = 46.8mL
NV$_{1}$ = 0.661N
V$_{2}$ = 25.0mL

N$_{2}$ = $\frac{V_{1}N_{1}}{V_{2}}$

N$_{2}$ = $\frac{46.8mL \times 0.661N}{25.0mL}$

N$_{2}$ = 1.24N H$_{2}$SO$_{4}$.