Home » Quality Control » Test » Preparation and Standardization of 0.1 M Ceric Ammonium Sulphate

# Preparation and Standardization of 0.1 M Ceric Ammonium Sulphate

Aim: The aim is to find out the normality of the given 0.1 M Ceric ammonium Sulphate.

## Requirements:

Equipment: A special tube with a tap, a cone-shaped flask, and a small tube.
Reagents: Ceric ammonium sulphate, Sulphuric acid, Arsenic trioxide, Sodium hydroxide, and some more sulphuric acid.

## Principle:

We test Ceric ammonium sulphate by mixing it with Arsenic trioxide solution in the presence of sodium hydroxide and osmic acid. This reaction is:

As2O3+ 2NaOH → 2NaAsO2+ H2O
NaAsO2+ 2H2O→ NaH2AsO4+ 2H ++4eCe 4+ + e- → Ce3+

## Procedure:

Preparation of 0.1 M Ceric ammonium sulphate:

First, we prepare a solution of Ceric ammonium sulphate. We mix 65 grams of it with 30 milliliters of sulphuric acid and 500 milliliters of water. We let it cool, then filter it, and add enough water to make 1000 milliliters.

Standardization 0f 0.1 M Ceric ammonium sulphate:

Next, we make sure our Arsenic trioxide is really dry. We take 0.2 grams of it and put it in a cone-shaped flask. Then we pour 25 milliliters of a special sodium hydroxide solution inside and mix it well. We add 100 milliliters of water, 30 milliliters of watery sulphuric acid, 0.15 milliliters of osmic acid, and 0.1 milliliters of ferroin sulphate. We pour in the Ceric ammonium sulphate until the color changes from pink to pale blue.

### Calculation:

We find the strength of Ceric ammonium sulphate like this:

Molarity of cerric ammonium sulphate = W x RM / V x E
Where W is the weight of arsenic trioxide
RM is the required molarity
E is the equivalent weight factor
V is the volume of Ceric ammonium sulphate.

Molarity of Ceric ammonium sulphate = 0.2 grams x 0.1 M / 34 milliliters x 0.004946 = [Result]

Each ml of 0.1 m Ce(SO4)2 is equivalent to= 0.004946 gm of AS2O3

Report: The strength of Ceric ammonium sulfate is …

Reference: Indian Pharmacopeia Volume-1