Preparation and standardization of 0.05 M Silver Nitrate EP Volumetric Solution
Dissolve 8.5 g of silver nitrate in 1000 mL carbon dioxide free water.
Dissolve 25 mg of Sodium chloride in water, add 5 mL of dilute nitric acid and dilute to 50 mL with water. Titrate with Silver nitrate solution, determining the end point potentiometrically.
1 mL of 0.05 M Silver Nitrate is equivalent to 0.5844 mg of NaCl.
Molarity (M) = [0.05 x Weight of NaCl (in mg)]/ [Titration reading (in mL) x 0.5844]
Limit: 0.05 M ± 5 % (0.0475 M – 0.0525 M)
3. PREPARATION OF INDICATORS / REAGENTS:
Dilute nitric acid:
Dilute 20 g of nitric acid to 100 mL with water
4. STORAGE CONDITION:
Store in amber glass stoppered bottles. (Protect from light)