Mohr Method by using Silver Nitrate

MOHR'S METHOD

Principle

Chlorides are estimated as sodium chloride by titration against silver nitrate using potassium chromate as indicator. Silver nitrate is added until all the chloride ions present in the SAMPLE WATER have combined with the silver ions. Any further silver nitrate added is now free to combine with the potassium chromate indicator to yield a red precipitate of silver chromate. The solution at this point suddenly changes from pale yellow to faint brick-red colour.

NaCl + AgNO3→AgCl + NaNO3

Reagents

1.

Standard silver nitrate solution (2.906 g of silver nitrate is dissolved in and made up to 1 litre with distilled water). Keep in an amber bottle and standardize against an accurately prepared sodium chloride solution containing 500 mg NaCl per 100 ml using the method described below. Check at frequent intervals, and determine the factor.

1 ml standard AgNO3 equal to 1 mg NaCl

2. 5 per cent potassium chromate solution.

Procedure : 

1. Pipette 1.0 ml SAMPLE WATER into a conical flask containing about 10 ml of distilled water and 2–3 drops of potassium chromate.

2. Slowly add the silver nitrate from a 10 ml burette and continuously rotate the flask, to ensure thorough mixing.

3. As the end-point nears, the silver nitrate solution should be added slowly and carefully to avoid adding excess of the reagent.

4. Note the titre and repeat the titration again if sufficient SAMPLE WATER is available.

5. Normal fluid usually requires between 7.0 and 7.6 ml of silver nitrate.

Reaction :
NaCl + AgNO3 = AgCl + NaNO3

From the above equation it can be seen that 58.5 g of sodium chloride are equivalent to 170 g of silver nitrate.

Therefore 58.5 mg NaCl ≡ 170 mg AgNO3

Now the standard solution of AgNO3 contains 2.906 g per litre or 2.906 mg per ml

Hence 2.906 mg or 1 ml of standard AgNO3 corresponds to
2.906 X  ( 58.5 / 170 mg of NaCl )
1 ml AgNO3 = 1 mg NaCl

If the titre was 7.3 ml of AgNO3 (or 7.3 ml × factor), then mg NaCl per 100 ml CSF =   titre X 1 X ( 100 / Amount Taken )
         7.3 X 1 X 100 = 730 mg per 100 ml

Notes—

(a) It is now desirable to express the NaCl value in terms of mmol per litre, in which case the following formula is used.

mmol per litre =  ( mg per 100 ml X 10 ) / molecular weight
                       = ( 730 X 10 ) / 58.5 = 124.8

(b) A solution of AgNO3 containing 5.812 g/litre can be used in which case 1 ml of AgNO3 = 2 mg NaCl.