Question

A sample of water on analysis contains 13.6 mg/ litre of calcium sulphate, 14.6 mg/ litre of magnesium bicarbonate, 12 mg/ litre of magnesium sulphate. Calculate the total hardness of the sample.

Answer 1

Answer:

The total hardness of the given sample is 39.95mg/L or ppm.

Explanation:

The equation to find the hardness of a given sample is given by

Mass of hardness producing substance × Chemical equivalence of CaCo3 ÷ chemical equivalence of hardness producing substance.

The chemical equivalence of CaCo3 is 50

• For calcium sulphate (CaSo4):

Mass of hardness producing substance

= 13.6mg/L

chemical equivalence of hardness producing substance

= 68

so applying the equation:

13.6 × 50 ÷ 68

= 10 mg/L or ppm

• For magnesium bicarbonate Mg(Hco3)2:

14.6 × 50 ÷ 36.585

= 19.95 mg/L

• For magnesium sulphate MgSo4 :

12 × 50 ÷ 60

= 10 mg/L

• The total hardness is given by

temporary hardness + permanent hardness

Temporary hardness is caused by bicarbonates

Permanent hardness is caused by Ca, Mg, Sulphates and any other heavy elements.

So Temporary hardness is

= 19.95mg/l

permanent hardness is

= 10mg/l + 10mg/l

therefore total hardness

= 19.95 + 10 + 10

= 39.95mg/l

Therefore Total hardness is 39.95mg/L