Question

water sample in analysis give the following results : Ca(HCo3)2 =105ppm, Mg(HCo3)2 = 125ppm, CaSo4 = 75ppm, CaCl2 = 82ppm, MgSo4 = 26ppm. Calculate the temporary hardness and permanent hardness in °Fr.​

Answer 1

Answer:

motor se Ambala City to PPM mgso4 26 PPM will be calculated in Fahrenheit

Answer 2

Given:

$Ca(HCO_3)_2$ $= 105ppm$

$CaSO_4 = 75ppm\\Mg(HCO_3)_2$$= 125ppm$

$CaCl_2= 82ppm\\MgSO_4 = 26ppm$

To find:

Temporary and permanent hardness of water

Explanation:

• Temporary hardness is due to the carbonate salts while permanent hardness is due to the non-carbonate salts.
• In this case ; temporary hardness will be given by :

Sum of hardness due to carbonates :

Calcium and magnesium carbonates : $105ppm + 125 ppm$ $= 230ppm$

To calculate the hardness in french degrees (°Fr) :

1°Fr $=10ppm$

$230ppm = 23.0$ °Fr ( Temporary hardness )

• Permanent hardness of water is given by :

sum of hardness due to non carbonates:

$75 ppm + 82 ppm + 26 ppm = 183 ppm$

$183 ppm = 18.3$ °Fr (permanent hardness)