Question

7. 15 g of CaCO3 was dissolved in HCl and the solution is made upto one litre with distilled water. 20mL of the above standard hard water required 25 mL of EDTA solution on titration.100 mL of hard water sample required 18 mL of same EDTA solution on titration.100 mL of this water, after boiling, cooling and filtering required 12 mL of EDTA solution on titration. Calculate the temporary and permanent and total hardness of water.

Answer 1

Solution:

Weight of $CaCO_3$ = 15g

Volume of solution = 1L

Concentration of solution of  $CaCO_3$  =  $\dfrac{15}{1000} g/mL$

Concentration of solution of  $CaCO_3$  = 15 mg/mL

20 mL of Standard Hard Water (S.H.W) requires volume of EDTA = 25mL

1 mL of EDTA is required for volume of S.H.W =  $\dfrac{20}{25} mL$

100 mL of hard water sample required 18 mL of same EDTA solution on titration.

$\therefore \text {Hardness of water per 100 mL} = 18\times \dfrac{20}{25}$

$\text {Hardness of water per Litre} = 18\times \dfrac{20}{25}\times \dfrac{1000}{100}$

Total harness of water = 144 ppm

100 mL of this water, after boiling, cooling and filtering required 12 mL of EDTA solution on titration.

$\therefore \text {Permanent Hardness of water per 100 mL} = 12\times \dfrac{20}{25}$

$\text {Permanent Hardness of water per Litre} = 12\times \dfrac{20}{25}\times \dfrac{1000}{100}$

Permanent hardness of water per Litre = 96 ppm

Temporary hardness = Total hardness - Permanent hardness

Temporary hardness = 144 - 96

Temporary hardness = 48 ppm