Question

50 mL of a water sample consumed 15 mL of 0.01M EDTA before boiling and 5 mL of

the same after boiling. 50 mL of the standard hard water (1000 mg/L) consumed 20 mL

of EDTA. Calculate the degree of permanent, temporary and total hardness in terms of

CaCO3 equiv.

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Answer 1

Answer:

50ml of water sample=15ml of 0.01M EDTA

=$\frac{50*100}{50}$ ml of 0.01 EDTA=300ml of 0.01M EDTA

=2*300ml of 0.01 N EDTA

=600 ml or 0.6 L of 0.01 eq. of CaCO3

=0.6*0.01*50g CaCO3 eq.

Then the total hardness=0.30g or 300mg of CaCO3 eq=300mg/L or ppm

now 50 ml of boiled water=5 ml of 0.01 M EDTA

1000ml of boiled water=$\frac{5*1000}{50}$ ml of 0.01 M EDTA

=100 mL of 0.01 M EDTA

=200ml or 0.2 L of 0.01 N EDTA

=0.2*0.01*50 g of CaCO3 eq

=0.1 g or 100mg of CaCO3 eq

permanent hardness=100mg/L or ppm

Temporary hardness=300-100= 200ppm

Answer 2

Answer:

50ml of water sample=15ml of 0.01M EDTA

=\frac{50*100}{50}

50

50∗100

ml of 0.01 EDTA=300ml of 0.01M EDTA

=2*300ml of 0.01 N EDTA

=600 ml or 0.6 L of 0.01 eq. of CaCO3

=0.6*0.01*50g CaCO3 eq.

Then the total hardness=0.30g or 300mg of CaCO3 eq=300mg/L or ppm

now 50 ml of boiled water=5 ml of 0.01 M EDTA

1000ml of boiled water=\frac{5*1000}{50}

50

5∗1000

ml of 0.01 M EDTA

=100 mL of 0.01 M EDTA

=200ml or 0.2 L of 0.01 N EDTA

=0.2*0.01*50 g of CaCO3 eq

=0.1 g or 100mg of CaCO3 eq

permanent hardness=100mg/L or ppm

Temporary hardness=300-100= 200ppm