50 mL of a sample water is titrated with 0.01 M EDTA and 15 mL of EDTA is consumed. Then, the same sample water (50 mL) is boiled and again titrated with 0.01 M EDTA. If consumption of EDTA is 5 mL for boiled water, calculate the degree of total hardness, permanent hardness and temporary hardness (as ppm CaCO3).
Answers
Answer:
50ml of water sample
=15ml of 0.01M EDTA
= 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= 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
permanent hardness=100mg/L or ppmTemporary hardness=300-100 = 200 ppm
Answer:
Explanation:
50ml of water sample
=15ml of 0.01M EDTA
= 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= 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
permanent hardness=100mg/L or ppmTemporary hardness=300-100 = 200 ppm