Chemistry, asked by tbedican, 5 months ago

) 50 mL of a sample water is titrated with 0.01 M EDTA and 15 mL of EDTA is consumed. Later, 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, permanenet hardness and temporary harness (as ppm CaCO3).

Answers

Answered by DreamBlackhole
0

Answer:

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

=\frac{50*100}{50}

=\frac{50*100}{50} 50

=\frac{50*100}{50} 5050∗100

=\frac{50*100}{50} 5050∗100 ml of 0.01 EDTA=300ml of 0.01M EDTA

=\frac{50*100}{50} 5050∗100 ml of 0.01 EDTA=300ml of 0.01M EDTA=2*300ml of 0.01 N EDTA

=\frac{50*100}{50} 5050∗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

=\frac{50*100}{50} 5050∗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.

=\frac{50*100}{50} 5050∗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

=\frac{50*100}{50} 5050∗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 ppmnow 50 ml of boiled water=5 ml of 0.01 M EDTA

=\frac{50*100}{50} 5050∗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 ppmnow 50 ml of boiled water=5 ml of 0.01 M EDTA1000ml of boiled water=\frac{5*1000}{50}

50

5∗1000

5∗1000 ml of 0.01 M EDTA

ml of 0.01 M EDTA=100 mL of 0.01 M EDTA

mL of 0.01 M EDTA=200ml or 0.2 L of 0.01 N EDTA

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

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

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 eqpermanent hardness=100mg/L or ppm

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 eqpermanent hardness=100mg/L or ppmTemporary hardness=300-100= 200ppm

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