Question

The water sample (40 mL) requires 9.6 mL of 0.01 M EDTA solution

Answer 1

Answer:

2

M1V1=M2V2

 50×M1=501×0.1

M1(HCO3−)=25×1031

mmoles of (HCO3−)=25×1031×50=5001

mmoles of Ca2+=10001

mmoles of CaCO3=10001

weight of CaCO3=10001×10−3×100=10−4g

Hardness = [weight of CaCO3 /weight of water] ×106=5010