An EDTA solution is made by dissolving $0.930 g$ of disodium EDTA dihydrate (molar mass $372$ ) in water to $250 mL$ . $50.0 mL$ of raw water needs $14.2 mL$ of this EDTA. Another $50.0 mL$ is boiled, the precipitate removed, the volume restored to $50.0 mL$ and then needs $6.8 mL$ . Find the temporary hardness in ppm of $CaCO X_{3}$ (molar mass 100; EDTA 1 : 1 with hardness ions), to the nearest integer.

Numerical answer type, answer in ppm, tolerance 1.

Answer: 148 ppm

Molarity of EDTA = (0.930/372)/0.250 = 0.0100 M. Temporary hardness corresponds to (14.2 - 6.8) = 7.4 mL, i.e. 7.400e-05 mol in 50.0 mL, or 1.480e-03 mol/L. Mass as CaCO3 = 0.1480 g/L = 148.0 mg/L, i.e. about 148 ppm.

Practice Equivalent Concept and Titration More on Back titration and hardness of water

Previous questionNext question

Answer: 148 ppm

More questions on Back titration and hardness of water