28.61cm^3 of 0.01010 molar EDTA solution was required to complex completely with Ca^2+ and Mg^2+ ions in 50 cm^3 ground water.The amount id EDTA required to complex only Ca^2+ ions was determined in a second experiment. 5-10 pellets KOH was added to a fresh 50 cm^3 ground water sample and the solution was allowed to stand for 5-10 minutes. In the subsequent titration ,6.62 cm^3 of the same EDTA was required. Calculate the concentration of Ca^2+( Mr=40.078g/mol) and Mg^2+ (Mr=24.305g/mol) in ppm

Question

28.61cm^3 of 0.01010 molar EDTA solution was required to complex completely with Ca^2+ and Mg^2+ ions in 50 cm^3 ground water.The amount id EDTA required to complex only Ca^2+ ions was determined in a second experiment. 5-10 pellets KOH was added to a fresh 50 cm^3 ground water sample and the solution was allowed to stand for 5-10 minutes. In the subsequent titration ,6.62 cm^3 of the same EDTA was required. Calculate the concentration of Ca^2+( Mr=40.078g/mol) and Mg^2+ (Mr=24.305g/mol) in ppm

Answer 1

First, we need to determine the amount of EDTA used to complex with both Ca^2+ and Mg^2+ ions in the first experiment.

Given:
Volume of EDTA solution used = 28.61 cm^3
Molarity of EDTA solution = 0.01010 mol/L
Volume of groundwater sample = 50 cm^3

Using the formula:
M1V1 = M2V2

(0.01010 mol/L) * V1 = (0.01010 mol/L) * 28.61 cm^3
V1 = 28.61 cm^3

Now, in the second experiment, we need to determine the amount of EDTA used to complex with only Ca^2+ ions. We can calculate this by finding the difference in the amount of EDTA used in the first and second experiments.

Volume of EDTA used in the second experiment = 6.62 cm^3
Amount of EDTA used to complex with only Ca^2+ ions = 28.61 cm^3 - 6.62 cm^3 = 21.99 cm^3

Since Ca^2+ has a molar mass of 40.078 g/mol, we can calculate the concentration of Ca^2+ ions in the groundwater sample.

Concentration of Ca^2+ ions = (0.01010 mol/L) * (21.99 cm^3) / 50 cm^3 = 0.00442 mol/L

To convert the concentration to ppm, we can use the formula:

Concentration in ppm = (Concentration in g/L / Molar mass) * 10^6

Concentration of Ca^2+ ions in ppm = (0.00442 g/L / 40.078 g/mol) * 10^6 = 110.21 ppm

To find the concentration of Mg^2+ ions in the groundwater sample, we can use the fact that Mg^2+ ions are only present in the excess of EDTA used in the first experiment.

Amount of EDTA used to complex with both Ca^2+ and Mg^2+ ions = 28.61 cm^3
Amount of EDTA used to complex with only Ca^2+ ions = 21.99 cm^3

Amount of EDTA used to complex with only Mg^2+ ions = 28.61 cm^3 - 21.99 cm^3 = 6.62 cm^3

Concentration of Mg^2+ ions = (0.01010 mol/L) * (6.62 cm^3) / 50 cm^3 = 0.00133 mol/L

Concentration of Mg^2+ ions in ppm = (0.00133 g/L / 24.305 g/mol) * 10^6 = 54.69 ppm

Therefore, the concentration of Ca^2+ ions in the groundwater sample is 110.21 ppm, and the concentration of Mg^2+ ions is 54.69 ppm.