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 required. Calculate the concentration of Ca^2+( Mr=40.078g/mol) and Mg^2+ (Mr=24.305g/mol) in ppm
First, we need to calculate the moles of EDTA used in the first experiment to complex with both Ca^2+ and Mg^2+ ions.
Given:
Volume of EDTA solution = 28.61 cm^3 = 0.02861 L
Concentration of EDTA solution = 0.01010 mol/L
Moles of EDTA = Volume x Concentration
Moles of EDTA = 0.02861 x 0.01010
Moles of EDTA = 2.88861 x 10^-4 mol
Since EDTA reacts in a 1:1 ratio with both Ca^2+ and Mg^2+ ions, the moles of Ca^2+ and Mg^2+ ions complexed are also 2.88861 x 10^-4 mol each.
Next, we need to calculate the concentration of Ca^2+ and Mg^2+ ions in the 50 cm^3 of ground water sample.
1. Concentration of Ca^2+ ions:
Molar mass of Ca^2+ = 40.078 g/mol
Concentration of Ca^2+ ions = Moles of Ca^2+ / Volume of solution
Concentration of Ca^2+ ions = 2.88861 x 10^-4 mol / 0.050 L
Concentration of Ca^2+ ions = 5.777 x 10^-3 mol/L
Concentration of Ca^2+ ions in ppm = 5.777 x 10^-3 mol/L x 40.078 g/mol x 10^6 ppm/g
Concentration of Ca^2+ ions in ppm = 231.9 ppm
2. Concentration of Mg^2+ ions:
Molar mass of Mg^2+ = 24.305 g/mol
Concentration of Mg^2+ ions = Moles of Mg^2+ / Volume of solution
Concentration of Mg^2+ ions = 2.88861 x 10^-4 mol / 0.050 L
Concentration of Mg^2+ ions = 5.777 x 10^-3 mol/L
Concentration of Mg^2+ ions in ppm = 5.777 x 10^-3 mol/L x 24.305 g/mol x 10^6 ppm/g
Concentration of Mg^2+ ions in ppm = 140.5 ppm
Therefore, the concentration of Ca^2+ ions in the ground water sample is 231.9 ppm and the concentration of Mg^2+ ions is 140.5 ppm.