004.8 9hol) in the sample. 7-28.) An EDTA solution was prepared by dissolving ap

Question

004.8 9hol) in the sample. 7-28.) An EDTA solution was prepared by dissolving approx- imately 4 g of the disodium salt in approximately 1 L of water. An average of 42.35 mL of this solution was required to titrate 50.00-mL aliquots of a standard that contained 0.7682 g of MgCO, per liter. Titration of a 25.00-mL sample of mineral water at pH 10 re- quired 18.81 mL of the EDTA solution. A 50.00-mL aliquot of the mineral water was rendered strongly alkaline to precipitate the magnesium at Mg(OH)2- Titration with a calcium-specific indicator required 31.54 mL of the EDTA solution. Calculate (a) the molar concentration of the EDTA solution.0.o (b) the concentration of CaCO, in the mineral water in ppm. (c) the concentration of MgCO, in the mineral wa ter in ppm.

Explanation / Answer

(a) Concentration of EDTA solution = moles/L = 9.11 x 10^-3 x 0.050/0.04235 = 0.26 M

(b) moles of EDTA required for titrating CaCO3 in solution = 0.26 x 0.03154 = 8.2 x 10^-3 mols

moles of CaCO3 in solution = 8.2 x 10^-3 mols

Concentration of CaCO3 in mineral water = 8.2 x 10^-3 x 100.0869 x 1000/0.05 = 16414.25 ppm

(c) Concentration of MgCO3

std MgCO3 concentration = 0.7682/84.3139 x 1 = 9.11 x 10^-3 M

moles of MgCO3 in 50 ml = 9.11 x 10^-3 x 0.050 = 4.55 x 10^-4 mols

moles of EDTA = 4.55 x 10^-4 mols

Molarity of EDTA in solution = 4.55 x 10^-4/0.04235 = 0.011 M

So molarity of EDTA in 1 L = 0.26 M

moles of EDTA used for 25 ml mineral water = 0.26 x 0.01881 = 4.891 x 10^-3 mols

So moles of MgCO3 = 4.891 x 10^-3 mols in solution

molarity of MgCO3 in 25 ml sample = 4.891 x 10^-3/0.025 = 0.1956 M

concentration of MgCO3 in mineral water = 0.1956 x 84.3139 x 1000 = 16492 ppm

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