For problem 5, the density of the sample is 0.9977 g/mL and the molar mass of calcium carbonate is 100.0 g/mol. 5. As described in Topic 2, the hardness of water is expressed in terms of amount of calcium carbonate according to the ppm expression. a. Calculate the mass of calcium carbonate present in a 50.00 mL sample of an aqueous calcium carbonate standard
For problem 5, the density of the sample is 0.9977 g/mL and the molar mass of calcium carbonate is 100.0 g/mol.
5. As described in Topic 2, the hardness of water is expressed in terms of amount of calcium carbonate according to the ppm expression.
a. Calculate the mass of calcium carbonate present in a 50.00 mL sample of an aqueous calcium carbonate standard, assuming the standard is known to have a hardness of 75.0 ppm (hardness du to CaCO3).
b.Calculate the number of moles of calcium ions present in this 50.00 mL sample of aqueous calcium carbonate standard.
c. The 50.00 mL sample (at pH 10) was titrated with a 0.00500 M EDTA solution,
i. Write the balance net ionic equation for the reaction of EDTA4- with a calcium ion.
ii. Calculate the number of moles of EDTA4- needed to completely react with calcium ions in the sample.
iii.What volume (in milliliters) of EDTA solution would be needed to reach the equivalence point?
This problem was partly answered but I am wondering what the answers/work are for problem c- ii and iii
Density of sample = 0.9977g/ml
mass of 50ppm sample = 0.9977g/ml × 50ml = 49.885g = 0.049885kg
Concentration of CaCO3(Hardness) = 75ppm = 75mg/kg
Molar mass of CaCO3 = 100g/mol
Molar mass of Ca = 40.08g/mol
CaCO3 to Ca conversion factor = 40.08/100= 0.4008
Concentration of Ca2+ = 0.4008 × 75= 30.06ppm = 30.06mg/kg
mass of Ca2+ = 30.06mg/kg × 0.049885kg = 1.4995mg = 0.0014995g
Number of mol of Ca2+ = 0.0014995g/40.08g = 3.74 ×10-5 mol
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