Question

the density of the sample is 0.9977 g/mL and the molar mass of calcium carbonate is 100.0 g/mol.
2. Calculate the number of moles of calcium ions present in a 50.00 mL water
sample that has a hardness of 75.0 ppm (hardness due to CaCO3).
3. If the 50.00 mL sample from problem 2 above was titrated with a 0.00500 M
EDTA, what volume (in milliliters) of EDTA solution would be needed to reach
the endpoint?

If someone could show their work and help me out that would be great. I always seem to get hung up on setting the problem up.

Thank you.

Answer 1

2)

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

No of mol of Ca2+ = 0.0014995g/40.08g = 3.74×10^-5mol

2) Stoichiometrically , 1mol of Ca2+ react with 1 mol of EDTA

we obtained no of mole o Ca2+ present in the 50ml samle = 3.74×10^-5mol

So, 3.74×10^-5mol of EDTA required

Molarity of EDTA = 0.0050M = 0.0050mol per Litre

Volume of EDTA required = (1000ml/0.0050mol)×0.0000374= 7.48ml