Chemistry HW question

Question

NOTE: You may never use the dilution equation in place of stoichiometry for a reaction.

1) A 10.00 mL aliquot of 0.010 M CaCO3 is titrated with 18.22 mL of EDTA solution. What is the EDTA molarity?

2) A 250.0 mL water sample requires 30.85 mL of the EDTA solution from question 1 to reach the calmagite endpoint. What was the molarity of the hard metal ions in the water sample?

3) If the metal ions in the water sample are assumed to be Ca2+ from CaCO3, express the concentration in ppm CaCO3. This is the same as the concentration in units of mg CaCO3 per liter of sample.

Answer 1

1) Assume a 1:1 mol ratio of EDTA : Ca²⁺

moles Ca²⁺ = 10.00 ml x 1 L / 1000 ml x 0.010 mol / L = 1.0x10^-4 mols Ca²⁺ = 1.0x10^-4 mol EDTA

[EDTA] = 1.0x10^-4 mol / 18.22 ml x 1000 ml / L = 5.49x10^-3 M EDTA

2) Assume a 1:1 mol ratio of EDTA : metal

mols EDTA used = 30.85 ml x 1 L / 1000 ml x 5.49x10^-3 mol / L = 1.69x10^-4 mols EDTA = moles metal

Molarity of metal = 1.69x10^-4 mols metal / 250 ml x 1000 ml / L = 6.77x10^-4 M metal

3) 6.77x10^-4 M Ca²⁺ = 6.77x10^-4 M CaCO₃

molar mass CaCO₃ = 100.0 g / mole

grams CaCO₃ present in 1 liter = 6.77x10^-4 mol / L x 100.0 g / mol = 0.0677 g CaCO₃ / liter

ppm = mg / L so converting 0.0677 g / L to mg / L we have...

67.7 mg / L = 67.7 ppm CaCO₃