Problem 1 : A 0.5118 g sample of CaCO3 is dissolved in 12 M HCl and the resultin

Question

Problem 1: A 0.5118 g sample of CaCO3 is dissolved in 12 M HCl and the resulting solution is diluted to 250.0 mL in a volumetric flask. (formula mass = 100.1)

25.00 mL aliquot of the solution from problem 1 is titrated with EDTA to the Eriochrome Black T end point. It requires 30.16 mL of the EDTA to reach the end point.

A.) How many moles of EDTA are there in the aliquot? (moles)

B.) What is the molarity of the EDTA solution? (M)

100 mL sample of hard water is titrated with the EDTA solution in above problem and the volume of the EDTA required is 17.64 mL.

C.) How many moles of EDTA are there is that volume? (moles)

D.) How many moles of Ca^2+ are there in the 100 mL of water? (moles)

E.) If the Ca^2+ comes from CaCO3, how many moles of CaCO3 are there in one liter of the water? (moles)

How many grams CaCO3 per liter? (grams)

F.) If 1 ppm CaCO3 = 1 mg per liter, what is the water hardness in ppm CaCO3? (ppm CaCO3)

Explanation / Answer

moles of CaCO3 = 0.5118/100.1 = 0.005 mols

molarity of CaCO3 solution = moles/L = 0.5118/100.1 x 0.250 = 0.02 M

A) moles of EDTA = moles of CaCO3

moles of CaCO3 = 0.02 M x 0.025 L = 0.0005 mols

moles of EDTA = 0.0005 mols

B) molarity of EDTA in 30.16 ml = 0.0005/0.03016 = 0.0166 M

moles = 0.0166 x 0.01764 = 0.0003 mols

Total volume of solution = 100 + 17.64 = 117.64 ml = 0.1176 L

Molarity of EDTA = 0.0003/0.1176 = 0.0025 M

C) moles of EDTA in the volume = 0.0003 mols

D) moles of Ca2+ in 100 ml water = 0.0003 mols

E) moles of CaCO3 in 100 ml = 0.0003 mols

so moles of Ca2+ in 1L = 0.003 mols

moles of CaCO3 = 0.003 mols/L

grams of CaCO3 = 0.003 x 100.1 = 0.3 g

F) water hardness in ppm = 0.3 x 1000/L = 3000 ppm

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