Part 1 1. A 0.5100-g sample of CaCO3 is dissolved in 12 M HCl and the resulting
ID: 1004867 • Letter: P
Question
Part 1
1. A 0.5100-g sample of CaCO3 is dissolved in 12 M HCl and the resulting solution is diluted to 250.0 mL in a volumetric flask.
a. How many moles of CaCO3 are used (MW=100.1 g/mol)? .5100g x (1 mol/100.1g)= 0.005095 mols
b. What is the molarity of the Ca^2+ in the 250 mL of solution? .005095 mols/.250L= 0.02038M
c. How many moles of Ca^2+ are in a 25.0-mL aliquot of the solution in b? .02038M x 0.025L= 5.095 x 10^-4 2.
Part 2
25.00-mL aliquots of the solution in problem 1 are titrated with EDTA to the calmagite end point. A blank containing a small measured amount of Mg^2+ requires 2.50 mL of the EDTA to reach the end point. An aliquot to which the same amount of Mg^2+ is added requires 28.55 mL of the EDTA to reach the end point.
a. How many mL of EDTA are needed to titrate the Ca^2+ ion in the aliquot? 28.55 mL-2.5 mL= 26.05 mL
b. How many moles of EDTA are there in the volume obtained in a? .005095 mols (same as mols CaCO3?)
c. What is the molarity of the EDTA solution? .005095 mols/.02605L= .1956M 3.
Part 3:
A 100-mL sample of hard water is titrated with the EDTA solution in problem 2. The same amount of Mg^2+ is added, and the volume of EDTA required is 22.51 mL.
a. What is the volume of EDTA used in titrating the Ca^2+?
b. How many moles of EDTA are there in that volume?
c. How many moles of Ca^2+ are there in 100 mL of water?
d. if Ca 2+ comes from CaCo how many moles if CaCo3 are there in one liter of water ? How many grams of Caco3 are present per liter of water?
e. if 1 ppm CAco3= 1mg per liter, wht is the water hardness in ppm CaCo3?
can someone help me solve part 3.
Explanation / Answer
Part 3:
a) Volume of EDTA used in titrating Ca2+ = (22.51 – 2.50)mL = 20.01 mL
b) From (c) of part 2 above, we have calculated the molarity of EDTA to be 0.1956 M.
Therefore, moles of EDTA present in 20.01 mL EDTA solution = (20.01 mL/1000 mL)*(1 L)*(0.1956 moles/L) = 3.914*10-4 mole
c) Ca2+ and EDTA react as
Ca2+ + H2EDTA2- ---------> CaEDTA2- + 2 H+
Ca2+ and EDTA have a 1:1 molar ration, hence the amount of Ca2+ present in the unknown solution = 3.914*10-4 mole
d) 100 mL of the unknown solution contains 3.914*10-4 mole Ca2+.
Therefore,1 L = (1000 mL) of the solution will contain (3.914*10-4 mole)*(1000 mL/100 mL) = 3.914*10-3 mole Ca2+.
The molar mass of CaCO3 is 100 gm/mole.
Therefore, weight of CaCO3 present in 1 L of the unknown solution = (3.914*10-3 mole)*(100 gm/1 mole) = 3.914*10-2 gm.
e) Mass of CaCO3 present in 1 L of the unknown solution = 3.914*10-2 gm = (3.914*10-2 gm/10-3 gm)*(1 mg) = 39.14 mg (since 1 mg = 10-3 gm)
Since 1 ppm CaCO3 = 1 mg CaCO3 per litre, the hardness of the unknown sample of water is 39.14 ppm (ans)