I think I have problem 2 it is just 3 and 4 that I really need help on. I am re-
ID: 633125 • Letter: I
Question
I think I have problem 2 it is just 3 and 4 that I really need help on. I am re-asking this question because I think I made a mistake the first time submitting it. I also made it easier to read
2.) A 0.5102 g sample of CaCO3 is dissolved in concentrated HCl (12M). The solution was diluted to 250.0mL in volumetric flask to obtain a solution for standardizing the EDTA titrant.
a.) How many moles of CaCO3 were used?
b.) what is the molarity of the CaCO3 in the flask?
c.) what is the molarity of the Ca2+ ions in flask?
d.) how many moles of Ca2+ are in a 25.00 mL aliquot of the solution in 2c?
3.) A blank containing a small measured amount of Mg2+ requires 2.10 mL of EDTA to reach the end point. A 25.00 mL aliquot of the solution from problem 2, which contains the same amount of Mg2+ as the blank, requires 13.48 mL of EDTA to reach end point.
a.) how many milliliters of EDTA are needed to titrate the Ca2+ ions in the aliquot?
b.) what is the mole ratio of Ca2+ ions to EDTA in solution?
c.) given answer in part b, how many moles of EDTA have been dispensed into the flask?
d.) what is the molarity of the EDTA solution?
4.) A 100.0 mL sample of water with unknown amount of calcium ions is titrated with the EDTA solution from problem 3. The same amount of Mg2+ is added as before. The water sample requires 12.48 mL of EDTA to reach end point.
a.) what volume of EDTA is used to titrate the Ca2+ in the sample?
b.) how many moles of EDTA are present in this volume?
c.) how many moles of Ca2+ are in the 100 mL sample of water?
d.) If the Ca2+ originates from calcium carbonate, how many moles of CaCO3 are present in one liter of water? How many grams of CaCO3 are in one liter? ____moles/L ___g/L
e.) If 1ppm = 1mg per L, what is the water hardness in ppm of CaCO3 ? Is this water considered "hard'?
Explanation / Answer
moles CaCO3 = 0.5118 g/ 100.1 g/mol=0.005113
[Ca2+]= 0.005113/ 0.250 L=0.02045 M
moles Ca2+ = 0.02045 x 0.02500 L=0.0005113