I\'ve already done these calculations, but I want to make sure they\'re complete
ID: 825890 • Letter: I
Question
I've already done these calculations, but I want to make sure they're completely right.
A. A solution using 0.2311 g of CaCO3 is made up in a 250.00 mL volumetric flask. If a 10.00 mL aliquot of this solution was titrated and 20.83 mL of EDTA was necessary to achieve the endpoint, what is the concentration of EDTA?
B. A chemist is given a piece of limestone and told that 50.00% of the sample exists as calcium oxide. To analyze the sample, the chemist wants to dissolve a portion of the sample in 100 mL and then titrate an aliquot (10 mL) using 0.005300 M EDTA. How many sample (in grams) is needed if a titration volume of 25.00 mL EDTA is desired
Explanation / Answer
(A) Ca2+ + EDTA4- => Ca(EDTA)2-
Total moles of Ca2+ = moles of CaCO3 = mass/molar mass of CaCO3
= 0.2311/100.0875 = 0.0023090 mol
Moles of EDTA = moles of Ca2+ in 10.00 mL aliquot
= 10.00/250.00 x total moles of Ca2+
= 10.00/250.00 x 0.0023089 = 9.2359 x 10^(-5) mol
Concentration of EDTA = moles/volume of EDTA
= 9.2359 x 10^(-5)/0.02083
= 4.434 x 10^(-3) M (= 0.004434 M)
(B) Ca2+ + EDTA4- => Ca(EDTA)2-
Moles of CaO in 10 mL alquot = moles of Ca2+ in 10 mL aliquot = moles of EDTA
= volume x concentration of EDTA
= 25.00/1000 x 0.005300 = 0.0001325 mol
Moles of CaO in sample = 100/10 x moles of CaO in 10 mL aliquot
= 100/10 x 0.0001325 = 0.001325 mol
Mass of CaO in sample = moles x molar mass of CaO
= 0.001325 x 56.078 = 0.0743 g
Mass of sample = 100/50.00 x mass of CaO
= 100/50.00 x 0.0743
= 0.1486 g