This question is about a chemistry lab experiment: Find the Ca 2+ concentration

ID: 905950 • Letter: T

Question

This question is about a chemistry lab experiment:

Find the Ca2+ concentration of kidney stones dissolved in HCl

Disolved 0.1g of kidney stone CaCO3 in 10mL HCl and then diluted the sample with distilled water till it reached 250mL in a volumentric flask.

Did the exact same except with kidney stone Ca3(PO4)2.

6 trials were titrated with EDTA. Each sample was 10mL from the respective volumentric flask.

Standardized EDTA: 0.077M

Titration Results:

Stone CaCO3: 0.90mL EDTA, 0.65mL EDTA, 1.05mL EDTA

Stone Ca3(PO4)2: 0.75mL EDTA, 0.40mL EDTA, 0.60mL EDTA

Find the Ca2+ concentration in ppm

The balanced equation for titration is

Ca+2 + EDTA-4-----> Ca(EDTA)-2

to calculate the ppm of Calcium

moles of Ca ion = [VEDTA(L) x MEDTA] = 0.90x 0.077 /1000 = 6.93 x 10-5 moles

6.93 x 10-5 moles of Ca in 10 ml solution.

thus in 250 mL the moles of ca is 250 x 6.93 x 10-5 moles/ 10 = 173.25 x10-5 = 1.73 x10-3 moles

weight of ca in 250 ml = 1.73 x 10-3 x 40 =0.692 g

0.692 g in 250 ml(g) hence in 106 g the amount of ca is 0.692 x106 / 250 =276.8 ppm

similarly for the other two

with 0.65ml Ca in ppm =[ 0.65x0.077x250x40x106] / [1000x 10x250] = 200.2

with 1.05mL

Ca in ppm = [1.05x0.077x250x40x106] / [1000x 10x250] = 323.4 ppm

Similarly for calcium phosphate soutions

Ca inppm is given by Ca3(PO4)2 + 3 EDTA-4 ----> 3 Ca(EDTA)-2 + 2 PO4-2
Thus moloes of calcium = Vx M/3

for 0.75mL

moles of ca = 0.75x 0.077 /1000x3 =1.925 x 10-5 mol

in ppm = 107.8

Similoarly for 0.40mL , Ca in ppm = 41.06

and for 0.60 ml Ca in ppm = 61.6

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