4. A chemist measured the amount of calcium carbonate present in an antacid tabl

Question

4. A chemist measured the amount of calcium carbonate present in an antacid tablet as follows. The tablet would not dissolve in water, but dissolved completely in 25.0 mL of 0.5 00 M HCl after boiling to remove the CO2 gas. The amount of HCl acid added was more than enough to react with all of the CaCO3 base. The amount by which the acid was in excess was determined by back titrating the remaining HCl with 0.1500 M NaOH, requiring 16.07 mL of the NaOH solution to reach the endpoint. How many milligrams of CaCO3 were in the antacid tablet?

Explanation / Answer

Moles of acid added initially to the antacid = 0.5*25/ 1000 = 0.0125 moles

moles of NaOH required for back titration = 0.15 *16.17/1000 = 0.0024 moles

amount of HCl used to titrate CaCO3 pressent in the antacid = 0.0125mol-0.0024 = 0.0101 moles

from the reaction 1 mole CaCO3 reacts with 2 moles of HCl

moles of CaCO3 present = 0.0101/2 = 5.05 *10-3 mol

amount of caCO3 present = moles * molar mass = 5.05 *10-3 mol * 100gmol-1 = 5.05 g

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