A chemist measured the amount of calcium carbonate present in an antacid tablet

Question

A chemist measured the amount of calcium carbonate present in an antacid tablet as follows. The tablet would not dissolve in water, but dissolve completely in 25.0mL of 0.500 M HCl after boiling to remove the CO2 gas.

CaCO3(s) + 2HCl (aq) --> CO2 (g) + H2O (l) + CaCl2(aq)

The amount of HCl acid added was more than enough to ract with all the CaCO3 base. The amount by which the acid was in excess was determined by backtitrating the remaining HCl with 0.1500 M NaOH, requiring 16.07 mL of the NaOH solution to reach the end point. How many milligrams of CaCO3 were in the antacid tabet?

HCl + NaOH --> H2O + Na+ + Cl-

Explanation / Answer

CaCO3 (s) + 2HCl (aq) --> CO2 (g) + H2O (l) + CaCl2(aq)

V = 25 ml of M = 0.5 HCL

second titration:

M = 0.15 NaOH

V = 16.07 ml

M = ?

V = 25 ml

find the concnetration of HCl in second titration first

1 mol of base : 1 mol of acid therefore:

MV acid = MV base

M = MV/V = (0.15*16.07/25) = 0.0964 M of Acid left

since original was 0.50 M of HCl

Calculate change

0.5 - 0.0964 = 0.4036 M

but we used 25 ml, let us find moles:

mol = M*V = 0.4036*25 = 10.09 mmol of HCl was used to neutralize

now...

ratio of CaCO3 and HCl

1 mol of CaCO3 needs 2 mol of HCl

therefore:

1/2 is the ration

1/2*10.09 = 5.05 mmol of CaCO3 present

MW of CaCO3 = 100.01 g/mol

mass = 5.05*100.01 = 505 mg of CaCO3

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