3198.6 mg CaCO3 in antacid?

Question

In this lab I was given an unknown antacid tablet and asked to determine the moles of base. First I crushed the tablet, which weighed 2.2650g and took 3 portions weighing 1)0.5880g, 2)0.5820g and 3)0.5737g. Then I added .1M HCl until the pH reached a point lower than 2.5 to neutralize the antacid. The total volumes of HCl added were 1)45.1mL, 2)45.0mL, and 3)49.0mL. Then I did a back titration of the excess HCl w/ .1M NaOH. The equivalence points were reached when the following amounts of NaOH were added: 1)4.30mL, 2)4.20mL, and 3)4.50mL. Then I had to calculate moles base in sample (3 trials averaged 0.00420), moles base in tablet (average of 0.0164), mg CaCO3 in sample (820.738), and mg CaCO3 in tablet. The average mg CaCO3 in tablet from the 3 trials I calculated to be 3198.6, but that seems really high. I'm curious if and where I made a mistake and if this amount of CaCO3 in an antacid tablet is even possible.

Answer 1

Of course it is not possible for a 2.2650g antacid tablet to contain 3198.6g of CaCO3. I have checked the formula of an antacid that I have at home, which contains CaCO3 680mg and MgCO3 80mg.Mass of each tablet 1.25g

Let us check your chemistry, concentrating only on series 1. You have

1) 0.5880g antacid

2.added 45.1ml 0.1M HCl

3. back titrated with 4.30ml 0.1M NaOH

HCl + NaOH → NaCl + H2O

Mol HCl : Mol NaOH = 1:1

From these figures, 45.1ml - 4.30 = 40.8ml 0.1M HCl was used to neutralise the antacid .That is (40.8/1000) *0.1* 36.461 = 0.1488g HCl.

2HCl + CaCO3 → CaCl2 + H2O + CO2

2mol of HCl required to neutralise 1 mol CaCO3

Molar mass HCl = 1.008 + 35.453 = 36.461g

Molar mass CaCO3 = 40.078+12.011+(5.999*3) = 100.086

72.922g HCl will react with 100.086g CaCO3

0.1488g HCl will react with 100.086/72.922*0.1488 = 0.2042g CaCO3

Your original sample of 0.5880g antacid contained 0.2042g CaCO3

The tablet, mass 2.265g will therefore contain:

2.265/0.588*0.2042 = 0.786g CaCO3

You can do this for the other 2 test series.

.