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I'm an IB Chemistry student and I had to do a back titration lab where 1g of a powdered antacid tablet were mixed with 50mL 1M HCl, and the mixture was titrated using 0.2 M NaOH; the NaOH was to neutralize the leftover HCl that hadn't reacted with the antacid. The active ingredient in the antacid is CaCO3.

It took me about 40mL of NaOH to titrate the leftover HCl to endpoint; 40 mL at 0.2 mol/L translate to 0.008 moles of NaOH. In the equation for reaction between NaOH and HCl the ratio of the two reactants is 1:1; therefore there were 0.008 mol of unreacted HCl in the mixture.

There were 50mL of HCl (1 mol/L) in the mixture originally; 50mL * 1 mol/1000mL = 0.05 mol HCl total. Subtracting from this the molar quantity of unreacted (leftover) HCl yields: 0.05 - 0.008 = 0.042 mol HCl that did react with the CaCO3 in the antacid.

The reaction between CaCO3 and HCl is as follows:CaCO3 + 2HCl --> CaCl2 + CO2 +H2O Therefore, molar ratio between CaCO3 and HCl is 1:2. 0.042 moles HCl reacted, therefore 0.021 mol CaCO3 were present in the sample. Multiplying this by the molar mass yields 0.021mol * (40+12+48)g/mol = 2.1 grams CaCO3.

That's a problem - i can't be having 210% composition by mass (the sample was 1g; I calculated 2.1 g CaCO3 present). So somewhere along the line i made a mistake, but I've checked my math so I don't know if it was a calculation error or if my method is dead wrong? Can anyone help me with this?

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  • $\begingroup$ The calculations seem correct and the methodology should work. Could it have been some sort of experimental mishap? Also, are you sure the compound in the antiacid tablet is calcium carbonate, and that there is no other active ingredient? There could be another compound with higher capacity to absorb protons per gram than $\ce{CaCO3}$ in the mixture, such as $\ce{Mg(OH)2}$. Other than that, I'm stumped. $\endgroup$
    – Nicolau Saker Neto
    Nov 10, 2013 at 16:48
  • $\begingroup$ That's what's annoying me, the lab is titled "to find the percent of caco3 in commercial antacid by using back titration", and our teacher told us to focus only on the CaCo3 . And everyone in my class got volumes of NaOH (to endpoint) around 40mL, so I don't think it was something wrong with my experiment $\endgroup$
    – mdan
    Nov 10, 2013 at 16:51
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    $\begingroup$ That's highly suspicious then. It is entirely possible for the antiacid bought to not have any $\ce{CaCO3}$, as it may have been purchased without due attention to that detail. Is there any chance you can see the package for the antiacid, or do you know the brand? $\endgroup$
    – Nicolau Saker Neto
    Nov 10, 2013 at 17:01

1 Answer 1

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In chemical analysis usually do three times the same experimet, so you can minimize experimental errors.

After that, in order to find where is the error you can check:

Has everyone in you class use the same HCl and NaOH solutions (taken from the same bottle i.e.?

The antiacid tablets are all the same (given by your teacher)? If don't, could you post the composition?

Have you measured correctly the volumes used? To obtain that 210%, it seems tha you have used around 25 mL of HCl solution instead 50 mL.

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  • $\begingroup$ Yeah we did several trials, still got around 40 mL. And yes, everyone used the same solutions, and the antacid tablets were supplied by the teacher, but he didn't give us the label on them. It's possible that the point of the lab is to conclude that CaCO3 isn't the only active ingredient - in IB we have to evaluate the experiment and how applicable it is, so perhaps there was a second reagent, such as Mg(OH)2 as @Nicolau suggested, that couldn't be determined from such a simple experiment. $\endgroup$
    – mdan
    Nov 10, 2013 at 20:31
  • $\begingroup$ Yes, but with a molar mass of near 57, does not fully explain the volume of HCl used. If all the tablet was magnesium hydroxide, maybe could obtain such titration values. $\endgroup$
    – RFG
    Nov 10, 2013 at 20:50
  • $\begingroup$ Another thought, if your teacher told to focus on CaCO3, maybe we can discard other HCl reactive compounds. But 1 g tablet of CaCO3 contains 0,01 mol of CaCO3 wich reacts with 0,02 mol HCl. 50 mL of HCl 1 M contains 0,05 mol, so 0,03 mol didn't react. 0,03 mol HCl needs 0,03 mol NaOH (150 mL). $\endgroup$
    – RFG
    Nov 10, 2013 at 21:05
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    $\begingroup$ @mdan Yes a mixture will work, but notice that even pure $\ce{MgO}$ will almost reach the $1\ g$ limit ($0.84\ g$ actually), so while there could be other compounds, they can't be present in a very large proportion relative to $\ce{MgO}$ or else their total mass would go over the limit. This is all of course assuming every other step of the experiment was performed perfectly, only then does it become mathematically impossible for the antiacid to be mostly $\ce{CaCO3}$, or indeed practically anything other than $\ce{MgO}$. $\endgroup$
    – Nicolau Saker Neto
    Nov 11, 2013 at 2:35
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    $\begingroup$ Well, this is embarrassing. The 0.008 moles of HCl that didn't react were for a 10 mL sample; I had forgotten that for each titration with the sodium hydroxide, a 10 mL sample of the completed reaction mixture (between HCl and antacid) was used. Therefore, total unreacted moles HCl in the 50 mL that were used overall = 5 * 0.008 = 0.04 mol, subtract that from the 0.05 mol present in 50 mL and get 0.01 mol HCl that reacted. Therefore 0.005 mol CaCO3 reacted, which in turn corresponds to 0.5 grams per 1 gram, so 50% of the tablet is calcium carbonate. Sorry for causing so much trouble! $\endgroup$
    – mdan
    Nov 12, 2013 at 3:40

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