Hello everyone. Well, I did a back titration lab a few days ago and I'm trying to do the calculations now and keep running into problems. I'm trying to find: The percentage of CaCO3 in an antacid sample.
The steps that it tells me to follow are as follows:
a) find the moles of HCl in the 100 mL added
b) find the moles of NaOH used to titrate the excess acid
c) find the moles of HCl which the NaOH in (b) neutralized (ie. the moles of HCl which were titrated)
d) By subtraction from the total moles of acid in the 100 mL calculate how much acid reacted with the carbonate, hence the moles of CaCO3 and mass of CaCO3.
e) find the percentage CaCO3 by mass in the tablet.
So, let's see...here's all the data I collected:
-The concentration of HCl is 0.1320 M (obtained from the container that held the HCl).
-The manufacturer's mass of the antacid (CaCO3) tablet is 500 mg/tablet
- I found that there was 0.1 M NaOH from a seperate calculation to find the molarily (n(HCl)=0.1320 M * 0.025 L = 0.0033 moles HCl, and since ratio of NaOH with HCl is 1:1, n(NaOH)=0.0033 moles. [NaOH]=0.0033 mol/0.03322 L (my average titre volume) = 0.1 M NaOH)
- for tablet 1, total volume of NaOH = 27.03 mL
- for tablet 1, mass of tablet = 1.2086 g
- for tablet 1, mass of tablet & container = 65.7656 g
- for tablet 1, mass of empty container = 64.5570 g
Now, for calculations:
a) So I have: moles of HCl in 100 mL that were added = 0.1320 M * 0.100 L = 0.0132 mol
b) moles of NaOH used to titrate excess acid = 0.1 M * 0.02703 L = 0.002703 mol
c) at this step and onwards, this is where I get stuck...I'm not really sure where to go from here. I'm pretty sure the steps above are right, but I might be wrong. To be honest, I only really need help with c) and d), even if it's just a general formula of some sort. Any help's greatly appreciated, and if I forgot any info to mention, please let me know. Thanks!
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Topic: Back Titration of CaCO3 (Read 23906 times)