Sean DeSilva
Chem 1A
Section3961
Locker Number A518
Dr. Cornett
Dec 11 2013
Experiment # 12
Titrimetric Analysis of an Antacid
Preparation
Objective
To find the number of moles of acid neutralized by a tums tablet. We
found the number of moles of acid neutralized in the titration by subtracting it
from the moles of acid in the initial solution. This was achieved by through a
standardization of a HCL and NaOH solution.
Procedure
The procedures of all the experiments that where preformed can be found in
the Santa Rosa Junior College Chem 1A laboratory manual from the fall of 2013. The
only deviation from these procedures was the doing three trails on section 12.6.1
and section 12.6.2instead of the recommended one.
Data
Please see the attacked sheet marked Data.
12.6.1:
Calculations and Results
Calculated molarity of HCL solution through Titration
Trial 1:
(Grams of Tris ) divided (Grams per mol of Tris) x (Ratio of HCL to Tris) = mols
HCL
Take mols HCL and divide it by amount of HCL added for titration to occur. That will
give you the Molarity of your HCL Solution.
1.0015g (Tris) x (1 mol/121.14 g (Tris per mol)) x (1 mol HCL/1 mol Tris)
= .008267 mols HCL
.008267 mols HCL / .039 L = .21198 mols HCL per Liter
Trial2: .2205 mols HCL per Liter
Trial3: .2195 mols HCL per Liter
Calculated molarity of HCL solution through given values of concentration.
Trial 1:
(Amount of HCL added) x (Mols of concentrate) = (mols of HCL)
Take (mols of HCL) divide (total volume of HCL+ H2O) = Molarity of Solution
(.014L of HCL) x (6 mols/1 L) = .084 mol HCL
(.084 mol HCL)/ (.414L of solution)= .2029 mols HCL per Liter
Error Calculations: Part 1
Standard Deviation
SD = SquareRoot( ((.21198 - .2029)^2 + (.2205 - .2029)^2 + (.2195 - .2029)^2 ) / 2)
SquareRoot( .000668/2) = .018272
.018272/.2029 = .09005 x 100 = 9%
Standard Deviation = 9%
Mean = (MolarityTrial1+ MolarityTrial2+ MolarityTrial3) / (number of trials)
(.21198+.2205+.2195) / 3 = .217331
Mean = .217331
RSD= (SD/Mean) x 100
(.018272/ .217331) = .08407
.08407 x 100 = 8.4%
RSD = 8.4%
12.6.2:
Calculated molarity of NaOH solution
Trial 1:
(Grams of KHP) divide (Grams per mol of KHP) x (Ratio of HCL to KHP) = mols
NaOH
Take mols NaOH and divide it by amount of NaOH added for titration to occur. That
will give you the Molarity of your NaOH Solution.
.8001g (KHP) x (1 mol/204.22 g (KHP per mol)) x (1 mol NaOH /1 mol KHP)
= .003918 mols NaOH
.003918 mols NaOH / .0374 L = .10476 mols NaOH per Liter
Trial2: .1051 mols NaOH per Liter
Trial3: .1056 mols NaOH per Liter
Calculated molarity of NaOH solution through given values of concentration.
Trial 1:
(Amount of NaOH added) x (Mols of concentrate) = (mols of NaOH)
Take (mols of NaOH) divide (total volume of NaOH + H2O) = Molarity of Solution
(.020L of NaOH) x (2 mols/1 L) = .04 mol NaOH
(.04 mol NaOH)/ (.395L of solution)= .10127 mols NaOH per Liter
Error Calculations: Part 2
Standard Deviation
SD = SquareRoot(.10476 - . 10127)^2 + (.1051 - . 10127)^2 + (.1056 - . 10127)^2 )
/ 2)
SquareRoot( .000046/2) = .000023
. 000023/1.0127 = .000023 x 100 = .0023%
Standard Deviation = .0023%
Mean = (MolarityTrial1+ MolarityTrial2+ MolarityTrial3) / (number of trials)
(.10476 +..1051 +.1056) / 3 = .105153
Mean = .105153
RSD= (SD/Mean) x 100
(.000023/ .105153) = .0002187
.0002187 x 100 = .022%
RSD = .022%
12.6.3
Total moles of HCL added
Liters of HCL added x (Molarity found in 12.6.1 for HCL (mol/L)) = Total moles of
HCL added
Trial 1:
.05L x (.21198 mols/1L) = .010599 moles of HCL added
Trial 2: .01103 moles of HCL added
Trail 3: .01098 moles of HCL added
Moles of HCL reacted with NaOH
NaOH L x (Molarity found in 12.6.2 for NaOH (mol/L)) x (1 mol HCL/1 mol NaOH)
Trial 1:
.0374L x (.104759 M) x (1 mol HCL/1 mol NaOH) = .00392 moles of NaOH react
Trial2: .00393
Trial3: .00393
Moles of HCL reacted with CaCO3
nacid neutralized by CaCO3 = nacid initially in flask – nacid neutralized by NaOH
CaCo3 react with HCL = .010599 moles of HCL added - .00392 moles of NaOH
react
CaCo3 react with HCL = .006679 moles
Trial2: .0071 moles
Trial3: .00705 moles
Mass of CaCO3 neutralized in g
Moles of CaCO3 x (1mol CaCo3 / 2 mols of HCL) x grams per mole of CaCO3
Trial1: .006679 moles x (1mol CaCo3 / 2 mols of HCL) x 100.81grams per mol of
CaCO3 = .3367 grams CaCO3
Trial2: .3578 grams CaCO3
Trial3: .3551 grams CaCO3
Mass Percent CaCO3 in antacid:
Mass Percent = (Grams of CaCO3 in above / Grams of antacid added) x 100
Trial1:
(.3367 grams CaCO3 / .865 (g) ) x 100 = 38.92%
Trial2: 41.36%
Trial3: 41.05%
Average Mass Percent = (38.92% + 41.36%+41.05%) / 3 = 40.44%
Average Mass of one tablet = (The mass of four tablets/4)
One Tablet = 5.206 g/ 4 tablets
= 1.3015g per tablet
Average Mass of CaCO3 per tablet
1.3015g per tablet x 40.44% = .526 g CaCO3 per tablet
Standard Deviation
On the bottle it says 500mg of active ingredient
Average weight of one tablet equals 1.3015 g or 1301.5mg
500mg / 1301.5mg = 38.4%
SD = SquareRoot(.3892- .384)^2 + (..4136- .384)^2 + (.4105 - .384)^2 ) / 2)
SquareRoot( .001605/2) = .0008
..0008/.384 = .00209 x 100 = .209%
Standard Deviation = .209%
Mean = (MolarityTrial1+ MolarityTrial2+ MolarityTrial3) / (number of trials)
(.3892 + .4136 + .4105) / 3 = .4044
Mean = . 4044
RSD= (SD/Mean) x 100
(..0008/ . 4044) = .001978
. 001978 x 100 = .1978%
RSD = .1978%
Day 1
Discussion
First we had to make a solution of HCL and NaOH this in its self was not
exciting, but knowing that this solution would be used in further steps added to its
importance. After the monkey work of mixing the solutions we need to find the
Molarity of that solution this was a very important step. It was nice to be able to get
a theoretical value of the solution and then match it up against the values that I
received through the process of titration. With my largest percent error being less
then 10% deviation that this was an efficient method for finding molarity. Can I just
mention the crazy process that we had to go through to find the exact amount of
Tris and KHP to use for Titration; we weighted the same substance three different
ways with in the three times we had to weight each. This is done of course to get the
most accurate results because some of the Tris or KHP could have stayed on our
weighting paper.
The second half of the day we made a solution of NaOH3 that we used to
titrate KHP. The titration process was pretty much identical in both of process for
finding molarity of each solution. The NaOH3 how every was much more accurate
according to my percent errors, getting a RSD of less then one percent. The molarity
figures for NaOH3 and HCL were in our Santa Rosa Junior College Lab manual and
each value was found on the container. With out these figure we would never be
able to find an actual molarity. After the completion we had a long two weeks to
wait until we could complete our experiment this lead to forgetting some of what
and why we did what we did on Day 1.
On the second day our goal was to calculate the grams of CaCO3 found in a
tablet of tums; we would then take our calculations and compare it to the amount of
active ingredient, which was 500mg according to the package that contained the
tablet. It was an interesting process back titrating to find the mols of the acid HCL,
which reacted with the base NAOH3. This process was very accurate seeing that my
Standard Deviation and my Relative Standard Deviation were both under a half of a
percent. The hardest part of this section was doing all of the calculations because
there were a lot of values that we had to keep track of. It doesn’t help that the lab
manual doesn’t give any help on the equations, luckily you showed us the procedure
but the number where a little different. We also had the milligrams of active
ingredient per tablet, unfortunately it took 20 minutes of calculations before we
could check our answer and then came the challenge of finding the mistake. I was
able to figure it out and had some good results.
Conclusion
Despite the fact that my calculation for the molarity of HCL was around eight
percent different from actual molarity, it did not affect my final results. The
molarity of HCL might have been thrown off because of the HCL dispenser it seemed
to stick a little bit. I remember having to add in a little extra after I thought it had
fully dispensed. I would have said that weighting out the TRIS and KHP might have
caused a mistake but we where very careful about those measurements. I think
processes like the weighting procedure, allowed for few chances to mess up the
chance to get good numbers.
The largest chance for error came in the titration processes; to stop adding
solution when the point of equilibrium had been reached was hard to do. I know
that during the titration of CaCO3 that I had three different hues of pink in each of
my trials. I could see the solution start to titrate so I would stop adding NaOH and
start gentle agitating the flask, the solution would immediately change from a faint
pink back to the white of our original solution. Just a drop or two extra of NaOH
would severely chance the color. I was able to the see the reaction that was
occurring between the Chemical solution added and the titrate solution. It was a
successful lab my numbers were very consistent, it was doing the math behind these
numbers that I think is going to be most beneficial going forward especially on the
exam. Being able to find out the amount to active ingredient in a substance through
calculating the molarity of a substance by titration is a process that is often used by
chemist and calculating mole will be important moving forward in chemistry.
1.
Post Lab Questions:
a. Heart burn is caused by excess acid that is in the stomach, that excess acid
travels out of the stomach and into the esophagus. As we saw in our lab CaCO3 is
very good at neutralizing acid.
b. When acid reacts with marble; the material breaks down into H2O, carbon
dioxide and calcium carbonate. This is why acid rain has lead to the erosions of
many famous statues throughout the world.
2.
There is no doubt a pipet is the most accurate way to measure a liquid, but I
can imagine being fairly accurate with a graduated cylinder. The last formal lab we
had to add 50ml of water and then had to weight the calorimeter, it was within .5
grams of 50g each time. Yes you will most likely get a 5 or 6 % add error on average
using a graduated cylinder.
3.
You can see for the equations below that the hydrogen carbonate is much
more effective at neutralizing acid.
1.7 grams NaHCO3 (1 mole NaHCO3/ 84.007 grams NaHCO3)(1 mole HCl/ 1 mole
NaHCO3) = .0202 moles HCl neutralized.
.5 grams CaCO3 (1 mole CaCO3/ 100.8 g CaCO3)(2 moles HCl/ 1 mole CaCO3)
=.00992 moles HCL neutralized
4.
NaOH is very soluble in water and dissociates almost completely
because it is such a strong base. I am sure that such a strong chemical would
not be good in anyone’s stomach