Marking Scheme for
DETERMINING THE AMOUNT OF STOMACH ACID THAT CAN BE
NEUTRALIZED BY AN ANTACID TABLET
Maximum = 13
Data Table
mass of 1st tablet = 1.81 g
burette reading at the start of 1st titration = 0.0 mL
burette reading at the end of 1st titration = 33.9 mL
mass of 2nd tablet = 1.80 g
burette reading at the start of 2nd titration = 0.0 mL
burette reading at the end of 2nd titration = 33.7 mL
mass of 3rd tablet = 1.78 g
burette reading at the start of 3rd titration = 0.0 mL
burette reading at the end of 3rd titration = 35.8 mL
CALCULATIONS
1. Calculate the total volume of HCl added in each titration (it equals the difference between
the starting and ending volumes).
total volume of HCl added in 1st titration = 33.9 – 0.0 = 33.9 mL
(1/2)
total volume of HCl added in 2nd titration = 33.7 – 0.0 = 33.7 mL
total volume of HCl added in 3rd titration = 35.8 – 0.0 = 35.8 mL
2. Use the total volume of HCl for each titration and the molarity of the HCl (0.500 M) to
calculate the total moles of HCl added in each of the three titrations. This is called the
“experimental total moles of HCl” for each tablet. Note: If the values of “experimental
total moles of HCl” for two of the tablets are very close to each other and the third value is
quite a bit different from the other two, state that you are omitting the value that is too far
from the others and average the remaining two.
experimental total moles of HCl in 1st titration= 0.0339 L x 0.500 mol/L = 0.01695 mol
(1)
experimental total moles of HCl in 2nd titration= 0.0337 L x 0.500 mol/L = 0.01685 mol
experimental total moles of HCl in 3rd titration= 0.0358 L x 0.500 mol/L = 0.0179 mol
Omitting the 3rd value because it is too far from the other two:
(1/2)
average experimental total moles of HCl =
0.01695 mol  0.01685 mol
= 0.0169 mol
2
3. The label on each package of Rolaids claims that each tablet contains 675 mg of CaCO 3
and 135 mg of Mg(OH)2 . Use this information to calculate
a) the moles of CaCO3 in each tablet, and
(1)
moles CaCO3 = 675 mg x
10 –3 g
1 mol
x
= 0.00674mol
1 mg
100.1 g
b) the moles of Mg(OH)2 in each tablet.
(1)
moles Mg(OH)2 = 135 mg x
10 –3 g
1 mol
x
= 0.00232 mol
1 mg
58.3 g
4. Use equations (1) and (2) in the INTRODUCTION, and the results of Calculation 3, to
calculate
a) the moles of HCl that react with the moles of CaCO3 in a tablet, and
(1)
moles of HCl (CaCO3 ) = 0.006743 mol CaCO3 x
2 mol HCl
= 0.01349 mol
1 mol CaCO 3
Page 2
b) the moles of HCl that react with the moles of Mg(OH) 2 in a tablet, and
(1/2)
moles of HCl (Mg(OH)2 ) = 0.00232 mol Mg(OH)2 x
2 mol HCl
= 0.004631 mol
1 mol Mg(OH) 2
c) the total moles of HCl that are expected to react with both the moles of CaCO3 and
moles of Mg(OH)2 in each tablet. This is called the “expected total moles of HCl”.
(1/2)
expected total moles of HCl = 0.01349 mol + 0.004631 mol = 0.01812 mol
5. Calculate the average percentage purity of the tablets from the following equation.
average percentage purity =
(1)
average percentage purity =
average experimental total moles of HCl
x 100%
expected total moles of HCl
0.0169 mol
x 100% = 93.3%
0.01812 mol
6. The CaCO3 and Mg(OH)2 referred to in Calculation 3 are called “active ingredients”.
Calculate the total mass of the active ingredients in a tablet. Express the total mass of the
active ingredients in “grams”. Using the “mass of 1st tablet” recorded in your Data Table,
calculate the percentage of the 1st tablet’s mass that was “active ingredients”. Repeat the
calculation for each of the other tablets and calculate the average of the percentages. As in
Calculation 5, if the percentages for two of the tablets are very close to each other and the
third value is quite a bit different from the other two, state that you are omitting the value
that is too far from the others and average the remaining two.
(1)
total mass of the active ingredients = 675 mg + 135 mg = 810 mg = 0.810 g
(1)
0.810 g
x 100% = 44.8%
1.81 g
0.810 g
Percentage of active ingredients in tablet #2 =
x 100% = 45.0%
1.80 g
0.810 g
Percentage of active ingredients in tablet #2 =
x 100% = 45.5%
1.78 g
(1/2)
Average percentage of active ingredients =
Percentage of active ingredients in tablet #1 =
44.8%  45.0%  45.5%
= 45.1%
3
7. a) Stomach acid has an average molarity of 0.010 M. Calculate the volume of stomach
acid which contains the “average experimental total moles of HCl”, using the average
molarity of stomach acid.
(1)
volume of stomach acid =
0.0169 mol
= 1.69 L
0.010 mol/L
b) Stomach acid has a density of 1000 g/L. Calculate the mass of stomach acid, using the
density of stomach acid and the volume of stomach acid calculated in 7(a).
(1)
mass of stomach acid = 1000 g/L x 1.69 L = 1690 g
c) Some antacids claim they consume many times their own weight in stomach acid.
Calculate the average mass of the antacid tablets you used. As in Calculation 6, if the
masses of two of the tablets are very close to each other and the third value is quite a
bit different from the other two, state that you are omitting the value that is too far
from the others and average the remaining two. How many times larger than the
average mass of the antacid tablets is the mass of stomach acid you calculated in
7(b)?
(1/2)
average mass of the antacid tablets =
1.81 g  1.80 g  1.78 g
= 1.80 g
3
(1)
# number of times larger =
1690 g
= 939
1.80 g