Transcript Free Response 2001 B Acids and Bases

Free Response 2001 C
Acids and Bases
Barb Fallon
AP Chemistry
June 2007
The Questions: A, B, and C
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A. The amount of acetylsalicylic acid (ASA) in a single
aspirin tablet is 325 mg, yet the tablet has a mass of
2.00 g. Calculate the mass percent of acetylsalicylic acid
in the tablet.
B. The elements contained in ASA are hydrogen, carbon,
and oxygen. The combustion of 3.000 g of the pure
compound yields 1.200 g of water and 3.72 L of dry
carbon dioxide, measure at 750. mmHg and 25 C.
Calculate the mass, in g, of each element in the 3.000 g
sample.
C. A student dissolved 1.625 g of pure ASA in distilled
water and titrated the resulting solution to the
equivalence point using 88.43 mL of 0.102 M NaOH(aq).
Assuming that ASA has only one ionizable hydrogen,
calculate the molar mass of the acid.
The Questions: D

D. A 2.00 x 10-3 mole sample of
pure ASA was dissolved in 15.00
mL of water and then titrated with
0.100 M NaOH(aq). The
equivalence point was reached
after 20.00 mL of the NaOH
solution had been added. Using
the data from the titration, shown
in the table below, determine:
 (i) the value of the acid
dissociation constant, pKa, for
ASA
 (ii) the pH of the solution after
a total volume of 25.00 mL of
the NaOH solution had been
added (assume that volumes
are additive).
Volume of 0.100 M
NaOH added (mL)
pH
0.00
2.22
5.00
2.97
10.00
3.44
15.00
3.92
20.00
8.13
25.00
?
Part A: Mass Percent
 A.
The amount of acetylsalicylic acid
(ASA) in a single aspirin tablet is 325
mg, yet the tablet has a mass of 2.00
g. Calculate the mass percent of
acetylsalicylic acid in the tablet.
 So, this question is straightforward:
find the mass percent of ASA in the
tablet.
Part A: Mass Percent
Mass percent of compound X is defined as
(mass of compound X) / (total mass) x
100%.
 Mass percent of ASA = (mass of ASA in
sample) / (total mass of sample) x 100%.
 Mass percent ASA = (.325 g ASA) / (2.00
g) x 100% = 16.25%
 Answer: the mass percent of ASA in the
tablet is 16.25%.

Part B: Percent Composition
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B. The elements contained in ASA are hydrogen,
carbon, and oxygen. The combustion of 3.000 g
of the pure compound yields 1.200 g of water
and 3.72 L of dry carbon dioxide, measure at
750. mmHg and 25 C. Calculate the mass, in g,
of each element in the 3.000 g sample.
So primarily, this is a percent composition
problem, with the twist that you have the amount
of each element produced from the compound’s
combustion, not actually in the compound itself.
Part B: Percent Composition
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The general formula for this reaction looks something
like this, unbalanced:
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CxHyOz + O2  CO2 + H2O
We know that 1.200 g of water and 3.72 L of CO2 at 750
mmHg and 25 C are produced. Now, let’s think logically.
All of the H in water must have come from the
compound. All of the C in CO2 must have also come
from the compound. However, the O in both comes from
both the compound and oxygen in the air. Thus, when
we work backwards (from the products), we can
calculate both the amount of C and H in the compound.
But because there is excess oxygen, we will subtract the
amount of C and H from the total mass of the tablet to
find the mass of O in the sample.
Part B: Percent Composition
First, convert the mass of water produced
to the amount of H in that water.
 1.200 g H2O x (1 mol H2O)/(18.02 g H2O)
X (2 mol H)/(1 mol H2O) x
(1.01 g H)/(1 mol H) = .1345 g H are in the
sample.

Part B: Percent Composition
Next, we’ll find the amount of C in the
sample by looking at the CO2 produced.
 PV = nRT
 (750 mmHg / 760 mmHg)(3.72 L) = (n
moles CO2)(.08206)(298.15 K)
 n = .150 mol CO2
 .150 mol CO2 x (1 mol C)/(1 mol CO2) x
(12.01 g C)/(1 mol C) = 1.802 g C are in
the sample.

Part B: Percent Composition
We have a 3.000 g sample of ASA, made
of C, H, and O.
 Total mass – mass of H – mass of C =
mass of O
 3.000 g - .1345 g – 1.802 g = 1.064 g O
 Rounded, there are .135 g H, 1.802 g C,
and 1.064 g of O in the sample.
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Part C: Titration
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C. A student dissolved 1.625 g of pure ASA in
distilled water and titrated the resulting solution
to the equivalence point using 88.43 mL of 0.102
M NaOH(aq). Assuming that ASA has only one
ionizable hydrogen, calculate the molar mass of
the acid.
Because ASA and NaOH each have one
equivalent, a 1:1 ratio of the compounds react.
We need to find the moles of OH- that reacted.
Part C: Titration
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(0.102 M NaOH)(0.08843 L NaOH) = 0.00902
moles NaOH, and therefore there are 0.00902
moles of OH-.
Now, we use our knowledge of the definition of
molar mass. Molar mass = (grams of a
substance)/(moles of a substance).
We have a 1.635 g sample of ASA. We also
know that this sample reacts in a 1:1 ratio with
NaOH. Therefore, there are 0.00902 moles ASA
in the sample.
Molar mass of ASA = (1.625 g ASA)/(0.00902
moles ASA) = 180. g/mol
Part D: More Titrations
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D. A 2.00 x 10-3 mole sample of
pure ASA was dissolved in 15.00
mL of water and then titrated with
0.100 M NaOH(aq). The
equivalence point was reached
after 20.00 mL of the NaOH
solution had been added. Using
the data from the titration, shown
in the table below, determine:
 (i) the value of the acid
dissociation constant, pKa, for
ASA
 (ii) the pH of the solution after
a total volume of 25.00 mL of
the NaOH solution had been
added (assume that volumes
are additive).
Volume of 0.100 M
NaOH added (mL)
pH
0.00
2.22
5.00
2.97
10.00
3.44
15.00
3.92
20.00
8.13
25.00
?
Part D: More Titrations
i. First things first: we want the pKa of ASA.
The pKa is equal to the pH halfway to the
equivalence point.
 If the equivalence point is reached after
20.00 mL of NaOH have been added, then
the pKa when 10.00 mL of NaOH have
been added equals 3.44, the pH.
 pKa = 3.44
 Ka of ASA = 10-3.44 = 3.6 x 10-4
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Part D: More Titrations
ii. Beyond the end point, there is excess
OH- in the solution. [OH-] determines the
pH. 5 mL of OH- were added past the end
point.
 Moles of excess OH- = (0.005 L)(0.100 M)
= 5.00 x 10-4 moles OH
Part D: More Titrations
Next we need to find the overall
concentration of OH.
 (5.00 x 10-4 moles OH-)/(0.040 L sol) =
1.24 x 10-2 M OH-.
 pOH = -log(1.24 x 10-2) = 1.90
 pH = 14 – pOH = 12.10
 pH of the solution = 12.10
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