Free Response 2001 B Acids and Bases
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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
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
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
The general formula for this reaction looks something
like this, unbalanced:
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.
Part C: Titration
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
(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
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
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