Vodcast Problem
Eric Liu
June 10, 2007
Chemistry AP
The Question: 1987 B
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Part A
Part B
Part C
The percentage by weight of nitric acid,
HNO3, in a sample of concentrated nitric
acid is to be determined.
This is a “calculations”-type
problem. Remember to use your
work and answers from earlier
parts to help out on later ones!
The Question: Part A
Initially a NaOH solution was standardized by titration with a sample of potassium
hydrogen phthalate (also known as KHP), KHC8H4O4, a monoprotic acid often used
as a primary standard. A sample of pure KHC8H4O4 weighing 1.518 grams was
dissolved in water and titrated with the NaOH solution. To reach the equivalence
point, 26.90 milliliters of base was required. Calculate the molarity of the NaOH
solution. (Molecular weight: KHC8H4O4 = 204.2)
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Things to keep in mind:
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monoprotic acid
Acid-base neutralization reaction
The question!
The How-to: Part A
Acid-base neutralization reaction
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2
K  (aq)  HC8 H 4 O4 (aq )  Na  (aq)  OH  (aq)  Na  (aq )  H 2 O(l )  K  (aq )  C8 H 4 O4 (aq)
1 to 1 mole ratio
1.518gKHC8 H 4 O4 1molKHC8 H 4 O4
1molNaOH
1
x
x
x
 0.2764MNaOH
1
204.2 gKHC8 H 4 O4 1molKHC8 H 4 O4 .02690LNaOH
The Question, cont.: Part B
A 10.00 milliliter sample of the concentrated nitric acid was diluted with water to a
total volume of 500.00 milliliters. Then 25.00 milliliters of the diluted acid solution
was titrated with the standardized NaOH solution prepared in Part A. The
equivalence point was reached after 28.35 milliliters of the base had been added.
Calculate the molarity of the concentrated nitric acid.
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Things to keep in mind:
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Info, work, answers from Part A
Dilution
Work Backwards
Titration
Units
The How-to: Part B
A
10.00
milliliter
sample
of the
concentrated
The
Then
equivalence
25.00
milliliters
pointofwas
the
reached
diluted
acid
after
nitric
wastitrated
diluted
withthe
water
to a total
28.35acid
solution
milliliters
was
of thewith
base
had been
volume
milliliters.
25.00
added. of 500.00
standardized
NaOH
solution Then
prepared
in  From Part A
milliliters
diluted acid solution…
Part A. of the
 .2764M NaOH
0.2764molNaOH
28.35mLNaOHx
1000mL
 .007836molNaOH
.007836molNaOH = .007836molHNO3
.007836molHNO3 1000mL
x
 .3134MHNO3
25.00mL
1L
.3134molHNO3
x.500L  .1567molHNO3
1L
.1567molHNO3 1000mL
x
 15.67MHNO3
10.00mL
1L
1 to 1 mole ratio
The Question, cont.: Part C
The density of the concentrated nitric acid used in this experiment was determined to
be 1.42 grams per milliliter. Determine the percentage by weight of HNO3 in the
original sample of concentrated nitric acid.
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Things to keep in mind:
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Info, work, answers from Part B
The question!
The How-to: Part C
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Density: 1.42 g/mL
From Part B
.1567mol HNO3 in the
gHNO3
9.875 gHNO3
10.00mL
sample

x100% 
x100%  69
.5%
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%Weight HNO3
10.00mL x
gtotal
14.2 gtotal
1.42 g
 14.2 g
1mL
MolarMassHNO3  1.01  14.01  16 x3  63.02
.1567molHNO3 x63.02
%Weight HNO3 
gHNO3
mol
gHNO3
 9.875gHNO3
molHNO3
gHNO3
9.875 gHNO3
x100% 
x100%  69.5%
gtotal
14.2 gtotal
The Conclusion
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You’ve just finished 1987 B of an actual
Chemistry AP Test!
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Congrats.