Ch 4 Review Problems
AP Chemistry
# 1- 1981 B
A 1.2516 gram sample of a mixture of CaCO3 and Na2SO4 was analyzed by dissolving the sample and
completely precipitating the Ca2+ as CaC2O4. The CaC2O4 was dissolved in sulfuric acid and the resulting
H2C2O4 was titrated with a standard KMnO4 solution.
(a) Write the balanced equation for the titration reaction, shown unbalanced below.
MnO4- + H2C2O4 + H+  Mn2+ + CO2 + H2O
Indicate which substance is the oxidizing agent and which substance is the reducing agent.
(b) The titration of the H2C2O4 obtained required 35.62 milliliters of 0.1092 molar MnO4- solution.
Calculate the number of moles of H2C2O4 that reacted with the MnO4(c) Calculate the number of moles of CaCO3 in the original sample.
(d) Calculate the percentage by weight of CaCO3 in the original sample.
#2- 1983 C
Ti3+ + HOBr  TiO2+ + Br- (in acid solution)
(a) Write the correctly balanced half-reactions and net ionic equation for the skeletal equation shown above.
(b) Identify the oxidizing agent and the reducing agent in this reaction.
#3- 1985 B
(a) Titanium can be reduced in an acid solution from TiO2+ to Ti3+ with zinc metal. Write a balanced
equation for the reaction of TiO2+ with zinc in acid solution.
(b) What mass of zinc metal is required for the reduction of a 50.00 millilitre sample of a 0.115 molar
solution of TiO2+?
#4 1996 B
Concentrated sulfuric acid (18.4-molar H2SO4) has a density of 1.84
grams per milliliter. After dilution with water to 5.20-molar, the solution
has a density of 1.38 grams per milliliter and can be used as an
electrolyte in lead storage batteries for automobiles.
(a) Calculate the volume of concentrated acid required to prepare 1.00
liter of 5.20-molar H2SO4.
(b) Determine the mass percent of H2SO4 in the original concentrated
solution.
(c) Calculate the volume of 5.20-molar H2SO4 that can be completely
neutralized with 10.5 grams of sodium bicarbonate, NaHCO3.
#5 2007 part B, question #4
A solution containing silver (I) ion (an oxidizing agent) is mixed with a
solution containing iron (II) ion (a reducing agent).
(i) Balanced equation:
(ii) If the contents of the reaction mixture described above are filtered,
what substance(s), if any, would remain on the filter paper.
#6 2009 form A
(a) A sample of solid iron(III) oxide is reduced completely with solid carbon.
(i) Balanced equation:
(ii) What is the oxidation number of carbon before the reaction, and what is the oxidation number of
carbon after the reaction is complete
#7 1983 B
The molecular weight of a monoprotic acid HX was to be determined. A sample of 15.126
grams of HX was dissolved in distilled water and the volume brought to exactly 250.00
millilitres in a volumetric flask. Several 50.00 millilitre portions of this solution were titrated
against NaOH solution, requiring an average of 38.21 millilitres of NaOH.
The NaOH solution was standardized against oxalic acid dihydrate, H 2C2O4.2H2O (molecular
weight: 126.066 gram mol–1). The volume of NaOH solution required to neutralize 1.2596
grams of oxalic acid dihydrate was 41.24 millilitres.
(a)
(b)
(c)
(d)
Calculate the molarity of the NaOH solution.
Calculate the number of moles of HX in a 50.00 millilitre portion used for titration.
Calculate the molecular weight of HX.
Discuss the effect of the calculated molecular weight of HX if the sample of oxalic acid
dihydrate contained a nonacidic impurity.
#8 1987 B
The percentage by weight of nitric acid, HNO3, in a sample of concentrated nitric acid is to be determined.
(a) Initially a NaOH solution was standardized by titration with a sample of potassium hydrogen phthalate,
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 millilitres of base was required. Calculate the molarity of the NaOH solution. (Molecular weight:
KHC8H4O4 = 204.2)
(b) A 10.00 millilitre sample of the concentrated nitric acid was diluted with water to a total volume of
500.00 millilitres. Then 25.00 millilitres of the diluted acid solution was titrated with the standardized
NaOH solution prepared in part (a). The equivalence point was reached after 28.35 millilitres of the base
had been added. Calculate the molarity of the concentrated nitric acid.
(c) The density of the concentrated nitric acid used in this experiment was determined to be 1.42 grams per
millilitre. Determine the percentage by weight of HNO3 in the original sample of concentrated nitric
acid.
#9 1998 D (Required)
[repeated in lab procedures section]
An approximately 0.1–molar solution of NaOH is to be standardized by titration. Assume that the following
materials are available.
• Clean, dry 50 mL buret
• 250 mL Erlenmeyer flask
• Wash bottle filled with distilled water
• Analytical balance
• Phenolphthalein indicator solution
• Potassium hydrogen phthalate, KHP, a pure solid monoprotic acid (to be used as the primary standard)
(a) Briefly describe the steps you would take, using the materials listed above, to standardize the NaOH
solution.
(b) Describe (i.e., set up) the calculations necessary to determine the concentration of the NaOH solution.
(c) After the NaOH solution has been standardized, it is used to titrate a weak monoprotic acid, HX. The
equivalence point is reached when 25.0 mL of NaOH solution has been added. In the space provided at
the right, sketch the titration curve, showing the pH changes that occur as the volume of NaOH solution
added increases from 0 to 35.0 mL. Clearly label the equivalence point on the curve.
Answers
#1
(a) 2 MnO4- + 5 H2C2O4 + 6 H+  2 Mn2+ + 10 CO2 + 8 H2O
oxidizing agent: MnO4- , reducing agent: H2C2O4
(b) 0.03562 L 
0.1092 mol 5 mol H2C2O4
3

 9.7210 mol H2C2O4
L
2 mol MnO4
(c) moles of H2C2O4 = moles CaCO3, therefore, 9.7210-3 mol H2C2O4 = 9.7210-3 mol CaCO3
100.1 g CaCO3
9.72103 mol CaCO3 
1 mol CaCO3
100 77.7% CaCO3
(d)
1.2516 g sample
#2 (a) Ti3+ + H2O  TiO2+ + 2 H+ + eH+ + HOBr + 2e-  Br- + H2O
2 Ti3+ + HOBr + H2O  2 TiO2+ + 3 H+ + Br(b) HOBr is the oxidizing agent and Ti3+ is the reducing agent.
#3(a) Zn  Zn2+ + 2e2 TiO2+ + 4 H+ + 2e-  2 Ti3+ + 2 H2O
-----------------------------------------------------------------------------2 TIO2+ + 4 H+ + Zn  Zn2+ + 2Ti3+ + 2 H2O
(b) 0.0500 L 
0.115 mol TiO
2+
1L

1 mol Zn
1 mol TiO
2+

65.4 g Zn
1 mo l Zn
= 0.188 g Zn
#4 (a) M1V1 = M2V2 ; (5.20 M)(1.00 L) = (18.4 M)(V2)
V2 = 0.283 L
(b) 18.4 mol/1 L  98.1 g H2SO4/1 mol = 1805 g H2SO4
1 L  1.84 g/1 mL  1000 mL/1 L = 1840 g solution
mass % = 1805 g/1840 g  100 = 98.1%
(c) 2 NaHCO3 + H2SO4  2 H2O + Na2SO4 + 2 CO2
10.5 g NaHCO3  1 mol NaHCO3/84.0 g NaHCO3 1 mol H2SO4/2 mol NaHCO3  1 L/5.20 mol
H2SO4 = 0.0120 L
#5 (i) Ag+ + Fe2+  Ag + Fe3+
(ii) solid silver
#6 (i) 2 Fe2O3 + 3 C  4 Fe + 3 CO2
(ii) before C = 0, after C = +4
#7 (a) mol H2C2O4.2H2O =
1.2596 g
= 9.991610–3 mol
g
126.066 mol
H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O
2 mol NaOH
9.991610–3 mol 
= 1.998310–2 mol
1 mol H 2C2 O 4
1.9983  10-3 mol
molarity of NaOH =
= 0.4846 M
0.04124 L
(b) mol HX = mol NaOH
0.03821 L  0.4846 M = 0.01852 mol HX
(c)
0.01852 mol
 250.00 mL = 0.09260 mol HX
50.00 mL
15.126 g
= 163.3 g mol
molec. wt. =
0.09260 mol
(d) The calculated molecular weight is smaller than true value, because:
measured g H2C2O4 is larger than true value,
calculated mol H2C2O4 is larger than true value,
calculated mol NaOH is larger than true value,
calculated M NaOH is larger than true value
calculated mol HX is larger than true value, therefore,
g HX (true value)
molec wt. =
mol HX (calc. & too large)
1 mol
#8 (a) 1.518 g 
= 7.43410-3 mol NaOH required to neut.
204.2 g
7.434  10 3 mol
= 0.2764 M NaOH
0.02690 L
(b)
28.35 mL NaOH 0.2764 mol 1 mol HNO 3


= 0.3134 M HNO3
25.00 mL HNO 3
1L
1 mol NaOH
MfVf=MiVi; (0.3134M)(500mL) = (M)(10.00mL)
M = 15.67 M HNO3
g
HNO 3
 100%
L sol'n
15.67 mol HNO 3 63.02 g

grams HNO3 in 1 L conc. sol’n =
= 987.5 g/mol
1L
1 mol
1.42 g sol'n 1000 mL

grams sol’n in 1 L conc. sol’n =
= 1420 g/L
1 mL
1L
(c) % HNO3 in conc. sol’n =
L
g
[HA] =
K=
(0.01523 L)(0.2211 mol L )
= 0.04810 M
0.07000 L
[OH- ][HA] K w 1.0  10 14
= 1.310-10


[A- ]
K a 7.7  10 5
7.789  10 3 mol
= 0.0914 M
0.08523 L
#9 Answer
(a) • exactly mass a sample of KHP in the Erlenmeyer flask and add distilled water to dissolve the solid.
• add a few drops of phenolphthalein to the flask.
• rinse the buret with the NaOH solution and fill.
• record starting volume of base in buret.
• with mixing, titrate the KHP with the NaOH solution until it just turns slightly pink.
• record end volume of buret.
• repeat to check your results.
[A-] =
(b)
mass of KHP
molar mass KHP
= moles of KHP
since KHP is monoprotic, this is the number of moles of NaOH
moles o f NaOH
L of titran t
(c)
= molarity of NaOH