AP CHEMISTRY
MRS. SPENCER
CHAPTER 4 TEST: SOLUTION STOICHIOMETRY KEY
1) Which of the following are strong electrolytes? HCl,
A) HCl, KCl
B) HCl, NH 3 , KCl
C) HCl, HC2 H 3 O 2 , NH 3 , KCl
Answer: A
Diff: 2
HC2 H 3 O 2 ,
NH 3 ,
KCl
D) HCl, HC2 H 3 O 2 , KCl
E) HC2 H 3 O 2 , KCl
Page Ref: Sec. 4.1
2) When H 2 SO 4 is neutralized by NaOH in aqueous solution, the net ionic equation is __________.
A) SO42 (aq)  2Na + (aq)  Na 2 SO 4 (aq)
B) SO42 (aq)  2Na + (aq)  Na 2 SO 4 (s)
C) H  (aq)  OH  (aq)  H 2 O (l)
D) H 2 SO 4 (aq)  2OH  (aq)  2H 2 O (l)  SO 4 2 (aq)
E) 2H  (aq)  2NaOH (aq)  2H 2 O (l)  2Na + (aq)
Answer: C: molecular equation: H2SO4 (aq) + 2NaOH(aq)  Na2SO4 (aq) + 2HOH (l)
complete ionic equation: 2H+(aq) + SO42-(aq) + 2Na+(aq) + 2OH-(aq) 2Na+(aq) + SO42- (aq) + 2HOH(l)
Diff: 2
Page Ref: Sec. 4.2
3) The spectator ions in the reaction between aqueous hydrochloric acid and aqueous ammonia are _________.
A) H + and NH 3
C) Cl and NH 4 
E) Cl only
D) H+ ,Cland NH 4 
B) H + ,Cl ,NH 3 and NH 4
Answer: E molecular equation: HCl(aq) + NH4OH(aq)  NH4Cl (aq) + HOH
Complete ionic equation: H+ + Cl- + NH4OH  NH4+ + Cl- + HOH
H+(aq) + Cl-(aq) + NH3(aq)  NH4+(aq) + Cl- (aq)
Diff: 2
Page Ref: Sec. 4.2
4) Which of the following are strong acids?
HI, HNO 3 , HF, HBr
A) HF , HBr
C) HI , HF , HBr
B) HI , HNO 3 , HF , HBr
D) HNO 3 , HF , HBr
Answer: E
E) HI , HNO 3 , HBr
5) A _________________reaction between an acid and a metal hydroxide produces salt and water.
A) neutralization
C) decomposition
E) combustion
B) single replacement
D) synthesis
Answer: A
6) Of the following elements, __________ is the most easily oxidized.
Oxygen, fluorine, nitrogen, aluminum, gold
A) oxygen
B) fluorine
C) nitrogen
D) aluminum
Answer: D
Diff: 3
E) gold
Page Ref: Sec. 4.4
7) What is the concentration (M) of NaBr in a solution made by mixing 35.0 mL of 0.250 M NaBr with 65.0
mL of 0.250 M NaBr?
A) 0.100
B) 0.0500
C) 0.0333
D) 0.2500
E) 125
Answer: D
Analysis: Both solutions are the same concentration of the same compound. If you combine them, you still
have the same concentration, so the answer is 0.250 M.
Diff: 3
Page Ref: Sec. 4.5
8) What is the concentration (M) of CH 3 OH in a solution prepared by dissolving 11.7 g of CH 3 OH in
sufficient water to give exactly 230 mL of solution?
A) 11.9
C) 0.0841
E) 11.9 10-3
D) 1.59
B) 1.59  10-3
Answer: D
Analysis: You are solving for molarity, which is moles/liter. You are given grams, from which you can
calculate moles. You are given mL, from which you can calculate liters. Do it, then divide moles by liters.
Diff: 3
Page Ref: Sec. 4.5
9) How many moles of Co2+ are present in 0.200 L of a 0.400 M solution of CoI 2 ?
A) 2.00
B) 0.500
C) 0.160
D) 0.0800
E) 0.0400
Answer: D
GIVEN: volume = 0.200L & molarity (mol/L)= 0.400mol/L
UNKNOWN: moles of Co.
Analysis: CoI2  Co2+ (aq) + 2 I-(aq). From this, we understand the molar ratio is 1:1:2. If you know how
many moles of CoI2 there are, you can determine the moles of Co present (1:1 ratio!)
moles (of CoI2) = M (of CoI2) x V (of CoI2)
= 0.400 mol CoI2 x 0.200 L of CoI2
1 L CoI2
= 0.0800 mol CoI2
Therefore, there are also 0.0800 mol Co2+ present.
Diff: 3
Page Ref: Sec. 4.5
10) The molarity of a solution prepared by diluting 43.72 mL of 1.005 M aqueous K 2 Cr 2 O 7 to 500. mL is ___.
A) 0.0879
B) 87.9
C) 0.0218
D) 0.0115
E) 0.870
Answer: A
Analysis: use M1V1 = M2V2
Diff: 2
Page Ref: Sec. 4.5
11) A 31.5 mL aliquot of HNO 3 (aq) of unknown concentration was titrated with 0.0134 M NaOH (aq). It took
23.9 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was __________.
A) 0.0102
B) 0.0051
C) 0.0204
D) 0.227
E) 1.02
Answer: A
Analysis: use MAVA = MBVB
Diff: 4
Page Ref: Sec. 4.6
12) Which of the following is insoluble in water at 25 °C?
A) Mg3(PO4)2
C) (NH4)2CO3
B) Na2S
D) Ca(OH)2
Bad question: Both A & E are insoluble!
13) In which species does bromine have the highest oxidation number?
A) H=+1 Br=-1
C)Br2 Br = 0
B) H=+1 O=2- Br=+1
D)HBrO4 H = +1 O=-2 Br = +7
Answer: D
Diff: 3
Page Ref: Sec. 4.4
E) BaSO4
E)BrF3 F = -1 Br = +3
14) Of the reactions below, only __________ will not occur.
A) F2(g) + 2NaCl (aq)  2NaF(aq) + Cl2(g)
D) 2Al (s)  6HBr (aq)  2AlBr 3 (aq)  3H 2 (g)
B) I2(s) + 2NaBr(aq)  2NaI (aq) + Br2 (l)
E) Ca(s) + 2HNO3(aq)  Ca(NO3)2(aq) + H2(g)
C) 2Ni (s)  H 2 SO 4 (aq)  Ni 2 SO 4 (aq)  H 2 (g)
Answer: B
Diff: 4
Page Ref: Sec. 4.4
15) Oxidation cannot occur without __________.
A) acid
B) oxygen
C) water
D) air
E) reduction
Answer: E
Diff: 1
Page Ref: Sec. 4.4
16) Pure acetic acid (HC 2 H 3 O 2 ) is a liquid and is known as glacial acetic acid. Calculate the molarity of a
solution prepared by dissolving 20.00 mL of glacial acetic acid at 25 °C in sufficient water to give 500.0 mL of
solution. The density of glacial acetic acid at 25 °C is 1.05 g/mL.
A) 0.700
C) 21.0
B) 1.26 × 103
E) 3.50 × 10-4
D) 0.0210
Answer: A
GIVEN: 20.00 mL HC2H3O2(l)
Analysis:
 We need to solve for molarity of aqueous HC2H3O2.
 Molarity of this solution is moles of HC2H3O2(l) /liter of HC2H3O2(aq).
 We have been given the volume of HC2H3O2(aq), which we can convert to liters (500.0 mL = 0.5000L)
 Now we need to solve for moles of HC2H3O2(l).
20.0mL HC2H3O2(l) x 1.05 g HC2H3O2(l) x 1 mol HC2H3O2(l) = ______ mol HC2H3O2(l)
1 mL HC2H3O2(l)
60 g HC2H3O2(l)
 Now we will divide moles by volume to solve for the molarity of the HC2H3O2(aq).
Molarity of HC2H3O2(aq) =
moles HC2H3O2(l)
0.5000L HC2H3O2(aq)
=
M
Diff: 4
Page Ref: Sec. 4.5
17) Silver ions can be precipitated from aqueous solutions by the addition of aqueous chloride:
Ag  (aq)  Cl (aq)  AgCl (s)
Silver chloride is virtually insoluble in water so that the reaction appears to go to completion. How many grams
of solid NaCl must be added to 25.0 mL of 0.366 M AgNO 3 solution to completely precipitate the silver?
C) 0.535
E) 6.39 × 103
A) 9.15 10-3
D) 0.157
B) 1.57 × 10-4
Answer: C
Diff: 4
Page Ref: Sec. 4.6
2010 Advanced Placement Chemistry Examination
Form B
Section II – Free Response Questions
Beginning with the 2007 examination, the numerical problems, 1, 2, and 3, are Part A. Students may use a calculator for this part (55
minutes). Part B (40 minutes) is the three reactions question (predict the products of a reaction, balance, and answer a short question
regarding the reaction) and the two theory questions. A laboratory question could be in either part A or B. NO calculator is allowed in
part B.
* * * Part A * * *
Time – 55 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A.
CLEARLY SHOW THE METHOD USED AND STEPS INVOLVED IN ARRIVING AT YOUR
ANSWERS. It is to your advantage to do this, because you may earn partial credit if you do and you will
receive little or no credit if you do not. Attention should be paid to significant figures. Be sure to write all
your answers to the questions on the lined pages following each question in this booklet.
Answer Question 1, 2, and 3. The Section II score weighting for each question is 20 percent.
Answer Question 4 below. The Section II score weighting for this question is 10 percent. (Note: I have
included below only 1 of the 3 reactions listed in this question.)
4. For (each of) the following (three) reaction(s), in part (i) write a balanced equation for the reaction and in
part (ii) answer the question about the reaction. In part (i), coefficients should be in terms of lowest whole
numbers. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solutions
as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are
unchanged by the reaction. You may use the empty space at the bottom of the next page for scratch work,
but only equations that are written in the answer boxes provided will be graded.
(a) Solid copper(II) sulfate pentahydrate is gently heated.
(i) Balanced equation:
heat
CuSO4•5H2O  CuSO4 + 5H2O
(ii) How many grams of water are present in 1.00 mol of copper(II) sulfate pentahydrate?
The ratio of units of hydrate: units of anhydrous compound: molecules of water is 1:1:5
If one mole of CuSO4•5H2O is present, then there will be five times as much water molecules, or 5 moles.
Answer Question 5 and Question 6. The Section II score weighting for these questions is 15 percent each.
Your responses to these questions will be graded on the basis of the accuracy and relevance of the information
cited. Explanations should be clear and well organized. Examples and equations may be included in your
responses where appropriate. Specific answers are preferable to broad, diffuse responses.
2010 B
5.
(a)
(b)
(c)
(d)
A solution of 0.100 M HCl and a solution of 0.100 M NaOH are prepared. A 40.0 mL sample of one of the
solutions is added to a beaker and then titrated with the other solution. A pH electrode is used to obtain the
data that are plotted in the titration curve shown above.
Identify the solution that was initially added to the beaker. Explain your reasoning.
On the titration curve above, circle the point that corresponds to the equivalence point.
At the equivalence point, how many moles of titrant have been added?
The same titration is to be performed again, this time using an indicator. Use the information in the table
below to select the best indicator for the titration. Explain your choice.
Indicator
pH Range of
Color Change
Methyl violet
0 - 1.6
Methyl red
4-6
Alizarin yellow
10 - 12
(e) What is the difference between the equivalence point of a titration and the end point of a titration?
(f) On the grid provided on the next page, sketch the titration curve that would result if the solutions in the
beaker and buret were reversed (i.e., if 40.0 mL of the solution used in the buret in the previous titration
were titrated with the solution that was in the beaker).