 Question 1 is question 4 in previous years, question 2 is question 1 in previous years and questions 3&4 are questions 2&3 in previous years.
 students are now allowed 10 minutes to answer question 1, after which they must seal
 that portion of the test.
 Go to Answers
Return to Additional Materials Menu
1) Give the formulas to show the reactants and the products for FIVE of the following chemical reactions. Each of the reactions occurs in aqueous solution unless otherwise indicated. Represent substances in solution as ions if the substance is extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. In all cases a reaction occurs. You need not balance.
Example: A strip of magnesium is added to a solution of silver nitrate.
Mg + Ag + ---> Mg 2+ + Ag
(a) Solid calcium carbonate is strongly heated.
(b) A piece of nickel metal is immersed in a solution of copper(II) sulfate.
(c) Equal volumes of equimolar solutions of disodium hydrogen phosphate and hydrochloric acid are mixed.
(d) Chlorine gas is bubbled into a solution of sodium bromide.
(e) Ammonia gas is bubbled into a solution of ethanoic (acetic) acid.
(f) Solid ammonium carbonate is added to a saturated solution of barium hydroxide.
(g) Drops of liquid dinitrogen trioxide are added to distilled water.
(h) Solutions of potassium permanganate and sodium oxalate are mixed.
2)
HOCl <===> OCl¯ + H +

Hypochlorous acid, HOCl, is a weak acid commonly used as a bleaching agent. The aciddissociation constant, K a
, for the reaction represented above is 3.2 x 10¯ 8
.
(a) Calculate the [H + ] of a 0.14-molar solution of HOCl.
(b) Write the correctly balanced net ionic equation for the reaction that occurs NaOCl is dissolved in water and calculate the numerical value of the equilibrium constant for the reaction.
(c) Calculate the pH of a solution made by combining 40.0 milliliters of 0.14-molar HOCl and
10.0 milliliters of 0.56-molar NaOH.
(d) How many millimoles of solid NaOH must be added to 50.0 milliliters of 0.20-molar HOCl to obtain a buffer solution that has a pH of 7.49? Assume that the addition of the solid NaOH results in a negligible change in volume.
(e) Household bleach is made by dissolving chlorine gas in water, as represented below.
Cl
2
(g) + H
2
O --> H
+
+ Cl¯ + HOCl(aq)
Calculate the pH of such a solution if the concentration of HOCl in the solution is 0.065 molar.
3)
C
2
H
2
(g) + 2 H
2
(g) ---> C
2
H
6
(g)
Information about the substances
Substance
C
2
H
2
(g)
H
2
(g)
C
2
H
6
(g)
S° (J/mol K)  H° f
(kJ/mol) Bond Bond Energy (kJ/mol)
200.9
130.7
--------
226.7
0
-84.7
C-C
C=C
C-H
H-H
347
611
414
436
(a) If the value of the standard entropy change,
 S°, for the reaction is -232.7 joules per mole
Kelvin, calculate the standard molar entropy, S°, of C
2
H
6
gas.
(b) Calculate the value of the standard free-energy change,
 G°, for the reaction. What does the sign of
 G° indicate about the reaction above?
(c) Calculate the value of the equilibrium constant, K, for the reaction at 298 K.
(d) Calculate the value of the C [triple bond] C bond energy in C
2
H
2
in kilojoules per mole.

4) Concentrated sulfuric acid (18.4-molar H
2
SO
4
) 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 H
2
SO
4
.
(b) Determine the mass percent of H
2
SO
4
in the original concentrated solution.
(c) Calculate the volume of 5.20-molar H
2
SO
4
that can be completely neutralized with 10.5 grams of sodium bicarbonate NaHCO
3
.
(d) What is the molality of the 5.20-molar H
2
SO
4
?
5)
Represented above are five identical balloons, each filled to the same volume at 25°C and 1.0 atmosphere pressure with the pure gas indicated.
(a) Which balloon contains the greatest mass of gas? Explain.
(b) Compare the average kinetic energies of the gas molecules in the balloons. Explain.
(c) Which balloon contains the gas that would be expected to deviate most from the behavior of an ideal gas? Explain.
(d) Twelve hours after being filled, all the balloons have decreased in size. Predict which balloon will be the smallest. Explain your reasoning.
6) A 0.500-gram sample of a weak, nonvolatile acid, HA, was dissolved in sufficient water to make 50.0 milliliters of solution. The solution was then titrated with a standard NaOH solution.
Predict how the calculated molar mass of HA would be affected (too high, too low, or not affected) by the following laboratory procedures. Explain each of your answers.
(a) After rinsing the buret with distilled water, the buret is filled with the standards NaOH solution; the weak acid HA is titrated to its equivalence point.
(b) Extra water is added to the 0.500-gram sample of HA.

(c) An indicator that changes color at pH 5 is used to signal the equivalence point.
(d) An air bubble passes unnoticed through the tip of the buret during the titration.
7)
Sr(s) + Mg
2+
<===> Sr
2+
+ Mg(s)
Consider the reaction represented above that occurs at 25°C. All reactants and products are in their standard states. The value of the equilibrium constant, K eq
, for the reaction is 4.2 x 10
17
25°C.
at
(a) Predict the sign of the standard cell potential, E°, for a cell based on the reaction. Explain your prediction.
(b) Identify the oxidizing agent for the spontaneous reaction.
(c) If the reaction were carried out at 60°C instead of 25°C, how would the cell potential change?
Justify your answer.
(d) How would the cell potential change if the reaction were carried out at 25°C with a 1.0-molar solution of Mg(NO
3
)
2
and a 0.10-molar solution of Sr(NO
3
)
2
? Explain.
(e) When the cell reaction in (d) reaches equilibrium, what is the cell potential?
8) The reaction between NO and H
2
is believed to occur in the following three-step process.
NO + NO <===> N
2
O
2
(fast)
N
2
O
2
+ H
2
---> N
2
O + H
2
O (slow)
N
2
O + H
2
---> N
2
+ H
2
O (fast)
(a) Write a balanced equation for the overall reaction.
(b) Identify the intermediates in the reaction. Explain your reasoning.
(c) From the mechanism represented above, a student correctly deduces that the rate law for the reaction is rate = k[NO]
2
[H
2
]. The student then concludes that (1) the reaction is third-order and
(2) the mechanism involves the simultaneous collision of two NO molecules and an H
2
molecule.
Are conclusions (1) and (2) correct? Explain.
(d) Explain why an increase in temperature increases the rate constant, k, given the rate law in
(c).

9) Explain each of the following in terms of the electronic structure and/or bonding of the compounds involved.
(a) At ordinary conditions, HF (normal boiling point = 20°C) is a liquid, whereas HCl (normal boiling point = -114°C) is a gas.
(b) Molecules of AsF
3
are polar, whereas molecules of AsF
5
are nonpolar.
(c) The N-O bonds in the NO
2
¯ ion are equal in length, whereas they are unequal in HNO
2
.
(d) For sulfur, the fluorides SF
2
, SF
4
, and SF
6
are known to exist, whereas for oxygen only OF
2
is known to exist.

Notes
 Question 1 is question 4 in previous years, question 2 is question 1 in previous years and
 questions 3&4 are questions 2&3 in previous years. students are now allowed 10 minutes to answer question 1, after which they must seal that portion of the test.


[delta] is used to indicate the capital Greek letter.
[square root] applies to the numbers enclosed in parenthesis immediately following


All simplifying assumptions are justified within 5%.
 One point deduction for a significant figure or math error, applied only once per problem.
No credit earned for numerical answer without justification.
 Return to Questions
 Return to Additional Materials Menu
1) Reaction question
(a) CaCO
3
---> CaO + CO
2
(b) Ni + Cu
2+
---> Ni
2+
+ Cu hydrated ions acceptable with correct charge
1 point for Ni(OH)
2
as product
(c) HPO
4
2 ¯ + H +
---> H
2
P0
4
¯ incorrect charge on H
2
P0
4
¯ when only one product occurs, 1 point only
1 product point for transfer if H
+
from an ionic reactant to product when a phosphate species is incorrectly but consistently written.
(d) Cl
2
+ Br- ---> Cl¯ + Br
2 no credit for monatomic Cl as reactant or Br as product
(e) NH
3
+ HC
2
H
3
O
2
---> C
2
H
3
O
2
¯ + NH
4
+
1 product point for NH
4
C
2
H
3
O
2
1 point for NH
3
+ H
+
---> NH
4
(f) (NH
4
)
2
CO
3
+ Ba 2+ + OH¯ ---> NH
3
+ BaCO
3
+ H
2
O

1 product point for either NH
3
or BaCO
3
2 product points for all three species correct
(g) N
2
O
3
+ H
2
O ---> HNO
2
1 product point for H
+
+ NO
2
¯
(h) MnO
4
¯ + C
2
O
4
2 ¯ ---> MnO
2
+ CO
2 no penalty for OH¯ or H
2
O in equation no point earned for Mn
2+
as product
2) a) two points total ; one point for correct substitutions; one point for computation
[H + ] = [OCl¯] = square root (0.14 x 3.2 x 10¯ 8 ) = 6.7 x 10¯ 5 M since K a
= ( [H
+
][OCl¯] ) / [HOCl]
= [H
+
]
2
/ c
HOCl
(b) two points total: one point each
OCl¯ + H
2
O <===> HOCl + OH¯ (or NaOCl + H
2
O ---> Na
+
+ HOCl + OH¯)
K b
= K w
/ K a
= 1 x 10¯
14
÷ 3.2 x 10¯
8
= 3.1 x 10¯
7
(c) two points total; one for concentrations and one for pH calc.
Concentrations before reaction:
[HOCl] = [(0.0400) (0.14)] / 0.050 = 0.11 M
[OH¯] = [(0.0100) (0.56)] / 0.050 = 0.11 M
Thus reaction is essentially complete and exactly equals a solution of NaOCl and [OCl¯] = 0.11
M (or reaction is at equivalence point).
Then

[OH¯] = [HOCl]
K b
= [OH¯]
2
/ 0.11 = 3.1 x 10¯
7
[OH¯] = square root [(0.11) (3.1 x 10¯ 7 )] = 1.8 x 10¯ 4 pOH = 3.73 pH = 14 - 3.73 = 10.27
(d) two points; one for half-neutralized; one for mmol calcs. pH = 7.49 therefore [H + ] = 3.2 x 10¯ 8 pH = pK a
, or [H
+
] = K a
.
So [OCl¯] / [HOCl] = 1 , or solution must be half neutralized. initial mmol HOCl = 50.0 x 0.20 = 10.0 mmol mmol NaOH required = 10.0 ÷ 2 = 5.0 mmol
(e) one point
From equation, 1 mol H + produced for each 1 mole of HOCl produced, thus [H + ] = [HOCl] =
0.065 therefore pH = 1.19
3)
(a) two points; one for line of answer
- 232.7 J/K = S° (C
2
H
6
) - (261.4 + 200.9) J./K
S° (C
2
H
6
) = 229.6 J/K units ignored; 1 point earned for 98.9 J/K; 1 point lost if stoichiometry is not implied in process
(b) three points total; one point each portion; any value for T (e.g., 273 K or 298 K) is allowable:
[delta]H° = (- 84.7 kJ) - (226.7 kJ) = -311.4kJ

= - 311.4kJ - (298 K) (- 0.2327kJ/K)
= - 311.4 kJ + 69.3 kJ
= - 242.1 kJ
Negative [delta]G° therefore reaction is spontaneous, or K eq
> 1 therefore reaction is spontaneous, or products are favored at equilibrium.
(c) two points ln K = 242.1 ÷ [(8.31 x 10¯ 3
) (298)] = 97.7
K = 3 x 10
42
(1,2,or 3 significant figures acceptable)
(d) two points; first point earned for correct substitution and correct number of bonds, second point earned for setting equal to [delta]H rxn
and correct calculation of answer; no points earned for "extrapolation" techniques to find carbon-carbon triple bond energy; E* is the energy of the carbon-carbon triple bond.
- 311.4 kJ = [(2) (436) + E* + (2) (414)] - [(347) + (6) (414)]
E* = 820 kJ
4)
(a) one point
V
1
M
1
= V
2
M
2
(1.00L) (5.20 mol/L) = (x) (18.4 mol/L) x = 5.2 mol / (18.4 mol/L) = 0.283 L (or 283 mL)
(b) two points mass 1 liter of concentrated H
2
SO
4
= 1 L x (1.84 g/mL) x (1,000 ml/L) = 1,840 g H
2
SO
4
18.4 mol H
2
SO
4
x 98.1 g/mol = 1,805 g H
2
SO
4

mass percent H
2
SO
4
= (1,805 g / 1,840 g) x 100 = 98.1%
(c) three points
Stoichiometric ratio of NaHCO
3
to H
2
SO
4
= 2:1
10.5 g NaHCO
3
x (1 mol NaHCO
3
/ 84.0 g NaHCO
3
) = 0.125 mol NaHCO
3
Since 1 mol H
2
SO
4
reacts with 2 mol NaHCO
3
, 0.125 mol NaHCO
3
reacts with 0.0625 mol
H
2
SO
4
0.0625 mol H
2
SO
4
= V x M = (V) (5.20 M)
V = 0.0625 mol / (5.20 mol/L) = 0.0120 L (or 12.0 mL)
(d) three points molality = moles solute / 1,000 g solvent = moles solute / 1 kg solvent mass of 1 L of 5.20 M H
2
SO
4
= 1 L x (1,000 mL / 1 L) x (1.38 g 1 mL) = 1,380 g mass of H
2
SO
4
in 1 L = (5.20 mol/L) (98.1 g.mol) = 510 g mass of H
2
O in 1 L = 1,380 - 510 = 870 g molality = (5.20 mol H
2
SO
4
/ 870 g) x (1,000 g / 1 kg) = 5.98 m
Note: no credit earned for 5.20 mol / 1.38 kg = 5.77 m
5)
(a) two points
CO
2 because all contain same number of molecules (moles), and CO
2
molecules are the heaviest
Note: total of 1 point earned if CO
2
not chosen but same number of molecules (moles) is specified

(b) two points
All are equal because same temperature, therefore same average kinetic energy
Note: just restatement of "same conditions, etc." does not earn second point
(c) two points
CO
2 either one: it has the most electrons, hence is the most polarizable it has the strongest intermolecular (London) forces
Note: also allowable are "polar bonds", "inelastic collisions"; claiming larger size or larger molecular volume does not earn second point
(d) two points
He
Any one: greatest movement through the balloom wall smallest size greatest molecular speed most rapid effusion (Graham's law)
6) for explanation point in 9 (a), (c), and (d), credit is earned at step indicted in boldface type.
(a) two points
Calculated M m
(HA) too low
M(NaOH) => V(NaOH) => n(NaOH) => n(HA) => Mm(HA)
(M = n ÷ V) and (M m
= m÷ n)

(b) two points
Calculated M m
(HA) not affected
Any one of the following reasons. Water: does not change n(HA) changes only M(HA) -- sense of dilution is not a reactant
(c) two points
Calculated M m
(HA) too high equivalence point => n(NaOH) => n(HA) => M m
(HA)
(expected pH higher)
Note: "no effect if NaOH standardized with same indicator" earns 2 points; no credit earned if pH=7 or neutral
(d) two points
Calculated M m
(HA) too low
V(NaOH) => n(NaOH) => n(HA) => M m
(HA)
Note: point earned for V(NaOH) only if:
(i) no explanation point is earned in (a)
(ii) explanation in (a) also includes V(NaOH)
7)
(a) two points
The sign of the cell potential will be positive because (any one is sufficient):

K is greater than 1 the reaction is spontaneous (occurs)
E° for Sr 2+ is more positive
Standard reduction potential for Sr more negative
E° = + 0.52 V
Note: only 1 point earned for just E° positive because K eq
positive.
(b) one point
The oxidizing agent is Mg
2+
(c) two point
The cell potential would increase
Since all ions are at 1 M, Q for the system is 1 and E° = (RT/nF) ln K so as T increases, so should E°
Note: no credit lost if student recognizes K eq
dependence on T. For temperature change in this problem, decrease in ln K term is small relative to the term RT/nF
OR
No change, because in the Nernst equation E cell
= E° - (RT/nF) ln Q ln Q = 0, and E cell
= E°
Note: this second approach earns 1 point only
(d) two points
E cell
will increase
In the equation E cell
= E° - (0.0592 / n) log Q
Q = 0.1 therefore log Q is negative therefore term after E° is positive therefore E cell
increases
OR

with the concentration of Mg
2+
larger than that of Sr
2+
, Le Chatelier's principle predicts the reaction will have a larger driving force to the right and a more positive E cell
(e) one point
At equilibrium, E cell
= 0
Note: "balanced", "neutral", or "no net reaction" not accepted
8)
(a) one point
2 NO + 2 H
2
---> N
2
+ 2 H
2
O
(b) two points
N
2
O
2
and N
2
O are intermediates because they appear in the mechanism but not in the overall products (or reactants)
(c) three points; one for each half of conclusion (1) answer, third for conclusion (2) answer
Student indicates conclusion (1) is correct, because the sum of the exponents in rate law is 2 + 1 = 3
Student indicates conclusion (2) is incorrect, because no step involves two NO molecules and a H
2
molecule
(d) two points; T goes up therefore k goes up: because increasing number of collisions between reactants are occuring with sufficient energy to form an activated complex
OR

T goes up therefore rate goes up because no change in concentration of reactants, therefore k must increase
OR from Arrhenius equation (not required in AP Chemistry curriculum, but noted in some student responses): as T goes up, k goes up
OR graph as below with proper explanation
9)
(a) two points
Hydrogen bonding (or dipole-dipole attraction) in HF is greater than it is in HCl
Note: only one point earned if simply stated that HF has greater intermolecular forces than HCl
(b) two points
AsF
3
has a trigonal pyramid shape and bond dipoles do NOT cancel (or asymmetric molecule)
AsF
5
has a trigonal bipyramid shape and bond dipoles cancel (or symmetric shape)
Notes: explanation must refer to shape in order to earn point; one point earned if only correct
Lewis structures are given.

(c) two points
NO
2
¯ has resonance structures
HNO
2
has no resonance structures
OR one N-O single bond, one N=O double bond
Note: one point earned if only correct Lewis structures, including resonance for NO
2
¯ given.
(d) two points
Sulfur uses d orbitals (or expanded octet), oxygen has no d orbitals in its valence shell
OR
Sulfur is a larger atom, can accomodate more bonds.