AP Chemistry Concept List – EQUILIBRIUM
All Problems are equilibrium problems because
All problems involve stoichiometry: soluble salts, strong acids, strong bases
Some problems involve equilibrium: “insoluble” salts, weak acids, weak bases
For chemical reactions – Keq, Kc, and Kp are the important quantities
For physical changes – Ka, Kb, Ksp, Kionize, and Kdissocation are the important quantities
Important points
1.
Law of mass action
aA + bB + …  rR +sS + xxx
Kc = [R]r [S]s … / [A]a [B]b …
2.
Kc for molarity for ions and gases
3.
Kp with atm, or mmHg for gases
Relationship / connection between these Kp = Kc (RT)Δn
4.
Orientation of collisions
5.
Shifting equilibrium – Le Chatlier’s Principle
a.
solid
b.
liquid
c.
catalyst
d.
inert gas added
e.
temperature changes (increasing T favors endothermic processes)
f.
only factors in equation constant will affect Keq eg. CaCO3(s)  CaO(s) + CO2(g)
g.
pressure / volume changes
6.
Important vocabulary
Driving force
Favors (reactants or products)
Shifts (in LeChatelier arguments
7.
K > 1 products favored
K < 1 reactants favored
8.
Excluded: solids, pure liquids, water (in aqueous solution)
9.
Typical question: Given Kc and the starting concentration of reactants, find the concentration (or pH !)
of products at equilibrium.
Example: Kc of acetic acid = 1.754 × 10-5. Find the pH of a 0.100 M solution of acetic acid.
1
10.
Equilibrium constant for a reverse reaction = 1 / K of the value of the forward reaction.
11.
When using Hess’s Law: Koverall = K1 × K2
12.
If out of equilibrium: Calculate the reaction quotient (Q) in a similar fashion to the way an equilibrium
constant would be found. If:
Q < K forward reaction occurs to reach equilibrium
Q > K reverse reaction occurs to reach equilibrium
13.
Problem solving: Learn when to make an approximation (needed for multiple choice and free response
questions!). 5% rules usually works when value of K is 10-2 or smaller than value of known
concentrations.
Example:
A  B + C
K = 3.0 × 10-6
If [A] = 5.0 M; find [C] at equilibrium
If greater than 5 % use the quadratic equation:
ax2 + bx + c = 0
x
 b  b 2  4ac
2a
2
Free Response Questions
2003B #1
After a 1.0 mole sample of HI(g) is placed into an evacuated 1.0 L container at 700. K, the reaction represented
occurs. The concentration of HI(g) as a function of time is shown below.
2 HI(g)  H2(g) + I2(g)
a.
Write the expression for the equilibrium constant, Kc, for the reaction.
b.
What is [HI] at equilibrium?
c.
Determine the equilibrium concentrations of H2(g) and I2(g).
d.
On the graph above, make a sketch that shows how the concentration of H2(g) changes as a function of
time.
e.
Calculate the value of the following equilibrium constants at 700. K.
f.
i.
Kc
ii.
Kp
At 1,000 K, the value of Kc for the reaction is 2.6 × 10-2. In an experiment, 0.75 mole of HI(g), 0.10
mole of H2(g), and 0.50 mole of I2(g) are placed in a 1.0 L container and allowed to reach equilibrium at
1,000 K. Determine whether the equilibrium concentration of HI(g) will be greater than, equal to, or
less than the initial concentration of HI(g). Justify your answer.
10 points total
3
a.
b.
c.
d.
e.
f.
Kc 
[H 2 ][I 2 ]
[HI]
1 point
From the graph, [HI] eq is 0.80 M
1 point
2HI(g) -->
H2(g) +
I2(g)
init
1.0 M
0M
0M
chang -0.20 M
+0.10 M
+0.10 M
final 0.80 M
0.10 M
0.10 M
[H2] = [I2] = 0.10 M
1 point for stoichiometric relationship between HI and H2 and I2
1 point for equilibrium concentrations of H2 and I2
From the graph, [H2] eq is 0.10 M. The curve should have the following characteristics: start at
0 M, increase to 0.10 M, reach equilibrium at the same time [HI] reaches equilibrium
1 point for any two characteristics, 2 points for all three characteristics
[H ][I ] (0.10)(0.10)
 0.016
i) K c  2 22 
[HI]
(0.80) 2
1 point for correct substitution which must agree with parts b and c
ii) KP = Kc = 0.016 because the number of moles of gaseous products is equal to the number of
moles of gaseous reactants
1 point
[H ][I ] (0.10)(0.50)
Q  2 22 
 0.089
[HI]
(0.75)2
Kc = 0.026, Q > Kc, therefore to establish equilibrium, the numerator must decrease and the
denominator must increase. Therefore, [HI] will increase.
1 point for calculating Q and comparing to Kc
1 point for predicting correct change in [HI]
2004B #1
N2(g) + 3 H2(g)  2 NH3(g)
For the reaction represented above, the value of the equilibrium constant, Kp is 3.1 × 10-4 at 700 K.
a.
Write the expression for the equilibrium constant, Kp, for the reaction.
b.
Assume that the initial partial pressures of the gases are as follows:
P(N2) = 0.411 atm, P(H2) = 0.903 atm, and P(NH3) = 0.224 atm.
i)
Calculate the value of the reaction quotient, Q, at these initial conditions.
ii)
Predict the direction in which the reaction will proceed at 700. K if the initial partial pressures
are those given above. Justify your answer.
c.
Calculate the value of the equilibrium constant, Kc, given that the value of Kp for the reaction at 700. K
is 3.1 × 10-4.
d.
The value of Kp for the reaction represented below is 8.3 × 10-3 at 700. K.
NH3(g) + H2S(g)  NH4HS(g)
4
Calculate the value of Kp at 700. K for each of the reactions represented below.
i)
NH4HS(g)  NH3(g) + H2S(g)
ii)
2 H2S(g) + N2(g) + 3 H2(g)  2 NH4HS(g)
10 points
a.
b.
c.
d.
KP 
2
PNH
3
PN 2 PH3 2
1 point for pressure expression
1 point for correct substitution
2
PNH
(0.224) 2
3

 0.166
i) Q 
PN 2 PH3 2 (0.411)(0.903)3
1 point for calculation of Q with correct mass action expression consistent with part a
ii) Since Q > Kp, the numerator must decrease and the denominator must increase, so the
reaction must proceed from right to left to establish equilibrium
1 point for direction or for stating that Q > Kp
1 point for explanation
Kp = Kc (RT)Δn, with Δn = 2 – 4 = -2
3.1 × 10-4 = Kc (0.0821 L atm mol-1 K-1 × 700 K)-2
3.1 × 10-4 = Kc (57.5)-2  3.1 × 10-4 = Kc(3.1 × 10-4 )
Kc = 1
1 point for calculating Δn
1 point for correct substitution and value of Kc
i) Kp = (8.3 × 10-3)-1 = 1.2 × 102
1 point
ii) 2 × [NH3 + H2S  NH4HS]
Kp = (8.3 × 10-3)2
N2 + 3 H2  2 NH3
Kp = 3.1 × 10-4
2 H2S + N2 + 3H2  2 NH4HS
Kp = (8.3 × 10-3)2(3.1 × 10-4) = 2.1 × 10-8
1 point for squaring Kp for NH4HS or for multiplying Kp’s
1 point for correct Kp
1988 #6
NH4HS(s)  NH3(g) + H2S(g)
For this reaction, ΔH° = + 93 kJ. The equilibrium above is established by placing solid NH4HS in an evacuated
container at 25 °C. At equilibrium, some solid NH4HS remains in the container. Predict and explain each of the
following.
a.
The effect on the equilibrium partial pressure of NH3 gas when additional solid NH4HS is introduced
into the container.
b.
The effect on the equilibrium partial pressure of NH3 gas when additional H2S gas is introduced into the
container.
c.
The effect on the mass of solid NH4HS present when the volume of the container is decreased.
d.
The effect on the mass of solid NH4HS present when the temperature is increased
5
Average score = 4.31
a) two points
The equilibrium pressure of NH3 gas would be unaffected Kp = (PNH3) (PH2S). Thus the amount of solid
NH4HS present does not affect the equilibrium.
b) two points
The equilibrium pressure of NH3 gas would decrease. In order for the pressure equilibrium constant, Kp,
to remain constant, the equilibrium pressure of NH3 must decrease when the pressure of H2S is increased.
Kp = (PNH3) (PH2S)
(A complete explanation based on Le Chatelier's principle is also acceptable.)
c) two points
The mass of NH4HS increases. A decrease in volume causes the pressure of each gas to increase. To
maintain the value of the pressure equilibrium constant, Kp, the pressure of each of the gases must
decrease. That decrease realized by the formation of more solid NH4HS.
Kp = (PNH3) (PH2S)
(A complete explanation based on Le Chatelier's principle is also acceptable.)
d) two points
The mass of NH4HS decreases because the endothermic reaction absorbs heat and goes nearer to
completion (to the right) as the temperature increases. (One point was assigned for each correct
prediction and one point for each correct explanation.)
1980 #6
NH4Cl(s)  NH3(g) + HCl(g) for this reaction, ΔH = +42.1 kilocalories
Suppose the substances in the reaction above are at equilibrium at 600 K in volume V and at pressure P. State
whether the partial pressure of NH3(g) will have increased, decreased, or remained the same when equilibrium
is reestablished after each of the following disturbances of the original system. Some solid NH4Cl remains in
the flask at all times. Justify each answer with a one- or two-sentence explanation.
a.
A small quantity of NH4Cl is added.
b.
The temperature of the system is increased.
c.
The volume of the system is increased.
d.
A quantity of gaseous HCl is added.
e.
A quantity of gaseous NH3 is added.
a. Partial pressure of ammonia will not change, ammonium chloride is a solid and does not effect
equilibrium position. If partial pressure of ammonia would increase the partial pressure so would
partial pressure of HCl and Keq value would change.
b. Increase the temp shifts equilibrium to the right to favor endothermic
NH3 would increase.
process, so partial pressure of
c. Increasing the volume (decrease the pressure) would not change the partial pressure because both NH3
and HCl would decrease and cause a change in the Keq value. Only temperature can cause this change.
6
d. The reaction would shift to the left so partial pressure of NH3 would decrease.
e. The partial pressure of NH3 would increase because the equilibrium would shift to the left, amount of
HCl decreases so amount of NH3 would need to increase to keep Keq constant.
Multiple Choice Questions
1999 #67
What is the molar solubility in water of Ag2CrO4? (The Ksp for Ag2CrO4 is 8 × 10-12.)
A) 8 × 10-12 M
B) 2 × 10-12 M
C) (4 × 10-12 M)1/2
D) (4 × 10-12 M)1/3
E) (2 × 10-12 M)1/3 (25 %)
2008 #59
The diagram above represents a mixture of NO2(g) and N2O4(g) in a 1.0 L container at a given temperature.
The two gases are in equilibrium according to the equation 2 NO2(g)  N2O4(g). Which of the following
must be true about the value of the equilibrium constant for the reaction at this temperature?
A)
B)
C)
D)
E)
2002 #42
K=0
0 < K < 1 (32%)
K=1
K>1
There is not enough information to determine the relative value of K.
H2(g) + Br2(g)  2 HBr(g)
At a certain temperature, the value of the equilibrium constant, K, for the reaction represented above is 2.0 ×
105. What is the value of K for the reverse reaction at the same temperature?
A)
B)
C)
D)
E)
2008 #35
-2.0 × 10-5
5.0 × 10-6 (24 %)
2.0 × 10-5
5.0 × 10-5
5.0 × 10-4
H2(g) + I2(g)  2 HI(g)
ΔH > 0
Which of the following changes to the equilibrium system represented above will increase the quantity of HI(g)
in the equilibrium mixture?
I.
Adding H2(g)
II.
Increasing the temperature
III.
Decreasing the pressure
7
A)
B)
C)
D)
E)
I only
III only
I and II only (56%)
II and III only
I, II, and III
1999 #54
2NO(g) + O2(g)  2 NO2(g)
H < 0
Which of the following changes alone would cause a decrease in the value of Keq for the reaction represented
above?
A) Decreasing the temperature
B) Increasing the temperature (35%)
C) Decreasing the volume of the reaction vessel
D) Increasing the volume of the reaction vessel
E) Adding a catalyst
2002 #37
HCO3-(aq) + OH-(aq)  H2O(l) + CO32-(aq)
ΔH = -41.4 kJ
When the reaction represented by the equation above is at equilibrium at 1 atm and 25 oC, the ratio
[CO 32- ]
can
[HCO 3- ]
be increased by doing which of the following?
A)
B)
C)
D)
E)
Decreasing the temperature (45%)
Adding acid
Adding a catalyst
Diluting the solution with distilled water
Bubbling neon gas through the solution
8
AP Chemistry Concept List – ACID - BASE
pH = - log [H+]
pOH = - log [OH-]
Kw = [H+] [OH-] = 1×10-14 at 25 oC
If you know one quantity, you know the other three
pH
[H+]
pOH
[OH-]
Definitions
Acid
Donates H+
Donates protons
Accepts e- pairs (AlCl3)
Base
Donates OHAccepts protons - {anions?}
Donates e- pairs (NH3)
Theory
Arrhenius
Bronsted – Lowry
Lewis
Conjugate Acid – Base Pairs
1.
HCl + H2O → H3O+ + Cl-
2.
NH3 + H2O  NH4+ + OH-
3.
HSO4- + H2O  H3O+ + SO42-
4.
CO32- + H3O+  HCO3- + H2O
A.
Ka
Ka 
Weak Acid
HCN  H+ + CN-
[H  ] [CN ]
 6.2  1010
[HCN]
What is the pH of a 0.5 M HCN solution?
B.
Kb
Kb 
Weak base NH3 + H2O  NH4+ + OH-
[ NH4 ] [OH ]
 1.8  10 5
[ NH3 ]
What is the pH of a 0.5 M NH2OH solution?
C.
Ksp
Insoluble Salts
MgF2(s)  Mg2+ + 2F-
Ksp = [Mg2+] [F-]2 = 6.6 × 10-9
9
What is the solubility of MgF2 in molarity?
D.
Buffers – a weak acid/base and its soluble salt (conjugate base or acid) mixture
pH  pK a  log
[base]
[acid ]
What is the pH of a 0.5 M HC2H3O2 in 2 M NaC2H3O2 solution? Ka = 1.8 × 10-5
E.
Salts of Weak Acids and Weak Bases
What is the pH of a 1 M NaC2H3O2 solution?
Titrations and Endpoints
At endpoint: acid moles = base moles or [H+] = [OH-]
Strong acid – strong base
endpoint pH = 7
Strong acid – weak base
endpoint pH < 7
Weak acid – strong base
endpoint pH > 7
The last two are important because of conjugate acid and base pairs
11.
Acid strength – know the 6 strong acids: HCl, HBr, HI, HNO3, HClO4, and H2SO4 (removal of the first
H+ only)
a)
binary acids – acid strength increased with increasing size and electronegativity of the “other
element”. (NOTE: Size predominates over electronegativity in determining acid strength.)
Example: H2Te > H2O and HF > NH3
b)
oxoacids – Acid strength increases with increasing:
1)
2)
3)
electronegativity
number of bonded oxygen atoms
oxidation state
of the “central atom”. However, need to show as electron withdrawing groups rather than trends
(trends need to be explained as a result of chemical principles rather than solely as a trend).
Example: HClO4 [O3Cl(OH)] is very acidic
NaOH is very basic
Acid strength also increases with DECREASING radii of the “central atom”
Example:
HOCl (bond between Cl and OH is covalent – acidic)
HOI (bond between I and OH is ionic – basic)
12.
Acid Ionization Constant (Ka):
10
HA + H2O  H3O + A
+
[A  ][H 3O  ]
Ka 
[HA ]
-
Example: HF + H2O  H3O+ + F-
Ka 
[F  ][ H 3O  ]
[HF]
What is the pH of 0.5 M HCN solution for which Ka = 6.2 × 10-10?
13.
Base Ionization Constant (Kb):
Kb 
B + H2O  BH+ + OH-
Example: F + H2O  HF + OH
-
-
[BH  ][OH ]
[B]
[HF][OH  ]
Kb 
[F  ]
What is the pH of a 0.5 M NH2OH solution for which Kb = 6.6 × 10-9?
Do equal number of Ka and Kb problems as they are equally likely!
14.
Ka × Kb = Kw = 10-14
ONLY applies for conjugate acids and bases at 25 oC!
15.
Percent ionization = [H+]equilibrium / [HA]initial × 100
16.
Lewis Acids and Bases:
Lewis acid – electron pair acceptor
Lewis bases – electron pair donor
In complex ions formation, metal ions are Lewis acids, and ligands are Lewis bases.
Example:
Cu2+ + 4 NH3  [Cu(NH3)4]2+
2+
Cu acts as an acid; NH3 acts as a base
17.
Buffers:
Similar concentrations of a weak acid and its conjugate base
-orSimilar concentrations of a weak base and its conjugate acid
If these concentrations are large in comparison to SMALL amounts of added acid or base, equilibrium
will be shifted slightly and the pH change resisted. Consider:
HA  H+ + A-
11
Ka 
[A  ][ H 3O  ]
[HA ]
[H+] = Ka [HA] / [A-]
pH = pKa – log [HA] / [A-] or
pH = pKa + log [A-] / [HA]
(Henderson-Hasselbach equation)
B + H2O  HB+ + OH-
Kb 
[BH  ][OH  ]
[B]
[OH-] = Kb [B] / [HB+]
or
pOH = pKb + log [HB+]/[B]
(Henderson-Hasselbach equation)
What is the pH of a solution which is 0.5 M HC2H3O2 in 2 M NaC2H3O2 for which Ka = 1.8 × 10-5?
18.
Polyprotic Acids: H3PO4, H2SO4, H2C2O4, etc.
19.
Equivalence Point – the point at which stoichiometric amounts of reactants have reacted.
NOTE: This only occurs at pH = 7 for the reaction of a strong acid with a strong base. The equivalence
point will occur ABOVE pH = 7 (more basic) for a weak acid / strong base titration. (the conjugate base of
the weak acid will react with water.) The equivalence point will occur BELOW pH = 7 for a weak base /
strong acid titration (the conjugate acid of the weak base with react with water).
20.
Indicators – select bases on the pH at the equivalence point.
21.
Titration curves:
a)
Weak acid / strong base
HA + OH-  A- + H2O
NOTE: Graph should have “pH” as the vertical axis and “added base” as the horizontal axis. The graph
should be in an “S” shape. The middle of the lower part of the “S” indicates the point of maximum
buffering where [HA] / [A-] = 1. The middle of the “S” is the equivalence point (above pH = 7) and
[HA] = 0. The top part of the “S” levels off at the pH of the base solution.
b)
Weak base / strong acid
B + H3O+  BH+ + H2O
NOTE: Graph should have “pH” as the vertical axis and “added base” as the horizontal axis. The graph
should be in a “backwards S” shape. The middle of the upper part of the “backwards S” indicates the
point of maximum buffering where [B] / [HB+] = 1. The middle of the “backwards S” is the equivalence
point (below pH = 7) and [B] = 0. The bottom part of the “backwards S” levels off at the pH of the acid
solution.
c)
Weak diprotic acid / strong base
H2A + OH-  HA- + H2O
12
HA- + OH-  A2- + H2O
NOTE: Graph should have “pH” as the vertical axis and “added base” as the horizontal axis. The graph
should be in a “double S” shape. The middle of the lower part of the “first S” indicates the point of
maximum buffering of the first buffering zone where [H2A] / [HA-] = 1. The middle of the “first S” is
the first equivalence point where [H2A] = 0. The top of the “first S” (i.e. the lower part of the “second
S”) indicates the point of maximum buffering of the second buffering zone where [HA-] / [A2-] = 1. The
middle of the “second S” is the second equivalence point where [HA-] = 0. The top part of the “second
S” levels off at the pH of the base solution.
22.
Solubility Product (Ksp)
Example 1: Co(OH)2(s)  Co2+ + 2OHKsp = [Co2+][OH-]2
(don’t forget – molar concentration of OH- is twice the solubility)
Example 2: Solubility of Ag2SO4 is 0.016 mol L-1 (5.0 g L-1). Find the Ksp of Ag2SO4. (Answer: Ksp =
1.5 × 10-5)
23.
Ion product (Qi) – equivalent to the “reaction quotient”
Qi < Ksp
Qi = Ksp
Qi > Ksp
24.
all ions in solution; more solid will dissolve
equilibrium – solution is saturated
precipitation will occur until Qi = Ksp
Solubility of any salt which contains a basic anion is influenced by pH:
Example: Consider a solution which is 1 M in each of Fe3+ and H+. Will Fe(OH)3 precipitate? Answer:
NO!
Fe(OH)3  Fe3+ + 3 OHQi = [Fe3+][OH-]
If [H+] = 1 M, then [OH-] = 10-14
Qi = (1) (10-14)3 = 10-42
Since Ksp for Fe(OH)3 = 3 × 10-39; precipitation does not occur.
However, what if [H+] = 10-5? Then [OH-] = 10-9
Qi = (1) (10-9)3 = 10-27
Since Qi > Ksp; Fe(OH)3 will precipitate!
13
Free Response Questions
2007 #1
1.
HF(aq) + H2O(l)  H3O+(aq) + F-(aq)
Ka = 7.2 × 10-4
Hydrofluoric acid, HF(aq), dissociates in water as represented by the equation above.
a.
Write the equilibrium constant expression for the dissociation of HF(aq) in water.
b.
Calculate the molar concentration of H3O+ in a 0.40 M HF(aq) solution.
HF(aq) reacts with NaOH(aq) according to the reaction represented below.
HF(aq) + OH-(aq)  H2O(l) + F-(aq)
A volume of 15 mL of 0.40 M NaOH(aq) is added to 25 mL of 0.40 M HF(aq) solution. Assume volumes are
additive.
c.
Calculate the number of moles of HF(aq) remaining in the solution.
d.
Calculate the molar concentration of F-(aq) in the solution.
e.
Calculate the pH of the solution.
9 points
a.
b.
c.
d.
e.
[ H 3O  ][ F  ]
1 point
Ka 
[ HF ]
[ H O  ][ F  ]
( x)( x)
Ka  3

 7.2  10 4
[ HF ]
0.40  x
Assume x << 0.40, then x2 = (0.40)(7.2 × 10-4), x = [H3O+] = 0.017 M
1 point is earned for the correct setup (or that consistent with part a)
1 point is earned for the correct concentration
mol HF = initial mol HF – mol NaOH added
= (0.025 L)(0.40 mol L-1)–(0.015 L)(0.40 mol L-1)=0.010–0.0060=0.0040 mol
1 point is earned for determining initial number of moles of HF and OH1 point is earned for setting up and doing correct subtraction
mol F- formed = mol NaOH added = 0.0060 mol F0.0060 mol F- / (0.015 + 0.025) L of solution = 0.15 M F1 point is earned for determining the number of moles of F1 point is earned for dividing the number of moles of F- by the correct total volume
[HF] = 0.004 mol HF / 0.040 L = 0.10 M HF
[ H 3O ][ F  ]
[ HF ]  K a
0.10  7.2  104

Ka 

 [ H 3O ] 
 4.8  10 4

[ HF ]
[F ]
0.15
-4
pH = - log (4.8 × 10 ) = 3.32
OR
Henderson – Hasselbach equation
1 point is earned for indicating that the resulting solution is a buffer (by showing a ration of [F-]
to [HF] or moles of F- to HF)
14
1 point is earned for the correct calculation of pH
2005B #1 Ka
[H3O ][OCl ]
Ka 
 3.2  108
[HOCl]
Hypochlorous acid, HOCl, is a weak acid in water. The Ka expression for HOCl is shown above.
a.
Write a chemical equation showing how HOCl behaves as an acid in water.
b.
Calculate the pH of a 0.175 M solution of HOCl.
c.
Write the net ionic equation for the reaction between the weak acid HOCl(aq) and the strong base
NaOH(aq)
d.
In an experiment, 20.00 mL of 0.175 M HOCl(aq) is placed in a flask and titrated with 6.55 mL of 0.435
M NaOH(aq).
i)
Calculate the number of moles of NaOH(aq) added.
ii)
Calculate [H3O+] in the flask after the NaOH(aq) has been added.
iii)
Calculate [OH-] in the flask after the NaOH(aq) has been added.
10 points
a.
HOCl + H2O  OCl- + H3O+
b.
0.175
0
0
-x
x
x change
0.175 –x
x
x final
Ka 
1 point
initial
[H3O ][OCl ]
xx

 3.2  108
[HOCl]
0.175  x
Assume that x <<< 0.175, x = 7.5 × 10-5 M, pH = 4.13
1 point is earned for calculating the value of [H3O+]
1 point is earned for calculating the pH
c.
HOCl + OH-  OCl- + H2O
1 point for both correct reactants
1 point for both correct products
d.
i) mol(NaOH)=6.55 mL(1L/1000 mL)(0.435 M)=2.85×10-31 point
ii) mol(HOCl) = 20.00 (1/1000) (0.175) = 3.50 ×10-3 mol 1 point
OH- is the limiting reactant, therefore all of it reacts
HOCl +
OH-  OCl-
+ H2O
0.00350
0.00285
0
-0.00285
-0.00285
0.00285
0.00065
0
0.00285
15
M(HOCl) = 0.00065 / 0.02655 = 0.0245 M
M(OCl-) = 0.00285 / 0.02655 = 0.107 M
HOCl + H2O -->
H3O+ +
0.0245
0
0.107
-x
x
x
0.0245
x
0.107 + x
Ka 
OCl-
1 point
[H3O ][OCl ] x (0.107  x)

 3.2  108
[HOCl]
(0.0245  x)
Assume 0.107 + x = 0.107 and 0.0245-x = 0.0245
x = 7.3 × 10-9 M
+
-
iii) [H3O ][OH ]= 10
1 point
-14
= Kw
[OH-] = 10-14 / [H3O+] = 10-14 / 7.3 × 10-9 = 1.4 × 10-6 M 1 point
2011 #1
1.
Each of three beakers contains 25.0 mL of a 0.100 M solution of HCl, NH3, or NH4Cl, as shown above.
Each solution is at 25 oC.
a.)
Determine the pH of the solution in beaker 1. Justify your answer.
b.)
In beaker 2, the reaction NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) occurs. The value of Kb
for NH3(aq) is 1.8 × 10-5 at 25 oC.
c.)
d.)
i.)
Write the Kb expression for the reaction of NH3(aq) with H2O(l).
ii.)
Calculate the [OH-] in the solution in beaker 2.
In beaker 3, the reaction NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq) occurs.
i.)
Calculate the value of Ka for NH4+ at 25 oC.
ii.)
The contents of beaker 2 are poured into beaker 3 and the resulting solution is stirred.
Assume that volumes are additive. Calculate the pH of the resulting solution.
The contents of beaker 1 are poured into the solution made in part c) ii). The resulting solution is
stirred. Assume that volumes are additive.
16
i.)
Is the resulting solution an effective buffer? Justify your answer.
ii.)
Calculate the final [NH4+] in the resulting solution at 25 oC.
a.)
pH = -log[H+] = -log(0.100) = 1.000
b.)
i)
Kb = [NH4+][OH-] / [NH3]
ii)
Let [OH−] = x, then Kb = (x)(x) / (0.100 - x) Assume that x << 0.100 M, then 1.8 × 10−5 = 0.100 / x ⇒
x = [OH−] = 1.3 × 10-3 M
1 point is earned for the correct setup. 1 point is earned for the
correct answer.
i)
Ka = Kw / Kb = 1.0 x 10-14 / 1.8 10-5 = 5.6 × 10-10 1 point is earned for the correct answer.
ii)
In the resulting solution, [NH3] = [NH4+]; Ka = 5.6 × 10-10 = [NH3][H3O+] / [NH4+] Thus [H3O+] =
5.6 × 10-10; pH = -log(5.6 × 10-10) = 9.25
1 point is earned for noting that the solution is a
buffer with [NH3] = [NH4+]. 1 point is earned for the correct pH.
c.)
d.)
1 point
1 point is earned for the correct expression.
i)
The resulting solution is not an effective buffer. Virtually all of the NH3 in the solution formed
in (c)(ii) will react with the H3O+ from solution 1: NH3 + H3O+ → NH4+ + H2O leaving mostly NH4+
in the final solution. Since only one member of the NH4+/NH3 conjugate acid-base pair is left, the
solution cannot buffer both base and acid.
1 point is earned for the correct response with an
acceptable justification.
ii)
moles = (volume)(molarity) moles H3O+ in sol. 1 = (0.0250)(0.100) = 0.00250 mol moles NH3 in sol.
2 = (0.0250)(0.100) = 0.00250 mol moles NH4+ in sol. 3 = (0.0250)(0.100) = 0.00250 mol When the
solutions are mixed, the H3O+ and NH3 react to form NH4+, resulting in a total of 0.00500 mol NH4+.
The final volume is the sum (25.0 + 25.0 + 25.0) = 75.0 mL. The final concentration of NH4+ =
(0.00500 mol/0.0750 L) = 0.0667 M.
1 point is earned for the correct calculation of moles of
NH4+. 1 point is earned for the correct calculation of the final volume and concentration.
1996 A/B lab #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 standard 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.
for explanation point in 9 (a), (c), and (d), credit is earned at step indicted in boldface type.
(a) two points
Calculated Mm(HA) too low
M(NaOH) => V(NaOH) => n(NaOH) => n(HA) => Mm(HA)
(M = n ÷ V) and (Mm = m÷ n)
17
(b) two points
Calculated Mm(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 Mm(HA) too high
equivalence point => n(NaOH) => n(HA) => Mm(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 Mm(HA) too low
V(NaOH) => n(NaOH) => n(HA) => Mm(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)
2000 #8 A/B Lab
A volume of 30.0 mL of 0.10 M NH3(aq) is titrated with 0.20 M HCl(aq). The value of the base-dissociation
constant, Kb, for NH3 in water is 1.8 × 10-5 at 25 oC.
a.
Write the net-ionic equation for the reaction of NH3(aq) with HCl(aq).
b.
Using the axes provided below, sketch the titration curve that results when a total of 40.0 mL of 0.20 M
HCl(aq) is added dropwise to the 30.0 mL volume of 0.10 M NH3(aq).
c.
From the table below, select the most appropriate indicator for the titration. Justify your choice
Indicator
Methyl Red
pKa
5.5
18
Bromothymol Blue
Phenolphthalein
d.
7.1
8.7
If equal volumes of 0.10 M NH3(aq) and 0.10 M NH4Cl(aq) are mixed, is the resulting solution acidic,
neutral, or basic? Explain
8 points
(a)
NH3(aq) + H+(aq) → NH4+(aq)
1 point
or
NH3(aq) + H3O+(aq) → NH4+(aq) + H2O(l)
Note: phase designations not required to earn point
(b)
Sketch of Titration Curve:
3 pnts
• 1st pt. → initial pH must be > 7 (calculated pH ≈ 11)
• 2nd pt. → equivalence point occurs at 15.0 mL ± 1 mL of HCl added (equivalence point must be
detectable from the shape of the curve or a mark on the curve)
• 3rd pt. → pH at equivalence point must be < 7 (calculated pH ≈ 5).
Note: a maximum of 1 point earned for any of the following:
- a line without an equivalence point
- a random line that goes from high pH to low pH
- an upward line with increasing pH (equivalence point MUST be at 15.0 mL)
(c)
Methyl Red would be the best choice of indicator,
1 point
Because the pKa for Methyl Red is closest to pH atequivalence point.
1 point
Notes: • explanation must agree with equivalence point on graph
• alternative explanation that titration involves strong acid and weak base (with product an acidic salt)
earns the point
(d)
The resulting solution is basic.
1 point
-5
+
-10
Kb for NH3 (1.8 × 10 ) and Ka for NH4 (5.6 × 10 ) indicate that NH3 is a stronger base than NH4+ is
an acid
or
[OH-] = Kb = 1.8 × 10-5 because of the equimolar and equivolume amounts of ammonium and ammonia
→ cancellation in the buffer pH calculation. Thus pOH ≈ 5 and pH ≈ 9 (i.e., recognition of buffer, so
 0.05 
that log 
  0 → pOH = pKb ≈ 5 → pH = 14 – pOH ≈ 9)
 0.05 
2011B #1
1.
Answer the following questions about the solubility and reactions of the ionic compounds M(OH)2 and
MCO3, where M represents an unidentified metal.
a.)
Identify the charge of the M ion in the ionic compounds above.
b.)
At 25 oC, a saturated solution of M(OH)2 has a pH of 9.15.
i.)
Calculate the molar concentration of OH-(aq) in the saturated solution.
ii.)
Write the solubility product expression for M(OH)2.
iii.)
Calculate the value of the solubility product constant, Ksp, for M(OH)2 at 25 oC.
19
c.)
For the metal carbonate, MCO3, the value of the solubility product constant, Ksp, is 7.4 × 10-14 at
25 oC. On the basis of this information and your results in part b), which compound, M(OH)2 or
MCO3, has the greater molar solubility in water at 25 oC? Justify your answer with a calculation.
d.)
MCO3 decomposes at high temperatures, as shown by the reaction represented below.
MCO3(s)  MO(s) + CO2(g)
A sample of MCO3 is placed in a previously evacuated container, heated to 423 K, and allowed
to come to equilibrium. Some solid MCO3 remains in the container. The value of KP for the
reaction at 423 K is 0.0012.
i.)
Write the equilibrium constant expression for KP of the reaction.
a.)
2+
b.)
i)
ii.)
Determine the pressure, in atm, of CO2(g) in the container at equilibrium at 423 K.
iii.)
Indicate whether the value of ΔGo for the reaction at 423 K is positive, negative, or zero.
Justify your answer.
1 point is earned for the correct charge.
pOH = 14 – pH, pOH = 14 − 9.15 = 4.85, [OH−] = 10−4.85 = 1.4 × 10−5 M
correct concentration.
1 point is earned for the
ii)
Ksp = [M2+] [OH−]2
iii)
[M2+] = (1/2)[OH−] = (1/2)(1.4 × 10−5 M) = 7.0 × 10−6 M
Ksp = [M2+] [OH−]2 = (7.0 × 10−6)(1.4
× 10−5)2
= 1.4 × 10−15 1 point is earned for the correct relationship between [M2+] and [OH−].
1 point is earned for the correct value.
1 point is earned for the correct expression.
c.)
For M(OH)2 : [M2+] and molar solubility = 7.0 × 10−6 M For MCO3: Ksp = 7.4 × 10−14 = [M2+][CO32−] [M2+]
and molar solubility = 2.7 × 10−7 M Because 7.0 × 10−6 M > 2.7 × 10−7 M, M(OH)2 has the greater molar
solubility.
1 point is earned for the molar solubility of MCO3. 1 point is earned for an answer consistent
with the calculated molar solubility.
d.)
i)
Kp = P(CO2)
ii)
P(CO2) = 0.0012 atm
iii)
ΔG° = −RT lnK K = 0.0012 < 1, thus lnK is negative; therefore ΔG° is positive.
earned for the correct answer with justification.
2010 #1
1 point is earned for the correct expression.
1 point is earned for the correct pressure.
1 point is
Several reactions are carried out using AgBr, a cream-colored silver salt for which the value of
the solubility-product constant, Ksp, is 5.0 × 10-13 at 298 K
(a) Write the expression for the solubility-product constant, Ksp, of AgBr
(b) Calculate the value of [Ag+] in 50.0 mL of a saturated solution of AgBr at 298 K.
20
Let x = equilibrium concentration of Ag+ (and of Br−). Then Ksp
= 5.0 × 10-13 = x2 ⇒ x = 7.1 × 10-7 M
One point is earned for the correct
value with supporting work (units not
necessary).
(c) A 50.0 mL sample of distilled water is added to the solution described in part (b), which is in a beaker with
some solid AgBr at the bottom. The solution is stirred and equilibrium is reestablished. Some solid AgBr
remains in the beaker. Is the value of [Ag+] greater than, less than, or equal to the value you calculated in part
(b) ? Justify your answer.
The value of [Ag+] after addition of distilled water is One point is earned for the correct
equal to the value in part (b). The concentration of
answer with justification.
ions in solution in equilibrium with a solid does not
depend on the volume of the solution.
(d) Calculate the minimum volume of distilled water, in liters, necessary to completely dissolve a 5.0 g sample
of AgBr(s) at 298 K. (The molar mass of AgBr is 188 g mol-1.)
5.0 g AgBr / (188 g AgBr mol-1 AgBr) = 0.0266 mol AgBr
One point is earned for the moles of dissolved AgBr.
0.0266 mol / V = 7.1 × 10-7 mol L-1  V = 3.7 × 104 L
One point for the correct answer for the volume of water
(e) A student mixes 10.0 mL of 1.5 × 10-4 M AgNO3 with 2.0 mL of 5.0 × 10-4 M NaBr and stirs the resulting
mixture. What will the student observe? Justify your answer with calculations.
[Ag+] = (10.0 mL) (1.5 × 10-4 M) / (12.0 mL) = 1.3 × 10-4 M
[Br-] = (2.0 mL) (5. 0 × 10-4 M) / (12.0 mL) = 8.3 × 10-5 M
Q = [Ag+][Br-] = (1.3 × 10-4 M) (8.3 × 10-5 M) = 1.1 × 10-8
1.1 × 10-8 > 5.0 × 10-13  a precipitate will form
One point is earned for calculation of
concentration of ions.
One point is earned for calculation of Q
and conclusion based on comparison
between Q and Ksp.
One point is earned for indicating the
precipitation of AgBr.
(f) The color of another salt of silver, AgI(s), is yellow. A student adds a solution of NaI to a test tube containing a small
amount of solid, cream-colored AgBr. After stirring the contents of the test tube, the student observes that the solid in the
test tube changes color from cream to yellow.
(i) Write the chemical equation for the reaction that occurred in the test tube.
AgBr(s) + I−(aq) → AgI(s) + Br−(aq)
OR
AgBr(s) + NaI(aq) → AgI(s) + NaBr(aq)
One point is earned for the correct equation.
(ii) Which salt has the greater value of Ksp : AgBr or AgI ? Justify your answer.
AgBr has the greater value of Ksp. The precipitate will
consist of the less soluble salt when both I−(aq) and
Br−(aq) are present. Because the color of the
precipitate in the test tube turns yellow, it must be
AgI(s) that precipitates; therefore Ksp for AgBr must
be greater than Ksp for AgI.
One point is earned for the correct choice with
justification.
21
OR
Keq for the displacement reaction is Ksp of AgBr / Ksp
of AgI. Because yellow AgI forms, Keq > 1; therefore
Ksp of AgBr > Ksp of AgI.
Multiple Choice Questions
2002 #64
Ascorbic acid, H2C6H6O6(s), is a diprotic acid with K1 = 7.9 × 10-5 and K2 = 1.6 × 10-12. In a
0.005 M aqueous solution of ascorbic acid, which of the following species is present in the lowest
concentration?
A)
B)
C)
D)
E)
H2O(l)
H3O+(aq)
H2C6H6O6(aq)
HC6H6O6-(aq)
C6H6O62-(aq) (38%)
2002 #33 Questions 33-34
The graph below shows the titration curve that results when 100. mL of 0.0250 M acetic acid is titrated with
0.100 M NaOH.
2002 #33.
A)
B)
C)
D)
E)
2002 #34.
pair?
A)
B)
Which of the following indicators is the best choice for this titration?
Indicator
Methyl Orange
Methyl Red
Bromothymol blue
Phenolphthalein
Alizarin
pH Range of
Color Change
3.2-4.4
4.8-6.0
6.1-7.6
8.2-10.0 (70%)
11.0-12.4
Which part of the curve corresponds to the optimum buffer action for the acetic acid / acetate ion
Point V (34%)
Point X
22
C)
D)
E)
Point Z
Along all of section WY
Along all of section YZ
2008 #4-6
A solution of a weak monoprotic acid is titrated with a solution of a strong base, KOH. Consider the points
labeled (A) through (E) on the titration curve that results, as shown below.
4.
5.
6.
The point at which the moles of the added strong base are equal to the moles of the weak acid initially
present (Answer C: 65%)
The point at which the pH is closest to that of a the strong base being added (Answer E: 69%)
The point at which the concentrations of the weak acid and its conjugate base are approximately equal
(Answer B: 38%)
23