2000 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
CHEMISTRY—SECTION II
(Total time—90 minutes)
Part A
Time—40 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 below. The Section II score weighting for this question is 20 percent.
2 H2S(g) →
← 2 H2(g) + S2(g)
1. When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g sample of H2S(g) is
introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.72 × 10−2 mol
of S2(g) is present at equilibrium.
(a) Write the expression for the equilibrium constant, Kc , for the decomposition reaction represented above.
(b) Calculate the equilibrium concentration, in mol L 1, of the following gases in the container at 483 K.
(i) H2(g)
(ii) H2S(g)
(c) Calculate the value of the equilibrium constant, Kc , for the decomposition reaction at 483 K.
(d) Calculate the partial pressure of S2(g) in the container at equilibrium at 483 K.
→ H S(g) at 483 K, calculate the value of the equilibrium constant, K .
(e) For the reaction H2(g) + 1 S2(g) ←
2
c
2
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®
2007 AP CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)
CHEMISTRY
Section II
(Total time—95 minutes)
Part A
Time—55 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A.
CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS.
It is to your advantage to do this, since you may obtain 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 the goldenrod
booklet. Do NOT write your answers on the lavender insert.
Answer Questions 1, 2, and 3. The Section II score weighting for each question is 20 percent.
1. A sample of solid U3O8 is placed in a rigid 1.500 L flask. Chlorine gas, Cl2(g), is added, and the flask is heated
to 862qC. The equation for the reaction that takes place and the equilibrium-constant expression for the reaction
are given below.
 3 UO Cl (g) + O (g)
U3O8(s) + 3 Cl2(g) Ž
2 2
2
Kp
( pUO2Cl2 )3 ( pO2 )
( pCl2 )3
When the system is at equilibrium, the partial pressure of Cl2(g) is 1.007 atm and the partial pressure
of UO2Cl2(g) is 9.734 u 104 atm.
(a) Calculate the partial pressure of O2(g) at equilibrium at 862qC.
(b) Calculate the value of the equilibrium constant, Kp , for the system at 862qC.
(c) Calculate the Gibbs free-energy change, 'Gq, for the reaction at 862qC.
(d) State whether the entropy change, 'Sq, for the reaction at 862qC is positive, negative, or zero. Justify
your answer.
(e) State whether the enthalpy change, 'Hq, for the reaction at 862qC is positive, negative, or zero. Justify
your answer.
(f) After a certain period of time, 1.000 mol of O2(g) is added to the mixture in the flask. Does the mass of
U3O8(s) in the flask increase, decrease, or remain the same? Justify your answer.
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2004 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
CHEMISTRY
Section II
(Total time—90 minutes)
Part A
Time—40 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A.
CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS.
It is to your advantage to do this, since you may obtain 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 the booklet with the
pink cover. Do NOT write your answers on the green insert.
Answer Question 1 below. The Section II score weighting for this question is 20 percent.
1. Answer the following questions relating to the solubilities of two silver compounds, Ag2CrO4 and Ag3PO4 .
Silver chromate dissociates in water according to the equation shown below.
→ 2 Ag+(aq) + CrO 2–(aq)
Ag2CrO4(s) ←
4
Ksp = 2.6 × 10–12 at 25°C
(a) Write the equilibrium-constant expression for the dissolving of Ag2CrO4(s).
(b) Calculate the concentration, in mol L−1, of Ag+(aq) in a saturated solution of Ag2CrO4 at 25°C.
(c) Calculate the maximum mass, in grams, of Ag2CrO4 that can dissolve in 100. mL of water at 25°C.
(d) A 0.100 mol sample of solid AgNO3 is added to a 1.00 L saturated solution of Ag2CrO4 . Assuming
no volume change, does [CrO42–] increase, decrease, or remain the same? Justify your answer.
In a saturated solution of Ag3PO4 at 25°C, the concentration of Ag+(aq) is 5.3 × 10– 5 M. The equilibriumconstant expression for the dissolving of Ag3PO4(s) in water is shown below.
Ksp = [Ag+] 3 [PO43−]
(e) Write the balanced equation for the dissolving of Ag3PO4 in water.
(f) Calculate the value of Ksp for Ag3PO4 at 25°C.
(g) A 1.00 L sample of saturated Ag3PO4 solution is allowed to evaporate at 25°C to a final volume
of 500. mL. What is [Ag+] in the solution? Justify your answer.
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2001 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
CHEMISTRY
Section II
(Total time—90 minutes)
Part A
Time—40 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A.
CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS.
It is to your advantage to do this, since you may obtain 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 the booklet with the
pink cover. Do NOT write your answers on the green insert.
Answer Question 1 below. The Section II score weighting for this question is 20 percent.
1. Answer the following questions relating to the solubility of the chlorides of silver and lead.
(a) At 10°C, 8.9 ! 10−5 g of AgCl(s) will dissolve in 100. mL of water.
(i) Write the equation for the dissociation of AgCl(s) in water.
(ii) Calculate the solubility, in mol L 1, of AgCl(s) in water at 10°C.
(iii) Calculate the value of the solubility-product constant, Ksp, for AgCl(s) at 10°C.
(b) At 25°C, the value of Ksp for PbCl2(s) is 1.6 ! 10−5 and the value of Ksp for AgCl(s) is 1.8 ! 10−10.
(i) If 60.0 mL of 0.0400 M NaCl(aq) is added to 60.0 mL of 0.0300 M Pb(NO3)2(aq), will a precipitate
form? Assume that volumes are additive. Show calculations to support your answer.
(ii) Calculate the equilibrium value of [Pb2+(aq)] in 1.00 L of saturated PbCl2 solution to which 0.250
mole of NaCl(s) has been added. Assume that no volume change occurs.
(iii) If 0.100 M NaCl(aq) is added slowly to a beaker containing both 0.120 M AgNO3(aq) and 0.150 M
Pb(NO3)2(aq) at 25°C, which will precipitate first, AgCl(s) or PbCl2(s)? Show calculations to
support your answer.
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2002 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)
CHEMISTRY
Section II
(Total time—90 minutes)
Part A
Time—40 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A.
CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS.
It is to your advantage to do this, since you may obtain 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 the booklet with the
goldenrod cover. Do NOT write your answers on the lavender insert.
Answer Question 1 below. The Section II score weighting for this question is 20 percent.
→ H+(aq) + C H O −(aq)
HC3H5O3(aq) ←
3 5 3
1. Lactic acid, HC3H5O3 , is a monoprotic acid that dissociates in aqueous solution, as represented by the equation
above. Lactic acid is 1.66 percent dissociated in 0.50 M HC3H5O3(aq) at 298 K. For parts (a) through (d)
below, assume the temperature remains at 298 K.
(a) Write the expression for the acid-dissociation constant, Ka , for lactic acid and calculate its value.
(b) Calculate the pH of 0.50 M HC3H5O3 .
(c) Calculate the pH of a solution formed by dissolving 0.045 mole of solid sodium lactate, NaC3H5O3 ,
in 250. mL of 0.50 M HC3H5O3 . Assume that volume change is negligible.
(d) A 100. mL sample of 0.10 M HCl is added to 100. mL of 0.50 M HC3H5O3 . Calculate the molar
concentration of lactate ion, C3H5O3−, in the resulting solution.
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2007 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
CHEMISTRY
Section II
(Total time—95 minutes)
Part A
Time—55 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A.
CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS.
It is to your advantage to do this, since you may obtain 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 the booklet with the
pink cover. Do NOT write your answers on the green insert.
Answer Questions 1, 2, and 3. The Section II score weighting for each question is 20 percent.
 H O+(aq) + F (aq)
HF(aq) + H2O(l ) Ž
3
Ka = 7.2 u 10 4
1. 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) o 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 that
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.
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ChemTeam: AP Free Response - 1987
http://www.gashalot.com/chem/dbhs.wvusd.k12.ca.us/AP-Chem/AP...
Advanced Placement Chemistry
1987 Free Response Questions
Go to the answers
Return to Additional Materials Menu
1)
NH3 + H2O <===> NH4+ + OH¯
Ammonia is a weak base that dissociates in water as shown above. At 25 °C, the base dissociation
constant, Kb, for NH3 is 1.8 x 10¯5.
(a) Determine the hydroxide ion concentration and the percentage dissociation of a 0.150-molar solution
of ammonia at 25 °C.
(b) Determine the pH of a solution prepared by adding 0.0500 mole of solid ammonium chloride to 100.
milliliters of a 0.150-molar solution of ammonia.
(c) If 0.0800 mole of solid magnesium chloride, MgCl2, is dissolved in the solution prepared in part (b)
and the resulting solution is well-stirred, will a precipitate of Mg(OH)2 form? Show calculation to
support your answer. (Assume the volume of the solution is unchanged. The solubility product constant
for Mg(OH)2 is 1.5 x 10¯11).
2)
2 HgCl2 (aq) + C2O42¯ --> 2 Cl¯ + 2 CO2 (g) + Hg2Cl2 (s)
The equation for the reaction between mercuric chloride and oxalate ion in hot aqueous solution is
shown above. The reaction rate may be determined by measuring the initial rate of formation of chloride
ion, at constant temperature, for various initial concentrations of mercuric chloride and oxalate as shown
in the following table.
Experiment
Initial
[HgCl2]
Initial
[C2O42¯]
Initial Rate
of formation
of Cl¯ (mole/liter-min)
(1)
0.0836 M
0.202 M
0.52 x 10¯4
(2)
0.0836 M
0.404 M
2.08 x 10¯4
(3)
0.0418 M
0.404 M
1.06 x 10¯4
(4)
0.0316 M
?
1.27 x 10¯4
(a) According to the data shown, what is the rate law for the reaction above?
(b) On the basis of the rate law determined in part (a), calculate the specific rate constant. Specify the
units.
1 of 4
9/6/06 6:23 PM
2009 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)
CHEMISTRY
Section II
(Total time—95 minutes)
Part A
Time— 55 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A.
CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS.
It is to your advantage to do this, since you may obtain 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. Do
NOT write your answers on the lavender insert.
Answer Questions 1, 2, and 3. The Section II score weighting for each question is 20 percent.
1. A pure 14.85 g sample of the weak base ethylamine, C2H5NH2 , is dissolved in enough distilled water to make
500. mL of solution.
(a) Calculate the molar concentration of the C2H5NH2 in the solution.
The aqueous ethylamine reacts with water according to the equation below.
! C H NH +(aq) + OH#(aq)
C2H5NH2(aq) + H2O(l) "
2 5
3
(b) Write the equilibrium-constant expression for the reaction between C2H5NH2(aq) and water.
(c) Of C2H5NH2(aq) and C2H5NH3+(aq), which is present in the solution at the higher concentration at
equilibrium? Justify your answer.
(d) A different solution is made by mixing 500. mL of 0.500 M C2H5NH2 with 500. mL of 0.200 M HCl.
Assume that volumes are additive. The pH of the resulting solution is found to be 10.93.
(i) Calculate the concentration of OH#(aq) in the solution.
(ii) Write the net-ionic equation that represents the reaction that occurs when the C2H5NH2 solution is
mixed with the HCl solution.
(iii) Calculate the molar concentration of the C2H5NH3+(aq) that is formed in the reaction.
(iv) Calculate the value of Kb for C2H5NH2 .
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!"®#$%&'()*+,#-...#!#Scoring Standards
Question 1
(10 points)
(a)
/0###=
[H 2S]2
1+
/0 =
(H 2 ) 2 (S2 )
1 pt.
(H 2S) 2
No point earned for a /2##expression or if an exponent is incorrectly used
•
(b)
[H 2 ]2 [S2 ]
(i)
[H2] = 2 [S2] =
(ii)
[H2S] =
(2.0)(3.72 " 10!2 mol)
1.25 L
0.10 mol ! (2.0)(3.72 " 10!2 mol)
1.25 L
= 5.95 " 10!2 3
= 2.05 " 10!2 3
1 pt.
2 pts.
or
[H2S] = 0.0800 3 – 0.0595 3 = 0.0205 3
Notes: One point is earned for getting the correct number of moles of H2S ; second
point is earned for dividing by 1.25 L and getting a consistent answer. Although not
correct, one point may be earned for calculating the initial [H2S] (see below) and using
that value as the equilibrium H2S concentration.
3440#g
[H2S] =
(c)
/0#
3441#g mol!1
= 040798 3
1425#L
& 3472 " 10!2 #
!
(5.95 " 10! 2 )2 $
$ 1425
!
!2 2
%
" = (5.95 " 10 ) (0.0298) = 0.250
=
( 2.05 " 10! 2 )2
(2.05 " 10! 2 )2
2 pts.
Notes: One point is earned for correctly using the molarity of the S2
(dividing the number of moles by the volume). The second point is earned
by correctly using the numbers generated in (b) in the expression shown in
(a) and getting the appropriate answer.
Copyright © 2000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP is a registered trademark of the College Entrance Examination Board.
!"®#$%&'()*+,#-...#!#Scoring Standards
Question 1
(015*(56&7)
"=
(d)
59:
(3.72 " 10!2 mol)(0.0821 L atm mol!1 K !1 )(483 K)
=
= 1.18 atm
1.25 L
8
or
2 pts.
" = [S2]9: = (0.02976 mol L!1)(0.0821 L atm mol!1 K!1 ) (483 K) = 1.18 atm
Note: The first point is earned for correctly setting up either of these expressions.
The second point is earned for the correct answer (or an answer consistent with the
numbers used). Also, combining "*1*;< 8 = 5*1*;< 9: with " S2 = = S 2 "*1*;< , where
= S 2 is the mole fraction of S2 , can earn the points.
(e)
/0# =
1
/0
=
1
0.250
= 2.00
2 pts.
Notes: One point is earned for recognizing that /0# must be an inverse related
to /0#. A second point is earned for recognizing that the square root of /0 is involved.
Only one point can be earned for the expression below, or for correctly calculating just
the inverse of /0 or just / 0 .
/0 =
[H 2S]
[H 2 ][S2 ]1 2
Copyright © 2000 College Entrance Examination Board and Educational Testing Service. All rights reserved.
AP is a registered trademark of the College Entrance Examination Board.
AP® CHEMISTRY
2007 SCORING GUIDELINES (Form B)
Question 1
A sample of solid U3O8 is placed in a rigid 1.500 L flask. Chlorine gas, Cl2(g), is added, and the flask is heated
to 862°C. The equation for the reaction that takes place and the equilibrium-constant expression for the reaction
are given below.
( pUO2 Cl2 )3 ( pO2 )
U3O8(s) + 3 Cl2(g) ! 3 UO2Cl2(g) + O2(g)
Kp !
( pCl2 )3
When the system is at equilibrium, the partial pressure of Cl2(g) is 1.007 atm and the partial pressure of
UO2Cl2(g) is 9.734 × 10− 4 atm.
(a) Calculate the partial pressure of O2(g) at equilibrium at 862°C.
U3O8(s) + 3 Cl2(g)
I
C
E
---
!
3 UO2Cl2(g) + O2(g)
?
1.007 atm
9.734 × 10− 4 atm UO2Cl2(g) ×
0
0
9.734 × 10− 4 atm
?
One point is earned for
the correct answer.
(1 mol O 2 )
= 3.245 × 10− 4 atm O2(g)
(3 mol UO 2 Cl2 )
(b) Calculate the value of the equilibrium constant, Kp , for the system at 862°C.
Kp =
(pUO 2Cl2 )3 ( pO 2 )
( pCl2 )3
=
(9.734 × 10−4 )3 (3.245 × 10−4 )
(1.007)3
= 2.931 × 10−13
One point is earned for the
correct substitution.
One point is earned for the
correct answer.
(c) Calculate the Gibbs free-energy change, ∆G°, for the reaction at 862°C.
∆G° = − RT ln Kp
= (−8.31 J mol−1 K−1)( (862+273) K)(ln (2.931 × 10−13))
= 272,000 J
mol−1
= 272 kJ
mol−1
One point is earned for
the correct setup.
One point is earned for the
correct answer with units.
© 2007 The College Board. All rights reserved.
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AP® CHEMISTRY
2007 SCORING GUIDELINES (Form B)
Question 1 (continued)
(d) State whether the entropy change, ∆S°, for the reaction at 862°C is positive, negative, or zero. Justify your
answer.
∆S° is positive because four moles of gaseous products are produced from
three moles of gaseous reactants.
One point is earned for
the correct explanation.
(e) State whether the enthalpy change, ∆H°, for the reaction at 862°C is positive, negative, or zero. Justify your
answer.
Both ∆G° and ∆S° are positive, as determined in parts (c) and (d).
Thus, ∆H° must be positive because ∆H° is the sum of two positive
terms in the equation ∆H° = ∆G° + T∆S°.
One point is earned
for the correct sign.
One point is earned for a
correct explanation.
(f) After a certain period of time, 1.000 mol of O2(g) is added to the mixture in the flask. Does the mass of
U3O8(s) in the flask increase, decrease, or remain the same? Justify your answer.
The mass of U3O8(s) will increase because the reaction is at equilibrium,
and the addition of a product creates a “stress” on the product (right) side of
the reaction. The reaction will then proceed from right to left to reestablish
equilibrium so that some O2(g) is consumed (tending to relieve the stress) as
more U3O8(s) is produced.
One point is earned for
a correct explanation.
© 2007 The College Board. All rights reserved.
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.//0$*%1,)23$34)5'6)2'*$
789:;<=>$0$
(10 points)
(a)
(i) AgCl(s) ! Ag+(aq) + Cl"(aq)
•
•
(ii)
1 point
Correct charges needed to earn credit.
Phases not necessary to earn credit.
8.9 # 10 "5 g
= 6.2 # 10"7 mol (in 100 mL)
143.32 g/mol
1 point
(6.2 # 10"7 mol/100 mL) (1,000 mL/1.000 L) = 6.2 ! 10"6 mol/L
1 point
Note: The first point is earned for the correct number of moles; the
second point is earned for the conversion from moles to molarity.
(iii) Ksp = [Ag+][Cl"] = (6.2 # 10"6)2 = 3.8 ! 10"11
1 point
Note: Students earn one point for squaring their result for molarity in (a) (ii).
(b)
(i) n Cl! = (0.060 L) (0.040 mol/L) = 0.0024 mol
1 point
[Cl"] = (0.0024 mol)/(0.120 L) = 0.020 mol/L = 0.020 M
n Pb2+ = (0.060 L) (0.030 mol/L) = 0.0018 mol
[Pb2+] = (0.0018 mol)/(0.120 L) = 0.015 mol/L = 0.015 M
Q = [Pb2+][Cl"]2 = (0.015)(0.020)2 = 6.0 ! 10"6
1 point
Q < Ksp , therefore no precipitate forms
1 point
Note: One point is earned for calculating the correct molarities; one point is
earned for calculating Q ; one point is earned for determining whether or not
a precipitate will form.
(ii) [Pb2+] =
K sp
[Cl " ] 2
=
1.6 # 10 "5
(0.25) 2
= 2.6 ! 10"4 M
Copyright © 2001 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.
2
1 point
!"#$%&'()*+,-$
.//0$*%1,)23$34)5'6)2'*$
789:;<=>$0$?@=>;AB$
"
(iii) for AgCl solution: [Cl ] =
"
for PbCl2 solution: [Cl ] =
AgCl
K sp
=
[Ag + ]
PbCl 2
K sp
[ Pb 2+ ]
=
1.8 # 10 "10
= 1.5 ! 10"9 M
0.120
1.6 # 10 "5
0.150
= 1.0 ! 10"2 M
The [Cl"] will reach a concentration of 1.5 # 10"9 M before it reaches
a concentration of 1.0 # 10"2 M, (or 1.5 # 10"9 << 1.0 # 10"2 ), therefore
AgCl(s) will precipitate first.
Note: One point is earned for calculating [Cl"] in saturated solutions with
the appropriate Ag+ and Pb2+ concentrations; one point is earned for
concluding which salt will precipitate first, based on the student’s calculations.
Copyright © 2001 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.
3
1 point
1 point
AP® CHEMISTRY
2002 SCORING GUIDELINES (Form B)
Question 1
10 points
HC3H5O3(aq) Æ̈ H+(aq) + C3H5O3–(aq)
1. Lactic acid, HC3H5O3 , is a monoprotic acid that dissociates in aqueous solution, as represented by the
equation above. Lactic acid is 1.66 percent dissociated in 0.50 M HC3H5O3(aq) at 298 K. For parts (a)
through (d) below, assume the temperature remains at 298 K.
(a) Write the expression for the acid-dissociation constant, Ka , for lactic acid and calculate its value.
_
Ka =
1 point earned for equilibrium
expression
[H + ][C 3 H 5 O 3 ]
[HC 3 H 5 O 3 ]
1 point earned for amount
of HC3H5O3 dissociating
0.50 M ! 0.0166 = 0.0083 M = x
HC3H5O3(aq) " H+(aq) + C3H5O3–(aq)
I
0.50
~0
0
C
–x
+x
+x
E
0.50 – x
+x
+x
[H + ][C3 H 5 O 3 ]
[0.0083][ 0.0083]
=
Ka =
[0.50 - 0.0083]
[HC 3 H 5 O 3 ]
__
1 point earned for
[H+] = [C3H5O3–] set up
and solution
Ka = 1.4 ! 10–4
(b) Calculate the pH of 0.50 M HC3H5O3 .
From part (a):
1 point earned for correctly
calculating pH
[H+] = 0.0083 M
pH = –log [H+] = –log (0.0083) = 2.08
Copyright © 2002 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.
2
AP® CHEMISTRY
2002 SCORING GUIDELINES (Form B)
Question 1 (cont’d.)
(c) Calculate the pH of a solution formed by dissolving 0.045 mole of solid sodium lactate, NaC3H5O3 , in
250. mL of 0.50 M HC3H5O3 . Assume that volume change is negligible.
0.045 mol NaC 3 H 5 O 3
= 0.18 M C3H5O3–
0.250 L
1 point earned for [C3H5O3–]
(or 0.250 L ! 0.50 mol/L = 0.125
mol HC3H5O3 and 0.045 mol C3H5O3–)
HC3H5O3(aq) " H+(aq) + C3H5O3–(aq)
I
0.50
~0
0.18
C
–x
+x
+x
E
0.50 – x
+x
0.18 + x
[H + ][C3 H 5 O 3 ] [ x ][0.18 + x ]
Ka =
=
[0.50 # x ]
[HC 3 H 5 O 3 ]
__
1 point earned for [H+]
(set up and calculation)
Assume that x << 0.18 M
Ka = 1.4 ! 10–4 =
[ x ][0.18]
[0.50]
1 point earned for calculating the value
of pH
x = 3.9 ! 10–4 M = [H+]
pH = – log [H+] = – log (3.9 ! 10#4) = 3.41
OR
pH = pKa + log 0.18 or 0.045 = 3.41
0.50 0.125
Copyright © 2002 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.
3
AP® CHEMISTRY
2002 SCORING GUIDELINES (Form B)
Question 1 (cont’d.)
(d) A 100. mL sample of 0.10 M HCl is added to 100. mL of 0.50 M HC3H5O3 . Calculate the molar
concentration of lactate ion, C3H5O3–, in the resulting solution.
1 point earned for initial [H+] and
[HC3H5O3]
100 mL)
0.50 M HC3H5O3 &%
=0.25 M HC3H5O3
$200 mL('
100 mL)
0.10 M HCl &%
= 0.050 M H+
$200 mL('
OR
(10 mmol H+; 50 mmol HC3H5O3)
HC3H5O3(aq) " H+(aq) + C3H5O3– (aq)
I
0.25
0.050
0
C
–x
+x
+x
E
0.25 – x
0.050 + x
+x
1 point earned for showing dilution or
moles of each
[H + ][C3 H 5 O 3 ]
[0.050 + x ][x ]
Ka =
=
[0.25 # x ]
[HC 3 H 5 O 3 ]
__
1 point earned for [C3H5O3–] setup
and calculation
Assume x << 0.050 M
[0.050][x]
Ka = 1.4 ! 10–4 =
[0.25]
–4
x = 7.0 ! 10 M = [C3H5O3–]
Copyright © 2002 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.
4
AP® CHEMISTRY
2007 SCORING GUIDELINES
Question 1
! H O+(aq) + F −(aq)
HF(aq) + H2O(l ) "
3
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.
Ka =
[H3O+ ][F − ]
[HF]
One point is earned for the correct expression.
(b) Calculate the molar concentration of H3O+ in a 0.40 M HF(aq) solution.
Ka =
[H3O+ ][F − ]
( x)( x)
=
= 7.2 # 10− 4
[HF]
0.40 − x
One point is earned for the correct setup
(or the setup consistent with part (a)).
Assume x << 0.40, then x2 = (0.40)(7.2 # 10− 4)
One point is earned for the correct concentration.
x = [ H3O+ ] = 0.017 M
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 that
volumes are additive.
(c) Calculate the number of moles of HF(aq) remaining in the solution.
mol HF(aq) = initial mol HF(aq) − mol NaOH(aq) added
= (0.025 L)(0.40 mol L−1) − (0.015 L)(0.40 mol L−1)
= 0.010 mol − 0.0060 mol = 0.004 mol
One point is earned for
determining the initial number
of moles of HF and OH− .
One point is earned for setting up
and doing correct subtraction.
(d) Calculate the molar concentration of F −(aq) in the solution.
mol
F −(aq)
formed = mol NaOH(aq) added = 0.0060 mol
0.0060 mol F − (aq)
= 0.15 M F −(aq)
(0.015 + 0.025) L of solution
F −(aq)
One point is earned for
determining the number
of moles of F −(aq).
One point is earned for dividing
the number of moles of F −(aq)
by the correct total volume.
© 2007 The College Board. All rights reserved.
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AP® CHEMISTRY
2007 SCORING GUIDELINES
Question 1 (continued)
(e) Calculate the pH of the solution.
[HF] =
Ka =
!
0.004 mol HF
= 0.10 M HF
0.040 L
[H3O+ ][F − ]
[HF] × K a
!
= [H3O+ ]
−
[HF]
[F ]
0.10 M (7.2 × 10
0.15 M
−4
)
= 4.8 × 10− 4
! pH = − log (4.8 × 10− 4) = 3.32
One point is earned for indicating
that the resulting solution is a buffer
(e.g., by showing a ratio of [F −] to [HF]
or moles of F − to HF ).
OR
pH = pKa + log
[F − ]
[HF]
= − log (7.2 × 10− 4) + log
One point is earned for the correct calculation
of pH.
0.15 M
0.10 M
= 3.14 + 0.18
= 3.32
© 2007 The College Board. All rights reserved.
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ChemTeam: AP Free Response Answers - 1987
http://www.gashalot.com/chem/dbhs.wvusd.k12.ca.us/AP-Chem/AP...
Advanced Placement Chemistry
1987 Free Response Answers
Notes
[delta], [sigma] and [gamma] are used to indicate the appropriate Greek letters.
[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)
a) three points
[NH4+ ] = [OH¯] = x
[NH3] = 0.150 mol/L - x
Kb = ([NH4+][OH¯]) ÷ [NH3]
1.8 x 10¯5 = [(x) (x)] ÷ (0.150 - x)
approximately equals x2 ÷ 0.150
x = [OH¯] = 1.6 x 10¯3 mol/L
% diss = [(1.6 x 10¯3) / (0.150)] x 100% = 1.1%
b) three points
[NH4+ ] = 0.0500 mol / 0.100 L = 0.500 mol/L NH4+
[NH3] = 0.150 mol/L
OR
mol NH4+ = 0.0500 mol NH4+
mol NH3 = 0.150 mol/L x 0.100 L = 0.0150 mol
THEN
1 of 9
9/6/06 6:23 PM
ChemTeam: AP Free Response Answers - 1987
http://www.gashalot.com/chem/dbhs.wvusd.k12.ca.us/AP-Chem/AP...
1.8 x 10¯5 = [(0.500) (x)] ÷ (0.150)
OR
pOH = 4.74 + log (0.500 / 0.150)
THEN
x = [OH¯] = 5.4 x 10¯6 mol/L
pOH = 5.27
pH = 14.00 - 5.27 = 8.73
c) three points
Mg(OH)2(s) <===> Mg2+ + OH¯
[Mg2+] = 0.0800 mol / 0.100 L = 0.800 mol/L (0.800 ± x = no credit)
([OH¯] = 5.4 x 10¯6 mol/L from b.)
Q= [Mg2+][OH¯]2 = (0.800) (5.4 x 10¯6)2
Q = 2.3 x 10¯11
Ksp = 1.5 x 10¯11
since Q > Ksp, Mg(OH)2 precipitates
(Q must be defined in the same way as Ksp)
OR
Ksp = 1.5 x 10¯11 = (0.800)[OH¯]2
[OH¯] = 4.3 x 10¯6 mol/L
since 5.4 x 10¯6 > 4.3 x 10¯6 mol/L, then Mg(OH)2 precipitates
2)
a) three points; one each for form of rate law, HgCl2 exponent, C2O42¯ exponent
Rate = k [HgCl2][C2O42¯]2
2 of 9
9/6/06 6:23 PM
AP® CHEMISTRY
2009 SCORING GUIDELINES (Form B)
Question 1 (10 points)
A pure 14.85 g sample of the weak base ethylamine, C2H5NH2 , is dissolved in enough distilled water to make
500. mL of solution.
(a) Calculate the molar concentration of the C2H5NH2 in the solution.
nC2H5NH2 = 14.85 g C2H5NH2 !
1 mol C2 H 5 NH 2
45.09 g C2 H 5 NH 2
= 0.3293 mol C2H5NH2
MC2H5NH2 =
One point is earned for the
correct number of moles.
One point is earned for the
correct concentration.
0.3293 mol C2 H 5 NH 2
= 0.659 M
0.500 L
The aqueous ethylamine reacts with water according to the equation below.
! C H NH +(aq) + OH#(aq)
C2H5NH2(aq) + H2O(l) "
2 5
3
(b) Write the equilibrium-constant expression for the reaction between C2H5NH2(aq) and water.
Kb =
[C2 H 5 NH3+ ][OH " ]
[C2 H 5 NH 2 ]
One point is earned for the correct expression.
(c) Of C2H5NH2(aq) and C2H5NH3+(aq), which is present in the solution at the higher concentration at
equilibrium? Justify your answer.
C2H5NH2 is present in the solution at the higher concentration at
equilibrium. Ethylamine is a weak base, and thus it has a small
Kb value. Therefore only partial dissociation of C2H5NH2 occurs in
water, and [C2H5NH3+] is thus less than [C2H5NH2].
One point is earned for the
correct answer with justification.
© 2009 The College Board. All rights reserved.
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AP® CHEMISTRY
2009 SCORING GUIDELINES (Form B)
Question 1 (continued)
(d) A different solution is made by mixing 500. mL of 0.500 M C2H5NH2 with 500. mL of 0.200 M HCl.
Assume that volumes are additive. The pH of the resulting solution is found to be 10.93.
(i)
Calculate the concentration of OH#(aq) in the solution.
pH = #log[H+]
[H+] = 10#10.93 = 1.17 $ 10#11
[OH#] =
Kw
#
[H ]
=
1.00 ! 10 "14
= 8.5 $ 10#4 M
"11
1.17 ! 10
One point is earned for
the correct concentration.
OR
pOH = 14 # pH = 14 # 10.93 = 3.07
pOH = #log[OH #]
[OH#] = 10#3.07 = 8.5 $ 10#4 M
(ii) Write the net-ionic equation that represents the reaction that occurs when the C2H5NH2 solution is
mixed with the HCl solution.
One point is earned for the correct equation.
C2H5NH2 + H3O+ ! C2H5NH3+ + H2O
(iii) Calculate the molar concentration of the C2H5NH3+(aq) that is formed in the reaction.
moles of C2H5NH2 = 0.500 L !
moles of H3O+ = 0.500 L !
initial value
change
final value
0.500 mol
= 0.250 mol
1.00 L
0.200 mol
= 0.100 mol
1.00 L
[C2H5NH2]
0.250
#0.100
0.150
[H3O+]
0.100
#0.100
~0
One point is earned for the
correct number of moles of
C2H5NH2 and H3O+.
[C2H5NH3+]
~0
+0.100
0.100
0.100 mol C2 H 5 NH3 #
= 0.100 M
[C2H5NH3+] =
1.00 L
One point is earned
for the correct concentration.
© 2009 The College Board. All rights reserved.
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AP® CHEMISTRY
2009 SCORING GUIDELINES (Form B)
Question 1 (continued)
(iv) Calculate the value of Kb for C2H5NH2 .
0.150 mol C2 H 5 NH 2
[C2H5NH2] =
= 0.150 M
1.00 L
Kb =
One point is earned for the
correct calculation of the
molarity of C2H5NH2 after
neutralization.
[C2 H 5 NH3+ ][OH " ]
(0.100)(8.5 ! 10 "4 )
=
= 5.67 $ 10#4
0.150
[C2 H 5 NH 2 ]
© 2009 The College Board. All rights reserved.
Visit the College Board on the Web: www.collegeboard.com.
One point is earned
for the correct value.