2000 AP CHEMISTRY FREE-RESPONSE QUESTIONScf.edliostatic.com/54nVRmb1OB7JmX62MNqn7rV1WbD4aTgp.pdf2007 AP® CHEMISTRY FREE-RESPONSE QUESTIONS ... Be sure to write all your answers to

2000 AP® CHEMISTRY FREE-RESPONSE QUESTIONS

Copyright © 2000 College Entrance Examination Board and Educational Testing Service. All rights reserved.AP is a registered trademark of the College Entrance Examination Board.

GO ON TO THE NEXT PAGE.-6-

CHEMISTRY—SECTION II(Total time—90 minutes)

Part ATime—40 minutes

YOU MAY USE YOUR CALCULATOR FOR PART A.

CLEARLY SHOW THE METHOD USED AND STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS. It isto your advantage to do this, because you may earn partial credit if you do and you will receive little or no credit ifyou do not. Attention should be paid to significant figures. Be sure to write all your answers to the questions on thelined 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.

(e) For the reaction H2(g) + 12

S2(g) !" H2S(g) at 483 K, calculate the value of the equilibrium constant, Kc .

2007 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)

© 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).

GO ON TO THE NEXT PAGE. -6-

CHEMISTRY Section II

(Total time—95 minutes)

Part A Time—55 minutes


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 862 C. The equation for the reaction that takes place and the equilibrium-constant expression for the reaction are given below.

U3O8(s) + 3 Cl2(g) 3 UO2Cl2(g) + O2(g) 2 2 2

2

3UO Cl O

3Cl

( ) ( )

( )p

p pK

p

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.

(b) Calculate the value of the equilibrium constant, Kp , for the system at 862 C.

(c) Calculate the Gibbs free-energy change, G , for the reaction at 862 C.

(d) State whether the entropy change, S , for the reaction at 862 C is positive, negative, or zero. Justify your answer.

(e) State whether the enthalpy change, H , for the reaction at 862 C 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.


Copyright © 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).

GO ON TO THE NEXT PAGE.

6





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.

Ag2CrO4(s) !" 2 Ag+(aq) + CrO42–(aq) 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 [CrO4

2–] 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 equilibrium-constant 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.


Copyright © 2001 by College Entrance Examination Board. All rights reserved.Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

GO ON TO THE NEXT PAGE.6

CHEMISTRYSection II


Part ATime—40 minutes


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 ifyou 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 thepink 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 precipitateform? 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 0.250mole 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.


Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

GO ON TO THE NEXT PAGE. 6





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.

HC3H5O3(aq) !" 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.

(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.



GO ON TO THE NEXT PAGE. -6-





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.

HF(aq) + H2O(l ) H3O+(aq) + F (aq) Ka = 7.2 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) 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.

ChemTeam: AP Free Response - 1987 http://www.gashalot.com/chem/dbhs.wvusd.k12.ca.us/AP-Chem/AP...

1 of 4 9/6/06 6:23 PM

Advanced Placement Chemistry

1987 Free Response Questions

Go to the answersReturn 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.

ExperimentInitial

[HgCl2]

Initial

[C2O42¯]

Initial Rateof 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.


© 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.

-6- GO ON TO THE NEXT PAGE.





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.

C2H5NH2(aq) + H2O(l) !" C2H5NH3+(aq) + OH#(aq)

(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.

!"®#$%&'()*+,#-...#!#Scoring Standards

Copyright © 2000 College Entrance Examination Board and Educational Testing Service. All rights reserved.

AP is a registered trademark of the College Entrance Examination Board.

Question 1

(10 points)

(a) /0###= 2

2

22

2

]SH[

]S[]H[1+ /0 =

22

22

2

S)(H

)(S)(H1 pt.

• No point earned for a /2##expression or if an exponent is incorrectly used

(b) (i) [H2] = 2 [S2] = L25.1

)mol 10 72.3)(0.2( 2!"

= 5.95 " 10!2 3 1 pt.

(ii) [H2S] = L25.1

)mol 1072.3)(0.2(mol 10.0 2!"!

= 2.05 " 10!2 3 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.

[H2S] = 34#4

#4

#4

07980L251

mol g134

g4031

=!

(c) /0# = 22

222

)1005.2(

251

10723)1055.9(

!

!!

"

!!

"

#

$$

%

& ""

4

4

= 22

22

)1005.2(

)0298.0()10 5.95(!

!

"

" = 0.250 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.

!"®#$%&'()*+,#-...#!#Scoring Standards

Copyright © 2000 College Entrance Examination Board and Educational Testing Service. All rights reserved.

AP is a registered trademark of the College Entrance Examination Board.

Question 1

(015*(56&7)

(d) " = 8

59: =

L25.1

K)483)(KmolatmL0821.0mol)(1072.3 112( !!!" = 1.18 atm

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 9:58" *1*;<*1*;< = with *1*;<"="22 SS = , where

2S= is the mole fraction of S2 , can earn the points.

(e) /0# = 0/

1 =

250.0

1 = 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/ .

21]S][H[

]SH[

22

2=0/

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.

U3O8(s) + 3 Cl2(g) ! 3 UO2Cl2(g) + O2(g) 2 2 2

2

3UO Cl O

3Cl

( ) ( )

( )p

p pK

p!

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) ! 3 UO2Cl2(g) + O2(g)

I --- ? 0 0 C E 1.007 atm 9.734 ! 10! 4 atm ?

9.734 ! 10! 4 atm UO2Cl2(g) !

2

2 2

(1 mol O )(3 mol UO Cl )

= 3.245 ! 10! 4 atm O2(g)

One point is earned for the correct answer.

(b) Calculate the value of the equilibrium constant, Kp , for the system at 862°C.

Kp = 2 2 2

2

3UO Cl O

3Cl

( ) ( )

( )

p p

p =

4 3 4

3(9.734 10 ) (3.245 10 )

(1.007)

! !" " = 2.931 ! 10!13

One point is earned for thecorrect 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.



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.

!"!#$%&'()*+,#

-../#)$0+(12#23(4&5(1&)#

!"#$%&'()*+*,--.*/$*!"001'1*23)%4351*2647&34)&"3*8"4%9:*;00*%&'()<*%1<1%=19:*

>&<&)*4#513)%40:5"001'1/"4%9:5"7*?@"%*;A*#%"@1<<&"340<B*439*CCC:5"001'1/"4%9:5"7D4#<)E913)<*?@"%*;A*<)E913)<*439*#4%13)<B:*

*

-#

*

6789:;<=#>#

** * ;3<C1%*)(1*@"00"C&3'*FE1<)&"3<*%104)&3'*)"*)(1*<"0E/&0&)&1<*"@*)C"*<&0=1%*5"7#"E39<G* ;',!%H.* 439* ;'IAH.:*

* *J&0=1%*5(%"74)1*9&<<"5&4)1<*&3*C4)1%*455"%9&3'*)"*)(1*1FE4)&"3*<("C3*/10"C:*

;',!%H.?!B**!" **,*;'K?"#B**K**!%H.

,L?"#B*******$!%**M**,:N*#*O-LO,*4)*,P°!*

?4B* Q%&)1*)(1*1FE&0&/%&E7R5"3<)43)*16#%1<<&"3*@"%*)(1*9&<<"0=&3'*"@* ;',!%H.?!B:*

$!%*M**S;'KT,S!%H.

,L*T*

O*#"&3)*@"%*5"%%15)*16#%1<<&"3*

*

?/B* !405E04)1*)(1*5"3513)%4)&"3G*&3*7"0*U$OG*"@* ;'K?"#B* &3*4*<4)E%4)19*<"0E)&"3*"@* ;',!%H.* 4)*,P°!:*

;',!%H.?!B**!" **,*;'K?"#B*K*!%H.

,L*?"#B*

V** L** -* -*

!** L** K,&** K&*

W** L** -*K*,&** -*K*&*

*

,:N*#*O-LO,**M**S;'KT,S!%H.,LT*

,:N*#*O-LO,**M**S,&T,S&T**M**.& I*X:Y*#*O-LP***M***6***M***S!%H.,LT*

S;'KT**M**,&**M**,*#*?X:Y*#*O-LP*' B**M**O:Y*#*O-L .*'**

*

*

*

O*#"&3)*@"%*5"%%15)*<)"&5(&"71)%$*"@*;'K?"#B*

439*!%H.,L?"#B*

*

*

*

*

O*#"&3)*@"%*<E/<)&)E)&3'*439*5405E04)&3'*S;'KT*

*

*

*

*

?5B* !405E04)1*)(1*746&7E7*74<<G*&3*'%47<G*"@* ;',!%H.* )(4)*543*9&<<"0=1*&3*O--:*7U*"@*C4)1%*4)*,P°!:*

*P

, .X:Y O- 7"01 ;' !%H

OU

$#*#*

, .

IIO:Y '

O7"01 ;' !%H*M* , .-:-,Z ' ;' !%H

OU*

*

, .-:-,Z ' ;' !%H

OU*#**-:O--*U**M**-:--,Z*'**;',!%H.

*

*

O*#"&3)*@"%*7"04%*74<<*"@*;',!%H.*

*

O*#"&3)*@"%*74<<*"@*;',!%H.*&3*O--*7U*

!"!#$%&'()*+,#

-../#)$0+(12#23(4&5(1&)#

!"#$%&'()*+*,--.*/$*!"001'1*23)%4351*2647&34)&"3*8"4%9:*;00*%&'()<*%1<1%=19:*

>&<&)*4#513)%40:5"001'1/"4%9:5"7*?@"%*;A*#%"@1<<&"340<B*439*CCC:5"001'1/"4%9:5"7D4#<)E913)<*?@"%*;A*<)E913)<*439*#4%13)<B:*

*

?#

*

!"#$%&'()*)+,'(%-./0)

*

?9B* ;*-:O--*7"0*<47#01*"@*<"0&9* ;'[HI* &<*49919*)"*4*O:--*U*<4)E%4)19*<"0E)&"3*"@* ;',!%H.*:*;<<E7&3'**

3"*="0E71*5(43'1G*9"1<*S!%H.,LT*&35%14<1G*915%14<1G*"%*%174&3*)(1*<471\*]E<)&@$*$"E%*43<C1%:

*

^(1*S!%H.,LT*C&00*915%14<1:*;99&3'*S;'KT*C&00*74_1*(*

?3"31FE&0&/%&E7*%145)&"3*FE")&13)B*'%14)1%*)(43*$:*^"*

%1R1<)4/0&<(*1FE&0&/%&E7G*)(1*%145)&"3*'"1<*@%"7*%&'()*)"*

01@)G*915%14<&3'*)(1*FE")&13)*)"*%1)E%3*)"*1FE&0&/%&E7:*

*

O*#"&3)*@"%*5"%%15)*#%19&5)&"3*439*16#0434)&"3*&3*

)1%7<*"@*()"%*U1!(`)10&1%a<*#%&35&#01*

V3*4*<4)E%4)19*<"0E)&"3*"@* ;'IAH.* 4)*,P°!G*)(1*5"3513)%4)&"3*"@* ;'K?"#B* &<*P:I*#*O-LP*':*^(1*1FE&0&/%&E7R5"3<)43)*16#%1<<&"3*@"%*)(1*9&<<"0=&3'*"@* ;'IAH.?!B* &3*C4)1%*&<*<("C3*/10"C:*

$!%**M**S;'KTISAH.

I$T*

?1B* Q%&)1*)(1*/4043519*1FE4)&"3*@"%*)(1*9&<<"0=&3'*"@* ;'IAH.* &3*C4)1%:*

;'IAH.?!B**!**I*;'K?"#B*K*AH.IL?"#B* O*#"&3)*@"%*5"%%15)G*/4043519*5(17&540*1FE4)&"3*

*

*

?@B* !405E04)1*)(1*=40E1*"@* $!%* @"%* ;'IAH.* 4)*,P°!:*

S;'KT**M**P:I*#*O-LP*'*

SAH.ILT**M**P:I*#*O-LP*');'K**#*

I.

K

O 7"0 AH

I 7"0 ;'

$**M**O:X*#*O-L P*'*

*

$!%**M**S;'KTISAH.

ILT*

* * M**?P:I*#*O-L PBI?O:X*#O-L PB**M**,:N*#*O-LOX**

*

O*#"&3)*@"%*5"%%15)* SAH.ILT*

*

*

O*#"&3)*@"%*$!%*

*

?'B* ;*O:--*U*<47#01*"@*<4)E%4)19* ;'IAH.* <"0E)&"3*&<*400"C19*)"*1=4#"%4)1*4)*,P°!*)"*4*@&340*="0E71**"@*P--:*7U:*Q(4)*&<* S;'KT* &3*)(1*<"0E)&"3\*]E<)&@$*$"E%*43<C1%:

*

S;'KT**M**P:I*#*O-L P)'*^(1* S;'KT* &3*4*<4)E%4)19*<"0E)&"3*"@* ;'IAH.* &<*&391#13913)*"@*

)(1*="0E71*"@*)(1*<"0E)&"3:*

O*#"&3)*@"%*5"%%15)*S;'KT*439*16#0434)&"3*

*

!"#$%&'()*+,-$

.//0$*%1,)23$34)5'6)2'*$

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

789:;<=>$0$

(10 points)

(a) (i) AgCl(s) ! Ag+(aq) + Cl"(aq) 1 point

• Correct charges needed to earn credit.

• Phases not necessary to earn credit.

(ii) g/mol 32.143

109.8 5 g"#

= 6.2 # 10"7 mol (in 100 mL) 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+] = 2

]Cl["

spK =

2

5

)25.0(

10 6.1 "#

= 2.6 ! 10"4 M 1 point

!"#$%&'()*+,-$

.//0$*%1,)23$34)5'6)2'*$



3

789:;<=>$0$?@=>;AB$

(iii) for AgCl solution: [Cl"] =

]Ag[

AgCl

+

spK =

120.0

10 8.1 10"#

= 1.5 ! 10"9

M 1 point

for PbCl2 solution: [Cl"] =

]Pb[ 2

PbCl2

+

spK =

150.0

10 6.1 5"#

= 1.0 ! 10"2

M

The [Cl"] will reach a concentration of 1.5 # 10"9 M before it reaches 1 point

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.

AP® CHEMISTRY

2002 SCORING GUIDELINES (Form B)



2

Question 1

10 points

HC3H5O3(aq) Æ¨ 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 =]OH[HC

]OH][C[H

353

_353

+

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

Ka =]OH[HC

]OH][C[H

353

353__+

= [0.0083][ 0.0083]

[0.50 - 0.0083]

Ka = 1.4 ! 10–4

1 point earned for equilibrium

expression

1 point earned for amount

of HC3H5O3 dissociating

1 point earned for

[H+] = [C3H5O3–] set up

and solution

(b) Calculate the pH of 0.50 M HC3H5O3.

From part (a):

[H+] = 0.0083 M

pH = –log [H+] = –log (0.0083) = 2.08

1 point earned for correctly

calculating pH

AP® CHEMISTRY




3

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.

L0.250

OHNaCmol0.045 353 = 0.18 M C3H5O3

–


–(aq)

I 0.50 ~0 0.18

C –x +x +x

E 0.50 – x +x 0.18 + x

Ka = ]OH[HC

]OH][C[H

353

353__+

= ]x50.0[

]x18.0][x[

#

+

Assume that x << 0.18 M

Ka = 1.4 ! 10–4 = ]50.0[

]18.0][x[

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

1 point earned for [C3H5O3–]

(or 0.250 L ! 0.50 mol/L = 0.125

mol HC3H5O3 and 0.045 mol C3H5O3–)

1 point earned for [H+]

(set up and calculation)

1 point earned for calculating the value

of pH

AP® CHEMISTRY




4

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.

0.50 M HC3H5O3 $%&

'()100 mL

200 mL =0.25 M HC3H5O3

0.10 M HCl $%&

'()100 mL

200 mL = 0.050 M H+


–(aq)

I 0.25 0.050 0

C –x +x + x

E 0.25 – x 0.050 + x + x

Ka = ]OH[HC

]OH][C[H

353

353__+

= ]x25.0[

]x][x050.0[

#

+

Assume x << 0.050 M

Ka = 1.4 ! 10–4 = [0.050][x]

[0.25]

x = 7.0 ! 10–4 M = [C3H5O3–]

1 point earned for initial [H+] and

[HC3H5O3]

OR

(10 mmol H+; 50 mmol HC3H5O3)

1 point earned for showing dilution or

moles of each

1 point earned for [C3H5O3–] setup

and calculation

AP® CHEMISTRY

2007 SCORING GUIDELINES


Question 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.

!=

+3[H O ][F ][HF]aK One point is earned for the correct expression.

(b) Calculate the molar concentration of H3O+ in a 0.40 M HF(aq) solution.

!

=+

3[H O ][F ][HF]aK =

( )( )0.40

x xx!

= 7.2 # 10! 4

Assume x << 0.40, then x2 = (0.40)(7.2 # 10! 4)

x = [ H3O+ ] = 0.017 M

One point is earned for the correct setup (or the setup consistent with part (a)).

One point is earned for the correct concentration.

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 F !(aq)

!

+0.0060 mol F ( )

(0.015 0.025) L of solutionaq

= 0.15 M 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.

AP® CHEMISTRY

2007 SCORING GUIDELINES



(e) Calculate the pH of the solution.

[HF] = 0.004 mol HF

0.040 L = 0.10 M HF

!

=+

3[H O ][F ][HF]aK !

!# = +

3[HF]

[H O ][F ]

aK

! !# 40.10 (7.2 10 )

0.15 M

M = 4.8 # 10!4

! pH = ! log (4.8 # 10!4) = 3.32

OR

pH = pKa + log ![F ]

[HF]

= ! log (7.2 # 10!4) + log0.15 0.10

MM

= 3.14 + 0.18

= 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 ).

One point is earned for the correct calculation of pH.

ChemTeam: AP Free Response Answers - 1987 http://www.gashalot.com/chem/dbhs.wvusd.k12.ca.us/AP-Chem/AP...

1 of 9 9/6/06 6:23 PM

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 followingAll 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 QuestionsReturn 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

ChemTeam: AP Free Response Answers - 1987 http://www.gashalot.com/chem/dbhs.wvusd.k12.ca.us/AP-Chem/AP...

2 of 9 9/6/06 6:23 PM

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



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.

2 5 2C H NHn = 14.85 g C2H5NH2 2 5 2

2 5 2

1 mol C H NH

45.09 g C H NH!

= 0.3293 mol C2H5NH2

2 5 2C H NHM = 2 5 20.3293 mol C H NH

0.500 L = 0.659 M

One point is earned for the

correct number of moles.


correct concentration.

The aqueous ethylamine reacts with water according to the equation below.

C2H5NH2(aq) + H2O(l) !" C2H5NH3+(aq) + OH#(aq)

(b) Write the equilibrium-constant expression for the reaction between C2H5NH2(aq) and water.

Kb =

+2 5 3

2 5 2

[C H NH ][OH ]

[C H NH ]

"

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].


correct answer with justification.




(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#] = [H ]

wK

# =

14

11

1.00 10

1.17 10

"

"

!

! = 8.5 $ 10

#4 M

OR

pOH = 14 # pH = 14 # 10.93 = 3.07

pOH = # log[OH#]

[OH#] = 10#3.07 = 8.5 $ 10#4

M

One point is earned for

the correct concentration.

(ii) Write the net-ionic equation that represents the reaction that occurs when the C2H5NH2 solution is

mixed with the HCl solution.

C2H5NH2 + H3O+ ! C2H5NH3

+ + H2O

One point is earned for the correct equation.

(iii) Calculate the molar concentration of the C2H5NH3+(aq) that is formed in the reaction.

moles of C2H5NH2 = 0.500 L !

0.500 mol

1.00 L = 0.250 mol

moles of H3O+ = 0.500 L ! 0.200 mol

1.00 L = 0.100 mol

[C2H5NH2] [H3O+] [C2H5NH3+]

initial value 0.250 0.100 ~ 0

change #0.100 #0.100 +0.100

final value 0.150 ~ 0 0.100

[C2H5NH3+] = 2 5 30.100 mol C H NH

1.00 L

#

= 0.100 M


correct number of moles of

C2H5NH2 and H3O+.

One point is earned

for the correct concentration.




(iv) Calculate the value of Kb for C2H5NH2.

[C2H5NH2] = 2 5 20.150 mol C H NH

1.00 L = 0.150 M

Kb =

+2 5 3

2 5 2

[C H NH ][OH ]

[C H NH ]

"

= 4(0.100)(8.5 10 )

0.150

"!

= 5.67 $ 10#4


correct calculation of the

molarity of C2H5NH2 after

neutralization.

One point is earned

for the correct value.

Documents

2000 AP CHEMISTRY FREE-RESPONSE QUESTIONScf.edliostatic.com/54nVRmb1OB7JmX62MNqn7rV1WbD4aTgp.pdf2007 AP® CHEMISTRY FREE-RESPONSE QUESTIONS ... Be sure to write all your answers to