2000 AP CHEMISTRY FREE-RESPONSE QUESTIONS

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1 000 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 0 percent. H S(g) H (g) + S (g) 1. When heated, hydrogen sulfide gas decomposes according to the equation above. A.40 g sample of H S(g) is introduced into an evacuated rigid 1.5 L container. The sealed container is heated to 48 K, and.7 10 mol of S (g) is present at equilibrium. (a) Write the expression for the equilibrium constant, K c, for the decomposition reaction represented above. (b) Calculate the equilibrium concentration, in mol L 1, of the following gases in the container at 48 K. (i) H (g) (ii) H S(g) (c) Calculate the value of the equilibrium constant, K c, for the decomposition reaction at 48 K. (d) Calculate the partial pressure of S (g) in the container at equilibrium at 48 K. (e) For the reaction H (g) + 1 S (g) H S(g) at 48 K, calculate the value of the equilibrium constant, K c. Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Examination Board. -6- GO ON TO THE NEXT PAGE.

2 007 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,, and. The Section II score weighting for each question is 0 percent. 1. A sample of solid U O 8 is placed in a rigid L flask. Chlorine gas, Cl (g), is added, and the flask is heated to 86 C. The equation for the reaction that takes place and the equilibrium-constant expression for the reaction are given below. ( p U O 8 (s) + Cl (g) UOCl ) ( p O ) UO Cl (g) + O (g) K p ( p ) When the system is at equilibrium, the partial pressure of Cl (g) is atm and the partial pressure of UO Cl (g) is atm. Cl (a) Calculate the partial pressure of O (g) at equilibrium at 86 C. (b) Calculate the value of the equilibrium constant, K p, for the system at 86 C. (c) Calculate the Gibbs free-energy change, G, for the reaction at 86 C. (d) State whether the entropy change, S, for the reaction at 86 C is positive, negative, or zero. Justify your answer. (e) State whether the enthalpy change, H, for the reaction at 86 C is positive, negative, or zero. Justify your answer. (f) After a certain period of time, mol of O (g) is added to the mixture in the flask. Does the mass of U O 8 (s) in the flask increase, decrease, or remain the same? Justify your answer. 007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents). -6- GO ON TO THE NEXT PAGE.

3 004 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 0 percent. 1. Answer the following questions relating to the solubilities of two silver compounds, Ag CrO 4 and Ag PO 4. Silver chromate dissociates in water according to the equation shown below. Ag CrO 4 (s) Ag + (aq) + CrO 4 (aq) K sp at 5 C (a) Write the equilibrium-constant expression for the dissolving of Ag CrO 4 (s). (b) Calculate the concentration, in mol L 1, of Ag + (aq) in a saturated solution of Ag CrO 4 at 5 C. (c) Calculate the maximum mass, in grams, of Ag CrO 4 that can dissolve in 100. ml of water at 5 C. (d) A mol sample of solid AgNO is added to a 1.00 L saturated solution of Ag CrO 4. Assuming no volume change, does [CrO 4 ] increase, decrease, or remain the same? Justify your answer. In a saturated solution of Ag PO 4 at 5 C, the concentration of Ag + (aq) is M. The equilibriumconstant expression for the dissolving of Ag PO 4 (s) in water is shown below. K sp [Ag + ] [PO 4 ] (e) Write the balanced equation for the dissolving of Ag PO 4 in water. (f) Calculate the value of K sp for Ag PO 4 at 5 C. (g) A 1.00 L sample of saturated Ag PO 4 solution is allowed to evaporate at 5 C to a final volume of 500. ml. What is [Ag + ] in the solution? Justify your answer. Copyright 004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 6 GO ON TO THE NEXT PAGE.

4 001 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 0 percent. 1. Answer the following questions relating to the solubility of the chlorides of silver and lead. (a) At 10 C, 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, K sp, for AgCl(s) at 10 C. (b) At 5 C, the value of K sp for PbCl (s) is and the value of K sp for AgCl(s) is (i) If 60.0 ml of M NaCl(aq) is added to 60.0 ml of M Pb(NO ) (aq), will a precipitate form? Assume that volumes are additive. Show calculations to support your answer. (ii) Calculate the equilibrium value of [Pb + (aq)] in 1.00 L of saturated PbCl solution to which 0.50 mole of NaCl(s) has been added. Assume that no volume change occurs. (iii) If M NaCl(aq) is added slowly to a beaker containing both 0.10 M AgNO (aq) and M Pb(NO ) (aq) at 5 C, which will precipitate first, AgCl(s) or PbCl (s)? Show calculations to support your answer. Copyright 001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 6 GO ON TO THE NEXT PAGE.

5 00 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 0 percent. HC H 5 O (aq) H + (aq) + C H 5 O (aq) 1. Lactic acid, HC H 5 O, 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 HC H 5 O (aq) at 98 K. For parts (a) through (d) below, assume the temperature remains at 98 K. (a) Write the expression for the acid-dissociation constant, K a, for lactic acid and calculate its value. (b) Calculate the ph of 0.50 M HC H 5 O. (c) Calculate the ph of a solution formed by dissolving mole of solid sodium lactate, NaC H 5 O, in 50. ml of 0.50 M HC H 5 O. 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 HC H 5 O. Calculate the molar concentration of lactate ion, C H 5 O, in the resulting solution. Copyright 00 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 6 GO ON TO THE NEXT PAGE.

6 007 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,, and. The Section II score weighting for each question is 0 percent. HF(aq) + H O(l ) H O + (aq) + F (aq) K a 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 H O + in a 0.40 M HF(aq) solution. HF(aq) reacts with NaOH(aq) according to the reaction represented below. HF(aq) + OH (aq) H O(l) + F (aq) A volume of 15 ml of 0.40 M NaOH(aq) is added to 5 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. 007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents). -6- GO ON TO THE NEXT PAGE.

7 Advanced Placement Chemistry 1987 Free Response Questions Go to the answers Return to Additional Materials Menu 1) NH + H O <> NH OH Ammonia is a weak base that dissociates in water as shown above. At 5 C, the base dissociation constant, K b, for NH is 1.8 x (a) Determine the hydroxide ion concentration and the percentage dissociation of a molar solution of ammonia at 5 C. (b) Determine the ph of a solution prepared by adding mole of solid ammonium chloride to 100. milliliters of a molar solution of ammonia. (c) If mole of solid magnesium chloride, MgCl, is dissolved in the solution prepared in part (b) and the resulting solution is well-stirred, will a precipitate of Mg(OH) form? Show calculation to support your answer. (Assume the volume of the solution is unchanged. The solubility product constant for Mg(OH) is 1.5 x ).

8 009 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,, and. The Section II score weighting for each question is 0 percent. 1. A pure g sample of the weak base ethylamine, C H 5 NH, is dissolved in enough distilled water to make 500. ml of solution. (a) Calculate the molar concentration of the C H 5 NH in the solution. The aqueous ethylamine reacts with water according to the equation below. C H 5 NH (aq) + H O(l)! " C H 5 NH + (aq) + OH # (aq) (b) Write the equilibrium-constant expression for the reaction between C H 5 NH (aq) and water. (c) Of C H 5 NH (aq) and C H 5 NH + (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 M C H 5 NH with 500. ml of 0.00 M HCl. Assume that volumes are additive. The ph of the resulting solution is found to be (i) Calculate the concentration of OH # (aq) in the solution. (ii) Write the net-ionic equation that represents the reaction that occurs when the C H 5 NH solution is mixed with the HCl solution. (iii) Calculate the molar concentration of the C H 5 NH + (aq) that is formed in the reaction. (iv) Calculate the value of K b for C H 5 NH. 009 The College Board. All rights reserved. Visit the College Board on the Web: GO ON TO THE NEXT PAGE.

9 !" #$%&'()+,#-...#!#Scoring Standards Question 1 (10 points) (a) / 0### [H ] [S ] [H S] 1+ / 0 (H ) (S (H S) ) 1 pt. No point earned for a / ## expression or if an exponent is incorrectly used (b) (i) [H ] [S ] (.0)(.7 " L! mol) 5.95 " 10! 1 pt. (ii) [H S] 0.10 mol! (.0)(.7 " L or! mol).05 " 10! pts. [H S] Notes: One point is earned for getting the correct number of moles of H S ; second point is earned for dividing by 1.5 L and getting a consistent answer. Although not correct, one point may be earned for calculating the initial [H S] (see below) and using that value as the equilibrium H S concentration. 440# g! # g mol [H S] # L (c) / 0#! 7 10 ( ) 4 " " 15 4! (.05" 10 )! (5.95" 10! ) (.05" 10 (0.098)! ) 0.50 pts. Notes: One point is earned for correctly using the molarity of the S (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 000 College Entrance Examination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Examination Board.

10 !" #$%&'()+,#-...#!#Scoring Standards Question 1 (015(56&7) 59: (.7 " 10! mol)(0.081 L atm mol! 1 K! 1)(48 K) (d) " 1.18 atm L or " [S ]9: ( mol L!1 )(0.081 L atm mol! 1 K!1 ) (48 K) 1.18 atm pts. 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 51;< 9: with " S " S 1;<, where S is the mole fraction of S, can earn the points. (e) / 0 # 1 / 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 [HS] [H ][S ] 1 Copyright 000 College Entrance Examination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Examination Board.

11 AP CHEMISTRY 007 SCORING GUIDELINES (Form B) Question 1 A sample of solid U O 8 is placed in a rigid L flask. Chlorine gas, Cl (g), is added, and the flask is heated to 86 C. The equation for the reaction that takes place and the equilibrium-constant expression for the reaction are given below. ( puocl ) ( p O ) U O 8 (s) + Cl (g)! UO Cl (g) + O (g) K p! ( p ) When the system is at equilibrium, the partial pressure of Cl (g) is atm and the partial pressure of UO Cl (g) is atm. (a) Calculate the partial pressure of O (g) at equilibrium at 86 C. Cl U O 8 (s) + Cl (g)! UO Cl (g) + O (g) I ---? 0 0 C E atm atm? One point is earned for the correct answer atm UO Cl (g) (1 mol O ) ( mol UO Cl ) atm O (g) (b) Calculate the value of the equilibrium constant, K p, for the system at 86 C. ( puocl ) ( p O ) K p ( p ) Cl 4 4 ( ) ( ) (1.007) 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 86 C. G RT ln K p ( 8.1 J mol 1 K 1 )( (86+7) K)(ln ( )) 7,000 J mol 1 7 kj mol 1 One point is earned for the correct setup. One point is earned for the correct answer with units. 007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents).

12 AP CHEMISTRY 007 SCORING GUIDELINES (Form B) Question 1 (continued) (d) State whether the entropy change, S, for the reaction at 86 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 86 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, mol of O (g) is added to the mixture in the flask. Does the mass of U O 8 (s) in the flask increase, decrease, or remain the same? Justify your answer. The mass of U O 8 (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 O (g) is consumed (tending to relieve the stress) as more U O 8 (s) is produced. One point is earned for a correct explanation. 007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents).

13 !"! #$%&'()+,# -../#)$0+(1#(4&5(1&)# 6789:;<#># ;',!%H. 49 ;' I AH. : J&01%5(%"74)19&<<"5&4)1<&C4)1%455"%9&')")(11FE4)&"<("C/10"C: ;',!%H.?!B,;' K?"#BK!%H.,L?"#B$!% M,:N O- LO, 4),P!?4B Q%&)1)(11FE&0&/%&E7R5"<)4)16#%1<<&"@"%)(19&<<"0&'"@ ;',!%H.?!B: $!% MS;' K T, S!%H.,L T O#"&)@"%5"%%15)16#%1<<&"?/B!405E04)1)(15"51)%4)&"G&7"0U O G"@ ;' K?"#B &4<4)E%4)19<"0E)&""@ ;',!%H. 4),P!: ;',!%H.?!B,;' K?"#BK!%H.,L?"#B V L - -! L K,& K& W L -K,& -K&,:N O- LO, MS;' K T, S!%H,L. T,:N O- LO, MS,&T, S&TM.& I X:Y O- LP M6MS!%H.,L T S;' K TM,&M,?X:Y O- LP ' BMO:Y O- L. ' O#"&)@"%5"%%15)<)"&5(&"71)%$"@;' K?"#B 49!%H.,L?"#B O#"&)@"%<E/<)&)E)&'495405E04)&'S;' K T?5B!405E04)1)(1746&7E774<<G&'%47<G"@ ;',!%H. )(4)549&<<"01&O--:7U"@C4)1%4),P!: P X:Y O- 7"01 ;'!%H OU,. -:-,Z ' ;',!%H. OU IIO:Y ' O 7"01 ;'!%H M -:-,Z ' ;',!%H. OU,. -:O--UM-:--,Z';',!%H. O#"&)@"%7"04%74<<"@;',!%H. O#"&)@"%74<<"@;',!%H. &O--7U!"#$%&'()+,--./$!"001'1)%451647&4)&"8"4%9:;00%&'()<%1<1%19: >&<&)4#51)%40:5"001'1/"4%9:5"7?@"%;A#%"@1<<&"40<B49CCC:5"001'1/"4%9:5"7D4#<)E91)<?@"%;A<)E91)<49#4%1)<B: -#

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15 !"#$%&'()+,-$.//0$%1,)$4)5'6)'$ 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) "5 g 8.9 # g/mol 6. # 10 "7 mol (in 100 ml) 1 point (6. # 10 "7 mol/100 ml) (1,000 ml/1.000 L) 6.! 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) K sp [Ag + ][Cl " ] (6. # 10 "6 ).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) mol 1 point [Cl " ] (0.004 mol)/(0.10 L) 0.00 mol/l 0.00 M n Pb + (0.060 L) (0.00 mol/l) mol [Pb + ] ( mol)/(0.10 L) mol/l M Q [Pb + ][Cl " ] (0.015)(0.00) 6.0! 10 "6 Q < K sp, therefore no precipitate forms 1 point 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) [Pb + ] K sp " [Cl ] 1.6 # 10 (0.5) " 5.6! 10 "4 M 1 point Copyright 001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

16 !"#$%&'()+,-$.//0$%1,)$4)5'6)'$ (iii) for AgCl solution: [Cl " ] AgCl K sp [Ag + ] 1.8 # "10 1.5! 10 "9 M 1 point for PbCl solution: [Cl " ] PbCl K sp + [Pb ] 1.6 # "5 1.0! 10 " M The [Cl " ] will reach a concentration of 1.5 # 10 "9 M before it reaches a concentration of 1.0 # 10 " M, (or 1.5 # 10 "9 << 1.0 # 10 " ), therefore AgCl(s) will precipitate first. 1 point Note: One point is earned for calculating [Cl " ] in saturated solutions with the appropriate Ag + and Pb + concentrations; one point is earned for concluding which salt will precipitate first, based on the student s calculations. Copyright 001 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

17 AP CHEMISTRY 00 SCORING GUIDELINES (Form B) Question 1 10 points HC H 5 O (aq) Æ H + (aq) + C H 5 O (aq) 1. Lactic acid, HC H 5 O, 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 HC H 5 O (aq) at 98 K. For parts (a) through (d) below, assume the temperature remains at 98 K. (a) Write the expression for the acid-dissociation constant, K a, for lactic acid and calculate its value. + [H ][CH5O ] K a [HC H O ] 5 _ 0.50 M! M x 1 point earned for equilibrium expression 1 point earned for amount of HC H 5 O dissociating HC H 5 O (aq) " H+ (aq) + C H 5 O (aq) I 0.50 ~0 0 C x +x +x E 0.50 x +x +x + [H ][CH5O K a [HCH5O ] K a 1.4! 10 4 ] [0.008][ 0.008] [ ] 1 point earned for [H + ] [C H 5 O ] set up and solution (b) Calculate the ph of 0.50 M HC H 5 O. From part (a): [H + ] M 1 point earned for correctly calculating ph ph log [H + ] log (0.008).08 Copyright 00 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

18 AP CHEMISTRY 00 SCORING GUIDELINES (Form B) Question 1 (cont d.) (c) Calculate the ph of a solution formed by dissolving mole of solid sodium lactate, NaC H 5 O, in 50. ml of 0.50 M HC H 5 O. Assume that volume change is negligible mol NaCH5O 0.50 L 0.18 M C H 5 O HC H 5 O (aq) " H+ (aq) + C H 5 O (aq) I 0.50 ~ C x +x +x E 0.50 x +x x 1 point earned for [C H 5 O ] (or 0.50 L! 0.50 mol/l 0.15 mol HC H 5 O and mol C H 5 O ) K a [H + ][C [HC H H 5 5 O O ] ] [x][ x] [0.50 # x] Assume that x << 0.18 M K a 1.4! 10 4 [x][0.18] [0.50] x.9! 10 4 M [H + ] 1 point earned for [H + ] (set up and calculation) 1 point earned for calculating the value of ph ph log [H + ] log (.9! 10 #4 ).41 OR ph pk a + log 0.18 or Copyright 00 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

19 AP CHEMISTRY 00 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 HC H 5 O. Calculate the molar concentration of lactate ion, C H 5 O, in the resulting solution M HC H 5 O $ %& 100 ml 00 ml ' () 0.5 M HC H 5 O 0.10 M HCl $ %& 100 ml 00 ml ' () M H+ HC H 5 O (aq) " H+ (aq) + C H 5 O (aq) I C x +x + x E 0.5 x x + x + [H ][CH5O ] [ x][x] K a [HCH5O ] [0.5 # x] Assume x << M K a 1.4! 10 4 [0.050][x] [0.5] x 7.0! 10 4 M [C H 5 O ] 1 point earned for initial [H + ] and [HC H 5 O ] OR (10 mmol H + ; 50 mmol HC H 5 O ) 1 point earned for showing dilution or moles of each 1 point earned for [C H 5 O ] setup and calculation Copyright 00 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 4

20 AP CHEMISTRY 007 SCORING GUIDELINES Question 1 HF(aq) + H O(l )! " H O + (aq) + F (aq) K a 7. # 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. + [HO ][F ] K a One point is earned for the correct expression. [HF] (b) Calculate the molar concentration of H O + in a 0.40 M HF(aq) solution. [HO ][F ] K a [HF] + ( x)( x) 7. # x Assume x << 0.40, then x (0.40)(7. # 10 4 ) x [ H O + ] 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) H O(l) + F (aq) A volume of 15 ml of 0.40 M NaOH(aq) is added to 5 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 (0.05 L)(0.40 mol L 1 ) (0.015 L)(0.40 mol L 1 ) mol mol 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 mol F (aq) mol F ( aq) ( ) L of solution 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. 007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents).

21 AP CHEMISTRY 007 SCORING GUIDELINES Question 1 (continued) (e) Calculate the ph of the solution. [HF] mol HF L 0.10 M HF [HO ][F ] K a! [HF]! M (7. 10 ) 0.15 M [HF] [F ] K a ! ph log ( ). + [H O ] 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 pk a + log [F ] [HF] One point is earned for the correct calculation of ph. log ( ) + log 0.15 M 0.10 M The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents).

22 ChemTeam: AP Free Response Answers Free Response Answers Notes Advanced Placement Chemistry [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 [NH + 4 ] [OH ] x [NH ] mol/l - x K b ([NH + 4 ][OH ]) [NH ] 1.8 x 10 5 [(x) (x)] ( x) approximately equals x x [OH ] 1.6 x 10 mol/l % diss [(1.6 x 10 ) / (0.150)] x 100% 1.1% b) three points [NH + 4 ] mol / L mol/l NH + 4 [NH ] mol/l OR mol NH mol NH + 4 mol NH mol/l x L mol THEN 1 of 9 9/6/06 6: PM

23 ChemTeam: AP Free Response Answers x 10 5 [(0.500) (x)] (0.150) OR poh log (0.500 / 0.150) THEN x [OH ] 5.4 x 10 6 mol/l poh 5.7 ph c) three points Mg(OH) (s) <> Mg + + OH [Mg + ] mol / L mol/l (0.800 ± x no credit) ([OH ] 5.4 x 10 6 mol/l from b.) Q [Mg + ][OH ] (0.800) (5.4 x 10 6) Q. x K sp 1.5 x since Q > K sp, Mg(OH) precipitates (Q must be defined in the same way as K sp ) OR K sp 1.5 x (0.800)[OH ] [OH ] 4. x 10 6 mol/l since 5.4 x 10 6 > 4. x 10 6 mol/l, then Mg(OH) precipitates ) a) three points; one each for form of rate law, HgCl exponent, C O 4 exponent Rate k [HgCl ][C O 4 ] of 9 9/6/06 6: PM

24 AP CHEMISTRY 009 SCORING GUIDELINES (Form B) Question 1 (10 points) A pure g sample of the weak base ethylamine, C H 5 NH, is dissolved in enough distilled water to make 500. ml of solution. (a) Calculate the molar concentration of the C H 5 NH in the solution. n CH5NH g C H 5 NH M CH5NH! 0.9 mol C H 5 NH 0.9 mol CH5NH L 1 mol CH5NH g C H NH M 5 One point is earned for the correct number of moles. One point is earned for the correct concentration. The aqueous ethylamine reacts with water according to the equation below. C H 5 NH (aq) + H O(l)! " C H 5 NH + (aq) + OH # (aq) (b) Write the equilibrium-constant expression for the reaction between C H 5 NH (aq) and water. K b + 5 [C H NH ][OH ] [C H NH ] 5 " One point is earned for the correct expression. (c) Of C H 5 NH (aq) and C H 5 NH + (aq), which is present in the solution at the higher concentration at equilibrium? Justify your answer. C H 5 NH is present in the solution at the higher concentration at equilibrium. Ethylamine is a weak base, and thus it has a small K b value. Therefore only partial dissociation of C H 5 NH occurs in water, and [C H 5 NH + ] is thus less than [C H 5 NH ]. One point is earned for the correct answer with justification. 009 The College Board. All rights reserved. Visit the College Board on the Web:

25 AP CHEMISTRY 009 SCORING GUIDELINES (Form B) Question 1 (continued) (d) A different solution is made by mixing 500. ml of M C H 5 NH with 500. ml of 0.00 M HCl. Assume that volumes are additive. The ph of the resulting solution is found to be (i) Calculate the concentration of OH # (aq) in the solution. ph #log[h + ] [H + ] 10 # $ 10 #11 [OH # K ] w [H # ] OR 1.00! ! 10 " 14 " 11 poh 14 # ph 14 # poh #log[oh # ] [OH # ] 10 # $ 10 #4 M 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 C H 5 NH solution is mixed with the HCl solution. C H 5 NH + H O +! C H 5 NH + + H O One point is earned for the correct equation. (iii) Calculate the molar concentration of the C H 5 NH + (aq) that is formed in the reaction. moles of C H 5 NH L! moles of H O L! mol 1.00 L 0.00 mol 1.00 L 0.50 mol mol [C H 5 NH ] [H O + ] [C H 5 NH + ] initial value ~ 0 change #0.100 # final value ~ [C H 5 NH + ] # mol C H NH 1.00 L M One point is earned for the correct number of moles of C H 5 NH and H O +. One point is earned for the correct concentration. 009 The College Board. All rights reserved. Visit the College Board on the Web:

26 AP CHEMISTRY 009 SCORING GUIDELINES (Form B) Question 1 (continued) (iv) Calculate the value of K b for C H 5 NH. [C H 5 NH ] mol CH5NH 1.00 L M One point is earned for the correct calculation of the molarity of C H 5 NH after neutralization. K b + 5 [C H NH ][OH ] [C H NH ] 5 " " 4 (0.100)(8.5! 10 ) $ 10 #4 One point is earned for the correct value. 009 The College Board. All rights reserved. Visit the College Board on the Web: