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Triad 5 Practice Test Key

Triad 5 Practice Test Key. West Valley High School AP Chemistry Mr. Mata. Chapter 14. 1. Which of the following does not fit the definition of an Arrhenius base? a . NH 3 b. NaOH c. KOH d. H 2 O e. not given Answer: a (no OH - ). Chapter 14.

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Triad 5 Practice Test Key

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  1. Triad 5 Practice Test Key West Valley High School AP Chemistry Mr. Mata

  2. Chapter 14 1. Which of the following does not fit the definition of an Arrhenius base? a. NH3 b. NaOH c. KOH d. H2O e. not given Answer: a (no OH- )

  3. Chapter 14 2. The conjugate acid of NH3 is • NH2- b. NHOH c. NH4+ d. NH2+ e. not given Answer: c (NH3 gained H+ = Bronsted Lowry base; resulting product = conjugate acid)

  4. Chapter 14 3. Which of the following is the correct equilibrium expression: HCN (aq) + H2O (l) <--> H3O + (aq) + CN - (aq) • [CN-] [H3O+] [H2O][HCN] b. [H3O+] [CN-] [HCN] c. [HCN] [H3O+] [CN-] d. [H2O][HCN] [CN-] [H3O+] Answer: b (H2O ignored in Keq set up)

  5. Chapter 14 4. What is Kb If Ka of an acid is 2.0 x 10 -3? • 5.0 x 10 11 b. 1 x 10 -3 c. 2.0 x 10 -11 d. 5.0 x 10 -12 e. not given Answer: d (Ka)(Kb) = 1.00 x 10 -14 Kb = 1.00 x 10 – 14 = 5.0 x 10 -12 2.0 x 10 -3

  6. Chapter 14 5. The pOH of a 2.0 M HCl solution is a. 12 b. 14 c. 13.7 d. 0.3 e. not given Answer: c pH = - log[H +] = - log [2.0M] = 0.3 pH + pOH = 14  pOH = 14 – 0.3 = 13.7 pOH = 13.7

  7. Chapter 14 6. Calculate the pH of a 0.25 M HCN solution. Ka= 6.2 x 10 -10. a. 4.7 b. 9.3 c. 2.5 d. 5.0 e. not given Answer: d HCN = weak acid (small Ka) HCN (aq) <-> CN – (aq) + H + (aq) I 0.25 0 0 C - x + x + x E 0.25 – x xx Ka = 6.2 x 10 – 10 = x2 = 1.24 x 10 -5 = [H +] 0.25 – x pH = - log [1.24 x 10 -5] = 4.9 = 5.0

  8. Chapter 14 7. Calculate pH of a 0.0075 M solution of a weak base, Kb = 1.6 x 10 -6. a. 10.04 b. 3.96 c. 0.00124 d. 13.98 e. not given Answer: a XOH = weak base (small ba) XOH (aq) <-> OH – (aq) + X + (aq) I 0.0075 0 0 C - x + x + x E 0.0075 – x xx Kb = 1.6 x 10 – 6= x2= 1.10 x 10 -4= [OH -] 0.0075 – x pOH= - log [1.10 x 10 -4] = 3.96  pH = 10.04

  9. Chapter 14 FRQ 8. What is the equilibrium constant for the reaction? N3–+ H3O +<-> HN3+ H2O The Ka value for HN3 = 1.9 x 10 –5. Answer: Ka for HN3 = weak acid Keq = [HN3] = [1.9 x 10 -5] [N3-] [H3O+] [1.9 x 10 -5 – x ] [1.9 x 10 -5 - x] Keq = 5.3 x 10 4

  10. Chapter 15 9. Calculate the pH of a solution prepared by mixing 40.0 mL of a 0.02 M HCl solution with 200.0 mL of 0.20 M HCN solution. Assume volumes to be additive. Ka for HCN = 1.0 x 10 -10. a. 2.4 b. 10 c. 2.0 d. 5.6 e. not given Answer: d mmoleHCl: 0.02 mmol/mL x 40.0 mL = 0.80 mmol= 0.0033 M 240 mL mmole HCN: 0.02 mmol/mL x 200.0 mL = 40.0 mmol= 0.167 M 240 mL Ka (HCN) = (1.0 x 10 -10)(0.167M) = x2 x = [H+] = 4.1 x 10 -6 Ka(HCl) = (1.0 x 10 -10)(0.0033) = x2  x = [H+] = 5.7 x 10 -7 Average [H+] = 2.3 x 10 -6M  pH = - log[2.3 x 10 -6] = 5.63

  11. Chapter 15 10. Calculate the pH of a solution prepared by mixing 60.0 mL of a 0.200 M NaOH solution with 60.0 mL of a 0.200 M CH3NH2 solution. Assume the volumes are additive. Kb for CH3NH2 = 4.3 x 10 -5. a. 11.8 b. 13.0 c. 1.0 d. 0.7 e. not given Answer: b mmoleNaOH: 0.200 mmol/mL x 60.0 mL = 12 mmol= 0.1 M 120 mL mmoleCH3NH2: 0.200 mmol/mL x 60.0 mL = 12 mmol= 0.1 M 120 mL Kb (NaOH) = (2.32 x 10 -10)(0.1M) = x2 x = [OH-] = 4.8 x 10 -6 Kb(CH3NH2) = (4.3 x 10 -5)(0.0033) = x2  x = [OH-] = 2.1 x 10 -3 Average [OH-] = 1.1 x 10 -3M  pOH= - log[1.1 x 10 -3] = 2.96 pH = 11.04

  12. Chapter 15 11. What is the pOH of a solution prepared by mixing 0.30 mol of HCN and 0.50 mol of NaCN? pKa = 9.40 a. 7.00 b. 10.9 c. 3.36 d. 4.38 e. not given Answer: e HCN  H + + CN – pKa = 9.40 = [H+] [CN-] = [H+][0.50] [HCN] [0.30] [H+] = (9.40)(0.30) = 5.64 0.50 pH = - log [H+] = - log [5.64] = 1.50 pOH = 12.5

  13. Chapter 15 12. By what factor is the [H+] of a solution lowered if the pH changes from 10.40 to 7.40? a. 100 b. 30.0 c. 3.00 d. 1000 e. not given Answer: d 7.40  8.40  9.40  10.40 Each 1 increment pH increase = 10x increase; 10 x 10 x 10 = 1000 factor increase from pH 7.40 to 10.40

  14. Chapter 15 13. 50.0 mL of 0.020 M HCN is titrated with 100.0 mL of 0.010 M LiOH. Calculate the pH of the resulting solution at the equivalence point. Ka= 4.9 x 10 -9. a. 9.9 b. 4.1 c. 7.7 d. 5.2 e. not given Answer: d MaVa = MbVb Ma = MbVb Va Ma = (0.010M)(100.0 mL) = 0.02 M 50.0 mL Ka = 4.9 x 10 -9 = x2 = 8.9 x 10 -6= [H +] 0.02 M pH = - log [H +] = - log [8.9 x 10 -6] = 5.1 ~ 5.2

  15. Chapter 15 14. How many mL of a 1.06 M Ca(OH)2 solution are required to titrate 70.0 mL of 0.88 M H3PO4 solution to the equivalence point? a. 87.2 mL b. 70.0 mL c. 22.6 mL d. 43.9 mL e. not given Answer: e MaVa = MbVb Vb= MaVa Mb Vb= (0.88M)(70.0 mL) = 58.1 mL 1.06 M

  16. Chapter 15 15. A 500.0 mL sample of solution saturated with AgCl, is allowed to evaporate to dryness. 0.966 mg of AgCl is recovered. Calculate Ksp for AgCl. a. 1.80 x 10 -10 b. 3.74 x 10 -8 c. 2.22 x 10 -7 d. 3.74 x 10 -6 e. not given Answer: e AgCl (s)  Ag + (aq) + Cl– (aq) Ksp = [Ag +] [Cl-] 0.966 mg x 1 x 10 -6 g x 1 molx 1000 mL = 1.35 x 10 -8 500 mL 1 mg 143.32 g 1 L Ksp = [1.35 x 10 -8] [1.35 x 10 -8] = 1.8 x 10 -16

  17. Chapter 15 FRQ 16. You are given 5.00 mL of an H2SO4 solution of unknown concentration. You divide the 5.00-mL sample into five 1.00-mL samples and titrate each separately with 0.1000 M NaOH. In each titration the H2SO4 is completely neutralized. The average volume of NaOH solution used to reach the endpoint is 15.3 mL. What was the concentration of H2SO4 in the 5.00-mL sample? Answer: MaVa = MbVb  Ma = MbVb = (0.1000 M)(15.3 mL) Va 1.00 mL Ma = 1.53 M

  18. Chapter 17 17. A state of higher entropy means: a. a lower number of possible arrangements b. a higher number of possible arrangements c. lower probabilities to reach a possible state d. lower probabilities to be reached e. not given Answer: b

  19. Chapter 17 18. Which of the following processes would result in a decrease in entropy? a. freezing b. melting c. evaporating d. an expanding gas e. not given Answer: a Solids have lower entropies

  20. Chapter 17 19. Heat is released during a particular process. This means that: a. the process is spontaneous under all conditions b. S surr > 0 c. the process tends to be spontaneous d. S > 0 e. not given Answer: b Exothermic; heat given off to surroundings > 0

  21. Chapter 17 20. Which of the following processes would you expect to be spontaneous? a. S surr = 25 J/K, S sys = - 27 J/K b. S surr = 25 J/K, S sys = 27 J/K c. S univ = - 20 J/K, S sys = - 20 J/K d. S surr = - 80 J/K, S sys = 20 J/K e. not given Answer: b + S for surroundings and system = high entropy and more likely to be spontaneous.

  22. Chapter 17 21. Which of the following conditions must be met for a process to be spontaneous? • G < 0 • H < 0 • S surr > 0 • S sys > 0 e. not given Answer: a G < 0 means – H and + S values

  23. Chapter 17 22. Calculate S for the following reaction: H3AsO4(aq)  3 H+ (aq) + AsO4-3 (aq) S(H+) = 0.00 J/mol K S (H3AsO4) = 44.0 J/mol K S ([AsO4]-3) = - 38.9 J/mol K • 5.10 J/molK b. – 82.9 J/mol K c. – 5.1 J/mol K • d. – 72.7 J/mol K e. not given Answer: b S = Sprod – Sreact S = [3(0 J/mol K) + - 38.9 J/mol K] – [44.0 J/mol K] S = - 82.9 J/mol K

  24. Chapter 17 23. Calculate the free energy change G, in kJ, for the following reaction at 298 K: N2 (g) + 2O2 (g)  2NO2 (g) 0.2 atm 0.2 atm 0.39 atm  a. 109 b. 37.32 c. 13.98 d. – 37.32 e. not given Answer: a (see page 808 in textbook)

  25. Chapter 17 FRQ 24. For the reaction Cl2O(g) + (3/2)O2(g) -> 2ClO2(g) H° = 126.4 kJ/mol ; S° = –74.9 J/K mol ; T = 377°C Calculate G° : Answer: G = H – T S G = (126.4 kJ/mol) – (650 K)(- 0.0749 kJ/mol) G = 175.1 kJ/mol

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