Role of The Weak Acid Component Questions

Chem 201 Exam 2 – Smith – BB Format page 1 Spring 2020 1a. What is the role of the weak acid component of an acid-base buffer solution? a. It minimizes the pH change caused by added strong base by reacting with the strong base to convert the strong base to its neutral conjugate. b. It minimizes the pH change caused by added strong acid by reacting with the strong acid to convert the strong acid to its neutral conjugate. c. It maximizes the pH change caused by added strong base by reacting with the strong base to convert the strong base to its strong acid conjugate. d. It maximizes the pH change caused by added strong base by reacting with the strong base to convert the strong base to its strong acid conjugate. 1b. What is the role of the weak base component of an acid-base buffer solution? a. It minimizes the pH change caused by added strong base by reacting with the strong base to convert the strong base to its neutral conjugate. b. It minimizes the pH change caused by added strong acid by reacting with the strong acid to convert the strong acid to its neutral conjugate. c. It maximizes the pH change caused by added strong base by reacting with the strong base to convert the strong base to its strong acid conjugate. d. It maximizes the pH change caused by added strong base by reacting with the strong base to convert the strong base to its strong acid conjugate. 2a. What is the pH of a buffer solution composed of 0.10 M (CH3)3N and 0.25 M [(CH3)3NH]Cl? a. 9.41 b. 10.20 c. 3.80 d. 4.59 e. 7.83 2b. What is the pH of a buffer solution composed of 0.25 M (CH3)3N and 0.10 M [(CH3)3NH]Cl? a. 9.41 b. 10.20 c. 3.80 d. 4.59 e. 7.83 Chem 201 Exam 2 – Smith – BB Format page 2 Spring 2020 3a. Carbonic acid, H2CO3, is a weak, diprotic acid (2 acidic H’s). For what pH range would a mixture of H2CO3 and NaHCO3 make a good buffer? a. 5.35 to 7.35 b. 6.35 to 10.33 c. 9.33 to 11.33 d. 5.35 to 11.33 3b. Carbonic acid, H2CO3, is a weak, diprotic acid (2 acidic H’s). . For what pH range would a mixture of Na2CO3 and NaHCO3 make a good buffer? a. 5.35 to 7.35 b. 6.35 to 10.33 c. 9.33 to 11.33 d. 5.35 to 11.33 4a. The pH of an aqueous solution of Na2CO3 is adjusted to pH 7.50 by adding HCl. Which of the following should be true about the resulting solution? The solution will contain (in addition to NaCl(aq))…. a. mainly H2CO3 b. equal amounts of H2CO3 and HCO3 − c. mainly HCO3 − d. equal amounts of HCO3 − and CO3 2− e. mainly CO3 2− 4b. The pH of an aqueous solution of Na2CO3 is adjusted to pH 10.33 by adding HCl. Which of the following should be true about the resulting solution? The solution will contain (in addition to NaCl(aq))…. a. mainly H2CO3 b. equal amounts of H2CO3 and HCO3 − c. mainly HCO3 − d. equal amounts of HCO3 − and CO3 2− e. mainly CO3 2− 4c. The pH of an aqueous solution of Na2CO3 is adjusted to pH 6.35 by adding HCl. Which of the following should be true about the resulting solution? The solution will contain (in addition to NaCl(aq))…. a. mainly H2CO3 b. equal amounts of H2CO3 and HCO3 − c. mainly HCO3 − d. equal amounts of HCO3 − and CO3 2− e. mainly CO3 2− Chem 201 Exam 2 – Smith – BB Format page 3 Spring 2020 4d. The pH of an aqueous solution of Na2CO3 is adjusted to pH 11.00 by adding HCl. Which of the following should be true about the resulting solution? The solution will contain (in addition to NaCl(aq))…. a. mainly H2CO3 b. equal amounts of H2CO3 and HCO3 − c. mainly HCO3 − d. equal amounts of HCO3 − and CO3 2− e. mainly CO3 2− 4e. The pH of an aqueous solution of Na2CO3 is adjusted to pH 5.25 by adding HCl. Which of the following should be true about the resulting solution? The solution will contain (in addition to NaCl(aq))…. a. mainly H2CO3 b. equal amounts of H2CO3 and HCO3 − c. mainly HCO3 − d. equal amounts of HCO3 − and CO3 2− e. mainly CO3 2− 5. Consider the following titration: 24.0 mL of 0.030 M HClO4(aq) (flask) is titrated with 0.045 M KOH(aq) (buret). What is the overall reaction occurring in the titration? a. HClO4(aq) + KOH(aq) → H2O(l) + K+ (aq) + ClO4 − (aq) b. HClO4(aq) + KOH(aq) → KH2ClO4(aq) c. HClO4(aq) + KOH(aq) → H2O(l) + K(aq) + ClO4(aq) d. HClO4(aq) + KOH(aq) → H2ClO4(aq) + KO(aq) e. HClO4(aq) + KOH(aq) → H2O(l) + KO(aq) + ClO3 − (aq) 6a. Consider the following titration (which is the same as in the previous problem): 24.0 mL of 0.030 M HClO4(aq) (flask) is titrated with 0.045 M KOH(aq) (buret). What is in the flask after the titration is started but before the equivalence point is reached? a. HClO4(aq) and KOH(aq) only b. HClO4(aq), KOH(aq), and all products c. HClO4(aq) and all products d. KOH(aq) and all products e. all products Chem 201 Exam 2 – Smith – BB Format page 4 Spring 2020 6b. Consider the following titration (which is the same as in the previous problem): 24.0 mL of 0.030 M HClO4(aq) (flask) is titrated with 0.045 M KOH(aq) (buret). What is in the flask at the equivalence point of the titration? a. HClO4(aq) and KOH(aq) only b. HClO4(aq), KOH(aq), and all products c. HClO4(aq) and all products d. KOH(aq) and all products e. all products 6c. Consider the following titration (which is the same as in the previous problem): 24.0 mL of 0.030 M HClO4(aq) (flask) is titrated with 0.045 M KOH(aq) (buret). What is in the flask past the equivalence point of the titration? a. HClO4(aq) and KOH(aq) only b. HClO4(aq), KOH(aq), and all products c. HClO4(aq) and all products d. KOH(aq) and all products e. all products 7a. Consider the following titration (which is the same as in the previous problem): 24.0 mL of 0.030 M HClO4(aq) (flask) is titrated with 0.045 M KOH(aq) (buret). What will be in the flask after 12.0 mL of KOH have been added? (In order to answer this question correctly, you will first need to calculate the equivalence point of the titration in order to determine where you are with respect to the equivalence point.) a. 2.30 b. 11.70 c. 1.52 d. 12.47 e. 3.58 f. 10.42 7b. Consider the following titration (which is the same as in the previous problem): 24.0 mL of 0.030 M HClO4(aq) (flask) is titrated with 0.045 M KOH(aq) (buret). What will be in the flask after 18.0 mL of KOH have been added? (In order to answer this question correctly, you will first need to calculate the equivalence point of the titration in order to determine where you are with respect to the equivalence point.) a. 11.33 b. 2.67 c. 12.28 d. 1.71 e. 4.72 f. 9.28 Chem 201 Exam 2 – Smith – BB Format page 5 Spring 2020 8a. A certain ionic compound with molecular formula M2X3 (where M is the cation and X is the anion) has a molar solubility in water at 25 °C of 1.5×10−5 M. What is Ksp of this ionic compound at 25 °C? a. 8.2×10−23 b. 4.3×10−2 c. 1.4×10−18 d. 1.4×10−14 e. 2.7×10−2 8b. A certain ionic compound with molecular formula MX3 (where M is the cation and X is the anion) has a molar solubility in water at 25 °C of 1.5×10−5 M. What is Ksp of this ionic compound at 25 °C? a. 8.2×10−23 b. 4.3×10−2 c. 1.4×10−18 d. 1.4×10−14 e. 2.7×10−2 8c. A certain ionic compound with molecular formula MX2 (where M is the cation and X is the anion) has a molar solubility in water at 25 °C of 1.5×10−5 M. What is Ksp of this ionic compound at 25 °C? a. 8.2×10−23 b. 4.3×10−2 c. 1.4×10−18 d. 1.4×10−14 e. 1.6×10−2 9. What is the Ksp expression for silver sulfide? a. [Ag+ ] 2 [S2−] b. [Ag+ ][S2−] 2 c. [Ag2+][S2−] d. [Ag+] 2 [SO4 2−] e. [Ag2+][SO4 − ] 2 f. [Ag2+] 2 [SO4 − ] 10a. What happens if some sodium sulfide is added to a saturated aqueous solution of silver sulfide? a. The silver sulfide solubility equilibrium shifts to the LEFT and more solid precipitates. b. The silver sulfide solubility equilibrium shifts to the RIGHT and more solid precipitates. c. The silver sulfide solubility equilibrium shifts to the LEFT and more solid dissolves. d. The silver sulfide solubility equilibrium shifts to the RIGHT and more solid dissolves. e. Nothing happens. Chem 201 Exam 2 – Smith – BB Format page 6 Spring 2020 10b. What happens if some solid silver sulfide is added to a saturated aqueous solution of silver sulfide? a. The silver sulfide solubility equilibrium shifts to the LEFT and more solid precipitates. b. The silver sulfide solubility equilibrium shifts to the RIGHT and more solid precipitates. c. The silver sulfide solubility equilibrium shifts to the LEFT and more solid dissolves. d. The silver sulfide solubility equilibrium shifts to the RIGHT and more solid dissolves. e. Nothing happens. 10c. What happens if some silver nitrate is added to a saturated aqueous solution of silver sulfide? a. The silver sulfide solubility equilibrium shifts to the LEFT and more solid precipitates. b. The silver sulfide solubility equilibrium shifts to the RIGHT and more solid precipitates. c. The silver sulfide solubility equilibrium shifts to the LEFT and more solid dissolves. d. The silver sulfide solubility equilibrium shifts to the RIGHT and more solid dissolves. e. Nothing happens 11a. What happens if some nitric acid is added to a saturated aqueous solution of silver sulfide? a. The silver sulfide solubility equilibrium shifts to the LEFT and more solid precipitates. b. The silver sulfide solubility equilibrium shifts to the RIGHT and more solid precipitates. c. The silver sulfide solubility equilibrium shifts to the LEFT and more solid dissolves. d. The silver sulfide solubility equilibrium shifts to the RIGHT and more solid dissolves. e. Nothing happens. 11b. Ammonia forms complex ions with many metal cations including silver. What happens if some ammonia is added to a saturated aqueous solution of silver sulfide? a. The silver sulfide solubility equilibrium shifts to the LEFT and more solid precipitates. b. The silver sulfide solubility equilibrium shifts to the RIGHT and more solid precipitates. c. The silver sulfide solubility equilibrium shifts to the LEFT and more solid dissolves. d. The silver sulfide solubility equilibrium shifts to the RIGHT and more solid dissolves. e. Nothing happens Chem 201 Exam 2 – Smith – BB Format page 7 Spring 2020 12a. Lead(II) forms a complex ion with three hydroxides. What is the reaction corresponding to the formation of this complex ion? a. Pb2+(aq) + 3 OH− (aq) ⇄ Pb(OH)3 − (aq) b. Pb(aq) + 3 OH(aq) ⇄ Pb(OH)3(aq) c. Pb2+(aq) + 3 O2− (aq) ⇄ Pb(O)3 4− (aq) d. Sn2+(aq) + 3 OH−(aq) ⇄ Sn(OH)3 − (aq) e. Sn(aq) + 3 OH(aq) ⇄ Sn(OH)3(aq) f. Sn2+(aq) + 3 O2−(aq) ⇄ Sn(O)3 4− (aq) 12b. Tin(II) forms a complex ion with three hydroxides. What is the reaction corresponding to the formation of this complex ion? a. Pb2+(aq) + 3 OH−(aq) ⇄ Pb(OH)3 − (aq) b. Pb(aq) + 3 OH(aq) ⇄ Pb(OH)3(aq) c. Pb2+(aq) + 3 O2−(aq) ⇄ Pb(O)3 4− (aq) d. Sn2+(aq) + 3 OH− (aq) ⇄ Sn(OH)3 − (aq) e. Sn(aq) + 3 OH(aq) ⇄ Sn(OH)3(aq) f. Sn2+(aq) + 3 O2−(aq) ⇄ Sn(O)3 4− (aq) 13a. What is the maximum amount of potassium chloride (mw = 74.55 g/mol ) that could be added to a 1.0 L of a 1.0×10−3 M solution of gold(III) nitrate that would not cause gold(III) chloride to precipitate? Ksp of gold(III) chloriide = 3.2×10−25 (FYI, Au is the elemental symbol for gold….although you don’t really need this info to solve this problem.) a. just under 5.1×10−6 g b. just under 2.2×10−8 g c. just under 2.6 g d. just under 7.4×0−3 g e. just under 0.045 g 13b. What is the maximum amount of potassium bromide (mw = 119.0 g/mol ) that could be added to a 25.0 L of a 1.0×10−2 M solution of gold(III) nitrate that would not cause gold(III) bromide to precipitate? Ksp of gold(III) bromide = 4.0×10−36 (FYI, Au is the elemental symbol for gold….although you don’t really need this info to solve this problem.) a. just under 5.1×10−6 g b. just under 2.2×10−8 g c. just under 2.6 g d. just under 7.4×0−3 g e. just under 0.045 g Chem 201 Exam 2 – Smith – BB Format page 8 Spring 2020 14a. Fill in the blanks. Physical processes tend to go towards ___________ entropy and ___________ energy. a. higher, lower b. lower, higher c. higher, higher d. lower, lower 14b. Fill in the blanks. Physical processes tend to go towards ___________ energy and ___________ entropy. a. higher, lower b. lower, higher c. higher, higher d. lower, lower 15a. Which one of the following reactions is most likely to have a ΔS° ~ 0? a. 2 CH4(g) ⇄ C2H2(g) + 3 H2(g) b. 2 Na2O2(s) + 2 CO2(g) ⇄ 2 Na2CO3(s) + O2(g) c. HNO3(l) + ClF(g) ⇄ ClONO2(g) + HF(g) d. CO2(g) + Ca(OH)2(s) ⇄ CaCO3(s) + H2O(g) 15b. Which one of the following reactions is most likely to have a ΔS° < 0? a. 2 CH4(g) ⇄ C2H2(g) + 3 H2(g) b. 2 Na2O2(s) + 2 CO2(g) ⇄ 2 Na2CO3(s) + O2(g) c. HNO3(l) + ClF(g) ⇄ ClONO2(g) + HF(g) d. CO2(g) + Ca(OH)2(s) ⇄ CaCO3(s) + H2O(g) 15c. Which one of the following reactions is most likely to have a ΔS° > 0? a. C2H2(g) + 3 H2(g) ⇄ 2 CH4(g) b. 2 Na2O2(s) + 2 CO2(g) ⇄ 2 Na2CO3(s) + O2(g) c. HNO3(l) + ClF(g) ⇄ ClONO2(g) + HF(g) d. CO2(g) + Ca(OH)2(s) ⇄ CaCO3(s) + H2O(g) Chem 201 Exam 2 – Smith – BB Format page 9 Spring 2020 16a. Calculate ΔS° for the following reaction: 4 NO(g) + 6 H2O(g) ⇄ 4 NH3(g) + 5 O2(g) S° of H2O(g) = 188.72 J/K; S° of H2O(l) = 69.94 J/K; S° of NH3(aq) = 110 J/K; S° of NH3(g) = 193 J/K; S° of NO(g) = 210.65 J/K; S° of O2(g) = 205.0 J/K a. −178 J/K b. +178 J/K c. −203 J/K d. +203 J/K e. +510 J/K f. −510 J/K 16b. Calculate ΔS° for the following reaction: 4 NH3(g) + 5 O2(g) ⇄ 4 NO(g) + 6 H2O(g) S° of H2O(g) = 188.72 J/K; S° of H2O(l) = 69.94 J/K; S° of NH3(aq) = 110 J/K; S° of NH3(g) = 193 J/K; S° of NO(g) = 210.65 J/K; S° of O2(g) = 205.0 J/K a. −178 J/K b. +178 J/K c. −203 J/K d. +203 J/K e. +510 J/K f. −510 J/K 16c. Calculate ΔS° for the following reaction: 4 NH3(aq) + 5 O2(g) ⇄ 4 NO(g) + 6 H2O(l) S° of H2O(g) = 188.72 J/K; S° of H2O(l) = 69.94 J/K; S° of NH3(aq) = 110 J/K; S° of NH3(g) = 193 J/K; S° of NO(g) = 210.65 J/K; S° of O2(g) = 205.0 J/K a. −178 J/K b. +178 J/K c. −203 J/K d. +203 J/K e. +510 J/K f. −510 J/K 16d. Calculate ΔS° for the following reaction: 4 NO(g) + 6 H2O(l) ⇄ 4 NH3(aq) + 5 O2(g) S° of H2O(g) = 188.72 J/K; S° of H2O(l) = 69.94 J/K; S° of NH3(aq) = 110 J/K; S° of NH3(g) = 193 J/K; S° of NO(g) = 210.65 J/K; S° of O2(g) = 205.0 J/K a. −178 J/K b. +178 J/K c. −203 J/K d. +203 J/K e. +510 J/K f. −510 J/K Chem 201 Exam 2 – Smith – BB Format page 10 Spring 2020 17a. According to the Second Law of Thermodynamics, what has to be true for any physical process, including a chemical reaction, to be spontaneous? a. The total disorder in the universe increases. b. The total disorder in the universe decreases. c. The total disorder in the system increases. d. The total disorder in the system decreases. 17b. According to the Second Law of Thermodynamics, what has to be true for any physical process, including a chemical reaction, to be nonspontaneous? a. The total disorder in the universe increases. b. The total disorder in the universe decreases. c. The total disorder in the system increases. d. The total disorder in the system decreases. 18a. A certain reaction has ∆S° = 80.0 J/K and ∆H° = 30.0 kJ. What is ∆G° of this reaction at −50.0 °C? Is this reaction overall favorable or unfavorable at this temperature? a. 12.1 kJ, unfavorable b. 12.1 kJ, favorable c. 34.0 kJ, unfavorable d. 34.0 kJ, favorable e. −31.9 kJ, favorable f. −31.9 kJ, unfavorable 18b. A certain reaction has ∆S° = 80.0 J/K and ∆H° = 30.0 kJ. What is ∆G° of this reaction at 500.0 °C? Is this reaction overall favorable or unfavorable at this temperature? a. 12.1 kJ, unfavorable b. 12.1 kJ, favorable c. −10.0 kJ, unfavorable d. −10.0 kJ, favorable e. −31.9 kJ, favorable f. −31.9 kJ, unfavorable 19a. A particular reaction is overall favorable at low temperatures, but unfavorable at high temperatures. Which of the following is true about this reaction? a. ∆S° > 0 and ∆H° > 0 b. ∆S° < 0 and ∆H° < 0 c. ∆S° > 0 and ∆H° < 0 d. ∆S° < 0 and ∆H° > 0 19b. A particular reaction is overall unfavorable at low temperatures, but favorable at high temperatures. Which of the following is true about this reaction? a. ∆S° > 0 and ∆H° > 0 b. ∆S° < 0 and ∆H° < 0 c. ∆S° > 0 and ∆H° < 0 d. ∆S° < 0 and ∆H° > 0 Chem 201 Exam 2 – Smith – BB Format page 11 Spring 2020 19c. A particular reaction is overall favorable at all temperatures. Which of the following is true about this reaction? a. ∆S° > 0 and ∆H° > 0 b. ∆S° < 0 and ∆H° < 0 c. ∆S° > 0 and ∆H° < 0 d. ∆S° < 0 and ∆H° > 0 19d. A particular reaction is overall unfavorable at all temperatures. Which of the following is true about this reaction? a. ∆S° > 0 and ∆H° > 0 b. ∆S° < 0 and ∆H° < 0 c. ∆S° > 0 and ∆H° < 0 d. ∆S° < 0 and ∆H° > 0 20a. Consider the following reaction. Calculate ∆G° for this reaction at 25 °C using information provided with the exam. C5H5N(aq) + H2O(l) ⇄ C5H5NH+(aq) + OH−(aq) a. 50.0 kJ b. −50.0 kJ c. 4.17 kJ d. −4.17 kJ e. 37.3 kJ f. −37.3 kJ 20b. Consider the following reaction. Calculate ∆G° for this reaction at 25 °C using information provided with the exam. HClO2(aq) + H2O(l) ⇄ ClO2−(aq) + H3O+ a. 11.2 kJ b. −11.2 kJ c. 0.94 kJ d. −0.94 kJ e. 26.5 kJ f. −26.5 kJ Chem 201 Exam 2 – Additional Information Spring 2020 Some Ka Values at 25 ̊C Ka1 Ka2 Ka3 CH3CO2H 1.8 ×10−5 CHOCO2H 3.5 ×10−4 H3BO3 5.4 ×10−10 HO2C(CH2)CO2H 1.4 ×10−3 2.0 ×10−6 HClO2 1.1 ×10−2 H2CO3 4.5 ×10−7 4.7 ×10−11 H3PO4 7.5 ×10−3 6.2 ×10−8 4.8 ×10−13 H2SO4 very large 1.2 ×10−2 HF 3.5 ×10−4 Some Kb values at 25 ̊C Kb NH3 1.8 ×10−5 CH3NH2 3.7 ×10−4 (CH3)3N 6.4 ×10−5 C5H5N 1.7 ×10−9 𝐾𝐾𝑎𝑎 𝐾𝐾𝑏𝑏 = 𝐾𝐾𝑤𝑤 = 1.008 × 10−14 𝑝𝑝𝑝𝑝 = 𝑝𝑝𝐾𝐾𝑎𝑎 + 𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑝𝑝𝐾𝐾𝑎𝑎 = −𝑙𝑙𝑙𝑙𝑙𝑙𝐾𝐾𝑎𝑎 ∆𝑆𝑆𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜 = ∑ 𝑆𝑆𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑜𝑜 − ∑ 𝑆𝑆𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜 ∆𝐻𝐻𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜 = ∑ 𝐻𝐻𝑓𝑓,𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑜𝑜 − ∑ 𝐻𝐻𝑓𝑓,𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜 ∆𝐺𝐺𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜 = ∑ 𝐺𝐺𝑓𝑓,𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑜𝑜 − ∑ 𝐺𝐺𝑓𝑓,𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜 ∆𝐺𝐺𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜 = ∆𝐻𝐻𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜 − 𝑇𝑇∆𝑆𝑆𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜 ∆𝐺𝐺𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜 = −𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝐾𝐾 = 𝑒𝑒−∆𝐺𝐺𝑜𝑜 𝑅𝑅𝑅𝑅 R = 8.314×10−3 kJ/(mol∙K) 0 o C = 273.15 K

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