++++Chem 162-2009 Chapter 15B-Application of acid and base equilibria practice problems
89 Pages

++++Chem 162-2009 Chapter 15B-Application of acid and base equilibria practice problems

Course Number: CHEM 162 162, Spring 2011

College/University: Rutgers

Word Count: 16143

Rating:

Document Preview

CHAPTER 15 2nd HALF APPLIC. OF ACID & BASE EQUILIBRIA (BUFFERS & TITRATIONS) Problems to prepare students for hourly exam II. Buffer concepts Buffer calculations Titrations and Indicators E. Tavss 4/06 1 BUFFER CONCEPTS 11 Chem 162-2008 Final Exam + Answers Chapter 15B- Acids, Bases and Acid-Base Equilibria Buffer concepts Which of the following mixtures does not result in a buffer? A. B. C. D....

Unformatted Document Excerpt
Coursehero >> New Jersey >> Rutgers >> CHEM 162 162

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

15 CHAPTER 2nd HALF APPLIC. OF ACID & BASE EQUILIBRIA (BUFFERS & TITRATIONS) Problems to prepare students for hourly exam II. Buffer concepts Buffer calculations Titrations and Indicators E. Tavss 4/06 1 BUFFER CONCEPTS 11 Chem 162-2008 Final Exam + Answers Chapter 15B- Acids, Bases and Acid-Base Equilibria Buffer concepts Which of the following mixtures does not result in a buffer? A. B. C. D. 0.100mol HCl with 0.300mol NH3 0.200mol NH4Cl with 0.400mol NH3 0.100mol NaOH with 0.200mol HBrO 0.100mol NaBrO with 0.200mol HBrO 0.100mol NaOH with 0.200mol NH3 E E. Assume everything is in a 1 L solution. A. NH3 + H3O+ NH4+ + H2O NH3 + H3O+ NH4+ + H2O 0 +0.1 0.2 Initial 0.3M 0.1M 0 Change -0.1 -0.1 +0.1 Equilibrium 0.2M 0.1M M of a base & 0.1M of its conjugate acid is a buffer. B. NH4+ + NH3 NR 0.2M of a conjugate acid & 0.4M of its base is a buffer. C. HBrO + OH- BrO- + H2O HBrO + OH Initial 0.2M 0.1M Change -0.1 -0.1 Equilibrium 0.1M 0M M weak acid & 0.1M conjugate base is a buffer. BrO0 +0.1 0.1M + H2O 0 +0.1 0.1M 0.1 D. HBrO + BrO- NR 0.2M of a weak acid & 0.1M of its conjugate base is a buffer E. NH3 + OH- NR A weak base & a strong base is not a buffer. 2 Chem 162-2008 Exam II + Answers Chapter 15B Applic. of Acid & Base Equilibria (Buffers & Titrations) Buffer concepts Definition of buffers; relationship between pH and pKa 20. Phosphoric acid, H3PO4, is a triprotic acid with: Ka1 = 7.5 x 10-3 Ka2 = 6.3 x 10-8 Ka3 = 4.3 x 10-13 Which of the following combinations would be best for preparing a pH 7 buffer? (a) (b) (c) (d) H3PO4 and Na3PO4 H3PO4 and NaH2PO4 Na2HPO4 and Na3PO4 H3PO4 and HCl NaH2PO4 and Na2HPO4 (e) Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) According to the Henderson-Hasselbalch equation, a good buffer is one which has approximately equal molar quantities of the weak acid and its conjugate base, resulting in pH = pKa + log(1) pH = pKa. H3PO4 + H2O H3O+ + H2PO4H2PO4- + H2O H3O+ + HPO42HPO42- + H2O H3O+ + PO43Ka1 = 7.5 x 10-3 Ka2 = 6.3 x 10-8 Ka3 = 4.3 x 10-13 pKa1 = -log(7.5 x 10-3) = 2.12 pKa2 = -log(6.3 x 10-8) = 7.20 pKa3 = -log(4.3 x 10-13) = 12.4 (a) False. A buffer is defined as a combination of a weak acid and its conjugate base; the conjugate base of H3PO4 is NaH2PO4, not Na3PO4. (b) False. Since the pKa of this combination is 2.12, it would make a good buffer at pH approximately 2. (c) False. Since the pKa of this combination is 12.4, it would make a good buffer at pH approximately 12. (d) False. A buffer is defined as a combination of a weak acid and its conjugate base; this is a combination of a weak acid and a strong acid. (e) True. Since the pKa of the acid is 7.20, it would make a good buffer at pH approximately 7. Chem 162-2008 Exam II + Answers Chapter 15B Applic. of Acid & Base Equilibria (Buffers & Titrations) Buffer concepts Given various mixtures of weak acids and strong bases, which one forms a buffer. 3 14. Which of the following combinations is a buffer? W. 1 mol CH3COOH, 0.5 mol NaOH, in 1 L water X. 0.5 mol CH3COOH, 1 mol NaOH, in 1 L water Y. 1 mol NH3, 0.5 mol HCl, in 1 L water Z. 0.5 mol NH3, 1 mol HCl, in 1 L water (a) (b) (c) (d) (e) W and Y only X and Z only W and X only Y and Z only all are buffers W: HA + OH- H2O + ASince the reactants contain OH-, bring reaction to completion. HA + H2O OH Initial 1 0.5 Change -0.5 -0.5 Equilibrium 0.5 0 A 1:1 mixture of a weak acid and its conjugate base is a buffer. X: HA + OH- H2O + ABring reaction to completion. HA + H2O OH Initial 0.5 1 Change -0.5 -0.5 Equilibrium 0 0.5 A 1:1 mixture of a strong base and a conjugate base is not a buffer. Y: NH3 + H3O+ H2O + NH4+ Bring reaction to completion. NH3 + H2O H3O+ Initial 1 0.5 Change -0.5 -0.5 Equilibrium 0.5 0 A 1:1 mixture of a weak base and its conjugate acid is a buffer. Z: NH3 + H3O+ H2O + NH4+ Bring reaction to completion. NH3 + H2O H3O+ Initial 0.5 1 Change -0.5 -0.5 Equilibrium 0 0.5 A 1:1 mixture of a conjugate acid and a strong acid is not a buffer. + A0 +0.5 0.5 + A0 +0.5 0.5 + NH4+ 0 +0.5 0.5 + NH4+ 0 +0.5 0.5 4 Chem 162-2008 Exam II + Answers Chapter 15B Applic. of Acid & Base Equilibria (Buffers & Titrations) Buffer concepts Given Ka of weak acid, or Kb of weak base, prepare buffer of pH9 9. Ammonia, NH3, is a weak base with Kb = 1.8 x 10-5 Acetic acid, CH3COOH, is a weak acid with Ka = 1.8 x 10-5 A group of students is trying out for a laboratory job. They are asked to prepare a buffer of pH 9. Archie says hed mix solutions of acetic acid and sodium acetate. Barbara says shed mix solutions of acetic acid and hydrochloric acid. Carla says shed mix solutions of ammonia and ammonium chloride. Dexter says hed mix solutions of ammonia and sodium hydroxide. Ethel says shed mix solutions of ammonia and acetic acid. Which student has the best procedure? (a) Archie (b) Barbara (c) Carla (d) Dexter (e) Ethel Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) A good buffer has approximately a 1:1 ratio of [A-] to [HA]; when this occurs, log([A-]/[HA]) = 0, and pH = pKa. Since we want a pH of 9, then the pKa should be approximately 9. NH3: Kw = Kca x Kb 1 x 10-14 = Kca x 1.8 x 10-5 Kca = 5.56 x 10-10 pKca = -log(5.56 x 10-10) = 9.25, which is certainly close to the desired pKca of 9. Hence, NH3 and a conjugate acid should provide a buffer of approximately pH 9. Acetic acid: Ka = 1.8 x 10-5 pKa = -log(1.8 x 10-5) = 4.74, which is certainly not close to the desired pKa of 9. Hence, a buffer using acetic acid is a very poor option. 5 (a) False. Archie: A buffer made with acetic acid wouldnt work. That would be for a pH of approximately 5, not 9. (b) False. Barbara: A buffer made with acetic acid wouldnt work. That would be for a pH of approximately 5, not 9. Furthermore, mixing acetic acid and hydrochloric acid wouldnt form any buffer. (c) True. Carla: A buffer made with ammonia and its conjugate base (e.g., ammonium chloride) would work. That would provide a buffer of approximately pH 9. (d) False. Dexter: A mixture consisting of ammonia and sodium hydroxide wouldnt work, because that wouldnt provide a buffer. A buffer, in this case, would be a mixture of a weak base and its conjugate acid, not a weak base and a strong base. (e) False. Ethyl: A buffer, in this case, would be a mixture of a weak base and its conjugate acid, not a mixture of a weak base and a weak acid. Acetic acid is not a conjugate acid of the weak base, ammonia. 6 10 Chem 162-2007 Final exam + answers Chapter 15B Applications of Acid & Base Equilibria Buffer concepts Which of the following solutions would be a buffer solution? A. 50 mL 1 M HCl + 50 mL 1 M NH3 B. 50 mL 1 M HCl + 50 mL 1 M NaOH C. 25 mL 1 M NaOH + 50 mL 1 M CH COOH 3 D. 25 mL 1 M NaOH + 50 mL 1 M NH3 E. 50 mL 1 M NaOH + 25 mL 1 M CH3COOH o The concentrations of the acids and bases are each diluted for Initial values. o A buffer must be a combination of a weak acid and its conjugate base, or a weak base and its conjugate acid. A. NH3 + H3O+ + NH4 + H2O Since this contains a strong acid, bring the reaction to completion. NH3 + NH4+ + H3O+ H2O Initial 0.5 0.5 0 Change -0.5 -0.5 +0.5 Equilibrium 0 0 +0.5 Since the equilibrium solution only contains a conjugate acid (i.e., no weak base), this isnt a buffer. B. H3O+ + OH- 2H2O Since this contains a strong acid (as well as a strong base), bring the reaction to completion. H3O+ + 2H2O OH- Initial 0.5 0.5 Change -0.5 -0.5 Equilibrium 0 0 equilibrium solution contains no buffer components. 0 +1.0 +1.0 The C. HA + OH- H2O + ASince this contains a strong base, bring the reaction to completion. HA + H2O + OH- A- Initial 0.67 0.33 0 0 Change -0.33 -0.33 +0.33 Equilibrium 0.33 0 0.33 the equilibrium solution contains a weak acid and a conjugate base, this is a buffer. D. NH3 + OH- No reaction Since the equilibrium solution contains only a weak base and a strong base, this is not a buffer. E. HA + OH H2O + A Since Since this contains a strong base, bring the reaction to completion. HA + H2O + OH- A- Initial 0.33 0.67 0 0 Change -0.33 -0.33 +0.33 Equilibrium 0 0.33 0.33 the equilibrium solution contains a conjugate base but no weak acid, this is not a buffer. Since 7 7 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CONCEPTS Consider a potential buffer solution containing 2.0 M HCN and 1.0 M NaCN (Ka for HCN = 6.2 10-10). Which one of the following statements is true? A. The solution is not a buffer, because [HCN] is not equal to [CN-]. B. The pH will be below 7.00, because the concentration of the acid is greater than that of the base. C. [OH-] > [H+] D. The buffer will be more resistant to pH changes from addition of strong acid than of strong base. E. The pH will equal 7.00, because the acid is very weak. A. This is a weak acid and its conjugate base in a 1:2 ratio. A buffer can be a weak acid and its conjugate base in a 1:10 ratio to a 10:1 ratio. Hence, this is a buffer. A is false. B. The pH depends on the Ka vs the Kcb. The Ka is 6.2 x 10-10. The Kcb is (1 x 10-14)/(6.2 x 10-10) = 1.61 x 10-5. Therefore, the acid is a weaker acid in formation of hydronium ions as compared to the strength of the base in the formation of hydroxide ions. Therefore, more hydroxide ions will form than hydronium ions, so the pH will be above 7. B is false. C. For the reasons stated in B, C is true. D. The buffer will be just as resistant in its reaction to strong acid or base, as long as the buffer system is not overwhelmed with strong acid or base. D is false. E. The pH will only equal 7 if the Ka of the acid and conjugate base were equal, i.e., 1 x 10-7. Several of the answers can also be determined quantitatively: HCN Initial Change Equilibrium 2.0 -X 2.0 - X + H2O H3O+ + 0 +X +X CN1.0 X 1.0 + X ([X] x [1.0 + X])/(2.0 X) = 6.2 x 10-10 Simplifying: ([X] x [1.0])/(2.0 ) = 6.2 x 10-10 X = 1.24 x 10-9 pH = -log(1.24 x 10-9) = 8.9 8 15 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CONCEPTS What is the net ionic equation, when a solution of potassium hydroxide is added to a solution of nitrous acid. A. B. C. HNO2 + K+ + OH- KNO2 + H2O HNO2 + H2O NO2- + H3O+ HNO2 + KOH K+ + NO2- + H2O HNO2 + OH- NO2- + H2O H+ + OH- H2O D. E. Complete equation: HNO2 + KOH HOH + KNO2 Ionic equation: HNO2 + K+ + OH- HOH + K+ + NO2 Net ionic equation: HNO2 + OH- HOH + NO2- 9 19. CHEM162-2000 HOURLY EXAM II CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB (BUFFERS & TITRATIONS) BUFFER CONCEPTS Which one of the following statements concerning buffers is false? A. A buffer can consist of a weak base and a salt containing its conjugate acid in the same solution. B. A buffer changes pH only slightly when small amounts of strong acid or strong base are added. C. The pH of a buffer solution should be within one pH unit of the pKa of the acid component of the buffer. D. The pH of a buffer will not change significantly if the buffer is diluted. E. The buffer capacity will not change significantly if the buffer is diluted. 8. CHEM162-2001 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) BUFFER CONCEPTS When dissolved in 1.0 L of water, which of the following would not produce a buffer solution? A buffer solution is a weak acid and the salt of the weak acid, or a weak base and the salt of the weak base. E is a strong acid and the salt of the strong acid. Hence, it is not a buffer. A. B. C. D. 1.0 mol HNO2 and 1.0 mol KNO2 1.0 mol NH3 and 1.0 mol NH4Br 1.0 mol HNO2 and 2.0 mol KNO2 1.0 mol HClO and 1.0 mol KClO E. 1.0 mol HCl and 1.0 mol KCl 10 CHEM162-2002 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) BUFFER CONCEPTS 14. Which of the following solutions will produce a buffer? Each has a volume of 1.0 Liter. A buffer contains a weak acid and its conjugate base, or a weak base and its conjugate acid. W. A mixture of 0.20 mol CH3COOH and 0.10 mol NaOH A mixture of 0.20 mol CH3COOH and 0.10 mol NaOH will go almost to completion to provide 0.10 mol CH3COOH and 0.10 mol CH3COO-Na+, a buffer solution. X. A mixture of 0.10 mol CH3COOH and 0.20 mol NaOH A mixture of 0.10 mol CH3COOH and 0.20 mol NaOH will go almost to completion to provide 0.10 mol CH3COO-Na+ and 0.10 mol NaOH, not a buffer solution. Y. A mixture of 0.20 mol NH3 and 0.10 mol HC1 A mixture of 0.20 mol NH3 and 0.10 mol HCl will go almost to completion to provide 0.10 mol NH3 and 0.10 mol NH4Cl, a buffer solution. Z. A mixture of 0.10 mol NH3 and 0.20 mol HC1 A mixture of 0.10 mol NH3 and 0.20 mol HCl will go almost to completion to provide 0.10 mol NH4Cl and 0.10 mol HCl, not a buffer solution. A. B. C. all will be buffers Y and Z only X and Z only W and Y only W and X only D. E. CHEM162-2003 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) 9. Which of the following combinations would be a buffer? W. 1.0 mol NaCl + 1.0 mol HCl in 1.00 liter solution X. 1.0 mol KNO3 + 1.0 mol HNO3 in 1.00 liter solution Y. 1.0 mol KNO2 + 1.0 mol HNO2 in 1.00 liter solution Z. 1.0 mol NH4Cl + 1.0 mol NH3 in 1.00 liter solution A. B. Y only W and X only 11 C. D. X and Y only all will be buffers E. Y and Z only A buffer is a weak acid and the salt of a weak acid, or a weak base and the salt of a weak base. W is a strong acid and the salt of a strong acid. X is a strong acid and the salt of a strong acid. Y is a weak acid and the salt of a weak acid. Z is a weak acid and the salt of a weak acid. CHEM162-2003 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) BUFFER CONCEPTS 23. Given the following information: H3PO4 Ka1 = 7.5 x 10-3 Ka2= 6.2x10-8 Ka3= 4.8x 10-13 H2SO4 Ka1 is large Ka2 = 1.2 x 10-2 Which one of the following combinations would be the best choice for preparing a buffer of pH 7.0? A. H3PO4 and NaH2PO4 B. Na2HPO4 and Na3PO4 C. D. E. NaH2PO4 and Na2HPO4 NaHSO4 and Na2SO4 H2SO4 and NaHSO4 pKa1 = -log(7.5x10-3) = 2.12 pKa1 = -log(6.2x10-8) = 7.21 pKa1 = -log(4.8x10-13) = 12.32 pKa1 = -log(large) = ~-7 pKa1 = -log(1.2x10-2) = 1.92 Ka1: H3PO4 + H2O H3O+ + H2PO4 Ka2: H2PO4- + H2O H3O+ + HPO42 Ka3: HPO42- + H2O H3O+ + PO43 Ka1: H2SO4 + H2O H3O+ + HSO4 Ka2: HSO4- + H2O H3O+ + SO42Henderson-Hasselbalch equation: pH = pKa + log([Base]/[Acid]) Use the Henderson-Hasselbalch equation for virtually all buffer problems. If we assume that the base and the acid are equal in concentration, then log([Base]/[Acid]) equals log(1). Since log(1) equals zero, then pH = pKa. In this case, we would want a buffer whose acid has a pKa of approximately 7. The acid that closest to that is H2PO4-, with pKa1 of 7.21. The conjugate base of H2PO4- is HPO42-. So the buffer system is H2PO4- is HPO42-. If we write it with the corresponding cations (spectator ions), it would be NaH2PO4 is Na2HPO4. (Correspondingly, if we want a buffer of pH 2.12, then H3PO4 and H2PO4- would be perfect because, at equal concentrations of H3PO4 and H2PO4-, log([A-]/[HA]) would be equal to log(1), which equals 12 zero, so since pKa1 = 2.12, then pH = 2.12.) 18. CHEM 162-2000 FINAL EXAM CHAPTER 15 - APPLIC OF ACID & BASE EQILIBR (BUFFERS & TITR) BUFFER CONCEPTS Which of the following would not produce a buffer solution when dissolved in 500 mL of water? A. B. C. D. E. 0.5 mol HCN and 0.2 mol KCN 0.5 mol HCN and 2.0 mol KCN 1.0 mol HCN and 0.5 mol KOH 0.5 mol HClO4 and 0.5 mol KClO4 0.5 mol CH3NH2 and 0.5 mol CH3NH3Cl D 13 11. CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB (BUFFERS & TITRATION) BUFFER CONCEPTS Which of the following would produce a buffer solution when dissolved in 500 mL of water? A buffer solution is a weak acid and a salt of a weak acid, or a weak base and a salt of a weak base. The ratios are usually from 1:100 to 100:1. I is a weak acid and a salt of a weak acid, in equal amounts. II is a weak acid and a salt of a weak acid, in a ratio acceptable for a buffer. In III the strong base will completely react with the weak acid producing H2O and 0.5 mol KF. This does not satisfy the definition of a buffer. In IV the strong base, which is the limiting reactant, will completely react with 0.3 mol of the weak acid, producing H2O, 0.3 mol KF, and leaving 0.2 mol HF. 0.3 mol KF and 0.2 mol HF constitute a buffer. I. 0.5 mol HF and 0.5 mol KF II. 0.5 mol HF and 0.3 mol KF III. 0.5 mol HF and 0.5 mol KOH IV. 0.5 mol HF and 0.3 mol KOH A. I only B. I and III only C. II and IV only D. I, II, and IV only E. I, II, III, and IV Chem 162-2003 Final exam + answers Chapter 15A-Applications of Aqueous Equilibria (of Acids and Bases) Buffer concepts (including Henderson-Hasselbalch equation) 23. Which of the following is false for a buffer solution? A. A buffer solution can be prepared by mixing appropriate amounts of a weak base and a strong acid. B. Exclusive of water, at least one component of a buffer solution must be neutral C. A buffer solution can consist of a mixture of a weak acid and its conjugate base. D. The solution resists changes in its [H+]. E. Added H+ ions will react with the conjugate base of the weak acid in solution A. True. This is one way to make a buffer. B. False. There is no reason to believe that a buffer cant consist of two charged species. For example, the combination of H2PO4- and HPO42- forms a buffer at pH of 7.2. C. True. Thats one of the definitions of a buffer. 14 D. True. Thats what buffers do. E. True. Thats how buffers work against acid addition. 17. Chem 162-2005 Exam II Applications of Aqueous Equlibria - Chapter 15A Buffers, common ions, Henderson-Hasselbalch equation Concept Which of the following would not produce a buffer solution when dissolved in 500 mL of water? A. B. 0.50 mol HCN and 0.20 mol KCN 0.20 mol HCN and 0.50 mol KCN 0.50 mol HClO4 and 0.50 mol KClO4 0.50 mol CH3NH2 and 0.50 mol CH3NH3Cl 1.0 mol HCN and 0.50 mol KOH C. D. E. A. A combination of a weak acid and its conjugate base is a buffer. B. A combination of a weak acid and its conjugate base is a buffer. C. A combination of a strong acid and its conjugate base is not a buffer. D. A combination of a weak base and its conjugate acid is a buffer. E. Reaction between 1.0 mol of a weak acid and 0.50 mol of a strong base results in 0.5 mol of the weak acid remaining, and 0.50 mol of the conjugate base formed. This is a buffer. 15 9. Chem 162-2005 Final Exam + Answers Chapter 15 - Applications of Aqueous Equilibria (of Acids and Bases) Buffers (including Henderson-Hasselbalch equation) Concepts Consider a 100 mL buffer solution of HA and NaA of pH = 5. If the solution is diluted to 500 mL by adding water, which one of the following will change? The ratio of [HA]/[A-]in Henderson-Hasselbalch equation The pKb of the conjugate base of the acid The capacity of a given volume of the buffer The pH of the solution The pKa of the acid A. B. C. D. E. HA + H2O H3O+ + ApH = 5 [H3O+] = 10-5 = 1 x 10-5 M HA + H3O+ + A- H2O Initial X 1 x 10-5 Z Change Equilibrium 0.2X 0.2Z Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) (A) Before dilution: [HA]/[A-] = X/Z After dilution: [HA]/[A-] = (0.2X)/(0.2Z) = X/Z, i.e., no change (B) The calculation of pKb is related to Kw and Ka, not to the concentration of HA or A-, so it is not related to a change in concentration of HA or A-. (C) See page 727 in Zumdahl. The capacity of a buffered solution is determined by the magnitudes of [HA] and [A-]. As an example, if [A-] is high, then the solution will buffer the effect of a lot of strong acid, but if [A-] is low, then as soon as the [A-] is depleted, the pH will drop sharply. (D) According to the Henderson-Hasselbalch equation, if the ratio of the concentrations of A- and HA doesnt change, and if the pKa doesnt change, then the pH wont change. If we dont use the Henderson-Hasselbalch equation for the calculation, but use the conventional ICE table type of equilibrium calculations, then well see that, in fact, the pH does change, However the pH change will be very small. The buffer capacity change is more meaningful than the pH change, and is therefore the better answer of the two. (E) The pKa of the acid is related to the Ka of the acid, which is unrelated to the concentration of the acid (or the concentration of the conjugate base). 16 BUFFER CALCULATIONS Chem 162-2008 Exam II + Answers Chapter 15B Applic. of Acid & Base Equilibria (Buffers & Titrations) Buffer calculations Henderson-Hasselbalch equation beginning with weak acid and hydroxide 11. Methanoic acid, HCOOH, has Ka = 1.8 x 10-4. Calculate the pH of a solution prepared by adding 0.60 mol HCOOH and 0.20 mol of NaOH to a total of 1.00 Liter aqueous solution. (a) (b) (c) (d) (e) 3.44 4.22 4.74 2.74 5.22 + OH0.20 M- HM + OH- H2O + MHM Initial 0.60 Change Equilibrium H2O + Since this is a reaction of a strong base, it goes to completion. HM + OH Initial 0.60 0.20 Change -0.20 -0.20 Equilibrium 0.40 Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) pH = -log(1.8 x 10-4) + log([0.20]/[0.40]) = 3.744 -0.301 = 3.44 H2O + M0 +0.20 0.20 CHEM 162-2007 EXAM II + ANSWERS CHAPTER 15B APPLICATIONS OF ACID & BASE EQUILIBRIA (BUFFERS & TITRATIONS) BUFFER CALCULATIONS 20. Acetic acid, CH3COOH, is a weak acid with Ka = 1.8 x 10-5. What amount of sodium acetate, CH3COONa, must be dissolved in 250.0 mL of 0.30 M CH3COOH, in order to produce a buffer with pH = 4.60? 17 (a) 0.12 mol (b) 0.054 mol (c) 0.022 mol (d) 0.015 mol (e) 0.078 mol 0.30 M HA Ka = 1.8 x 10-5 [A ] = ? pH = 4.60; [H+] = 10-4.60 = 2.51 x 10-5 HA + H2O H3O+ + AHA Initial Change Equilibrium 0.30 2.51 x 10-5 HA Initial Change Equilibrium X = 2.51 x 10-5 HA + H2O H3O+ + AInitial 0.30 0 Y -5 -5 Change -2.51 x 10 +2.51 x 10 +2.51 x 10-5 Equilibrium 0.30-2.51 x 10-5 2.51 x 10-5 Y + 2.51 x 10-5 ([H3O+][A-])/[HA] = Ka ([2.51 x 10-5][Y + (2.51 x 10-5)])/[0.30 (2.51 x 10-5)] = 1.8 x 10-5 Due to small Ka, assume that 2.51 x 10-5 is very small and can be dropped. Therefore, [HA] = 0.30; [A-] = Y ([2.51 x 10-5][Y])/[0.30] = 1.8 x 10-5 (((([2.51 x 10-5] x [Y])/[0.30]) = (1.8 x 10-5)),X) Y = 0.215mol/L = 0.0538 mol/0.250L 0.30 -X 0.30 - X + H2O H3O+ 0 +X 2.51 x 10-5 + AY +X Y+X + H2O H3O+ + AY 18 2 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CALCULATIONS Which one of the following solutions will be the best buffer at a pH of 9.26? (Ka for HC2H3O2 is 1.8 10-5, Kb for NH3 is 1.8 10-5). A. B. C. 0.10 M HC2H3O2 and 0.10 M NaC2H3O2 5.0 M HC2H3O2 and 5.0 M NaC2H3O2 0.10 M NH3 and 0.10 M NH4Cl 5.0 M NH3 and 5.0 M NH4Cl 5.0 M HC2H3O2 and 5.0 M NH3 Register to View Answer D. E. pH = pKa + log([A-]/[HA]) or pH = pKca + log([B]/[BH+]) A. pH = -log(1.8x10-5) + log(0.1/0.1) = 4.74 B. pH = -log(1.8x10-5) + log(5.0/5.0) = 4.74 C. pH = -log((1x10-14)/(1.8x10-5)) + log(0.10/0.10) = 9.26 D. pH = -log((1x10-14)/(1.8x10-5)) + log(5.0/5.0) = 9.26 This will be a better buffer than C because it has a higher buffer capacity. E. A buffer cannot be made with a weak acid and a weak base, because they are two weak to form enough of the conjugate acid or conjugate base to satisfy the 1:10 to 10:1 rule for making buffers. 19 6 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CALCULATIONS HF is a weak acid with Ka = 7.2 10-4. What is the pH of a solution that contains 0.50 M HF and 0.60 M NaF? A. 1.72 3.22 3.44 5.53 8.46 B. C. D. E. Equilibrium 0.50-X - +X 0.60+X pH = pKa + log([A ]/[HA]) Simplify, by dropping the X and +X. pH = -log(7.2x10-4) + log(0.60/50) pH = 3.22 20 21 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CALCULATIONS Given the following data: H2CO3 HCO3- + H+ Ka = 4.0 10-7 H2PO4- HPO42- + H+ Ka = 6.3 10-8 A buffer of pH 6.4 may be prepared by using either a mixture of H2CO3 and HCO3- or a mixture of H2PO4- and HPO42-. Which one of the following statements is true concerning pH 6.4 buffers? A. [H2CO3] > [HCO3-] and [H2PO4-] > [HPO42-] [H2CO3] = [HCO3-] and [H2PO4-] > [HPO42-] [H2CO3] = [HCO3-] and [HPO42-] > [H2PO4-] [HCO3-] > [H2CO3] and [HPO42-] > [H2PO4-] [H2CO3] > [HCO3-] and [HPO42-] > [H2PO4-] B. C. D. E. pH = pKa + log([A-]/[HA]) For the H2CO3/HCO3- system, pKa = -log(4.0 x 10-7) = 6.40 For the H2CO3/HCO3- buffer system, according to HH: 6.4 = 6.4 + log([A-]/[HA]) Therefore, log([A-]/[HA]) = 0 Therefore, [A-]/[HA] = 1 Therefore, [A-] = [HA]. For the H2PO4-/HPO42- system, pKa = -log(6.3x10-8) = 7.20 6.4 = 7.20 + log([A-]/[HA]) For the H2PO4-/HPO42- buffer system, according to HH the base concentration must be < the acid concentration in order to lower the 7.20 pH to 6.40. [H2PO4-] > [HPO42-]. B meets these requirements. 21 22 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CALCULATIONS H2A is a diprotic acid with Ka1 = 2.5 10-5, Ka2 = 3.1 10-9. If 100.0 mL of 1.00 M NaOH is added to 200.0 mL of 1.00 M H2A, calculate the pH of the resultant solution. A. B. C. D. 13.4 11.0 8.51 6.56 4.60 This is a very difficult problem. I spent more time on it then on any other problem, and Im still not certain how to do it. E.T. H2A+H2O H3O+ + HAKa1 = 2.5x10-5 + 2 HA- + H2O Ka2 = 3.1x10-9 H3O + A Equilibrium Bringtocompletion Equilibrium +0.0333 0 pH = pKa + log([A ]/[HA]) pH=log(2.5x105)+log([0.0333]/[0.0333])=4.60 pOH=14.04.60=9.40 [OH]=109.40=3.98x1010 Fromfirstreaction,pH=4.60.Therefore,[H3O+]=2.51x105 Equilibrium 0.0333-X (2.51x10-5)+X [H3O+] = 2.51x10-5 pH = -log(2.51x10-5) = 4.60 E. +0.0333 +X 22 24 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CALCULATIONS HOCl is a weak acid with Ka = 3.5 10-8. A buffer solution is prepared which contains 0.50 M HOCl and 0.40 M NaOCl. What is the pH after 10.0 mL of 1.0 M NaOH has been added to 100.0 mL of this buffer solution? A. B. C. D. 6.45 6.64 7.36 7.45 7.55 E. 10.0 mL of 1M NaOH added to 100 mL of buffer solution = 100 mL of 1M NaOH added to 1L of buffer solution = 0.100M NaOH 0.100M [OH-] diluted to 0.1mol/(1.1L) = 0.0909 M 0.50M [HOCl] diluted to 0.50 mol/(1.1L) = 0.455L 0.40M [OCl-] changed to 0.40 mol/1.1L = 0.364L The equilibrium equation should be the main reaction. Since OH- is a stronger base than H2O, the HOCl will preferentially react with the OH- rather than the H2O. Hence, HOCl + OH- H2O + OCl- is the main reaction. HOCl + H2O + OCl OH Initial 0.455 M 0.0909 M 0.364 Change Equilibrium To avoid quadratic equation, bring reaction to completion since all reactions with strong base go to completion. HOCl + + OCl H2O OH Initial 0.455 M 0.0909 M 0.364 Change -0.0909 -0.0909 +0.0909 Equilibrium 0.364M 0 0.455M pH = pKa + log([A ]/[HA]) pH = -log(3.5 x 10-8) + log(0.455/0.364) = 7.55 CHEM162-2003 5TH WEEK RECITATION CHAPTER 14 - ACID AND BASE EQUILIBRIA ACID AND BASE EQUILIBRIA CALCULATIONS 27 Calculate the pH after 0.020 mol HCl is added to 1.00 L of 0.1 M sodium propanoate. (HC3H5O2, Ka = 1.3 x 10-5) 23 (b) Approach I. This approach is not recommended, because it requires the use of a quadratic equation; i.e., X is too large to be dropped. P- + H3O+ HP + H2O PInitial 0.100 Change -X Equilibrium 0.100 - X H3O+ 0.020 -X 0.020-X HP 0 +X +X X/((0.100 - X)(0.020 - X)) = 1/(1.3 x 10-5) = 7.69 x 104 X = 0.0199968 H3O+ = 3.2 x 10-6 pH = -log (3 x 10-6) = 5.49 Approach II. This approach is better than approach I, because it doesnt require the use of a quadratic equation, but it is still not the best approach to solving this problem. P- + H3O+ HP + H2O PH3O+ HP Initial 0.100 0.020 0 Change -0.020 -0.020 +0.020 Equilibrium 0.080 0 0.020 P- + H2O HP + OHPNew Initial 0.080 Change -X Equilibrium 0.080 - X HP 0.020 +X 0.020+X OH0 +X +X ((0.020+X) x (X))/(0.080-X) = (1 x 10-14)/(1.3 x 10-5) X = +3.077 x 10-9 = [OH-] H+ = (1 x 10-14)/(3.077 x 10-9) = 3.25 x 10-6 pH = -log(3.25 x 10-6) = 5.49 Approach III. Approach dumbest, but by far the fastest, and is therefore the one of choice. Henderson-Hasselbach equation: pH = pKa + log([A-]/[HA]) pH = -log (1.3x10-5) + log(0.080/0.020) = 5.49 24 27d. Calculate the pH after 0.020 mol HCl is added to 1.00 L of a mixture containing 0.100 M HC3H5O2 and 0.100 M NaC3H5O2. Ka = 1.3 x 10-5 P- + H3O+ HP + H2O P+ H3O+ HP + H 2O Initial 0.1 0.020 0.1 Change Equilibrium Which way will the reaction go? The reaction is the reverse of the Ka reaction. Hence, K is the inverse of Ka. [HP]/([P-][H3O+]) = 1/Ka = 1/(1.3 x 10-5) = 7.69 x 104 Q = [HP]/([P-][H3O+]) = 0.1/([0.1][0.020]) = 50 << 7.69 x 104 Since Q is smaller than K, then the reaction goes to the right. (We really dont have to find Q. Since the reaction involves a strong acid, then it must go to the right.) P- + H3O+ HP + H2O P+ H3O+ HP + H 2O Initial 0.1 0.020 0.1 Change -X -X +X Equilibrium 0.1-X 0.020-X 0.1+X [0.1+X]/([0.1-X][0.020-X]) = 7.69 x 104 This is a quadratic equation. To avoid a quadratic equation, bring the reaction to completion (since reactions of strong acids go to completion, or since K is so large [7.69 x 104]). P+ H3O+ HP + H 2O Initial 0.1 0.020 0.1 Change -0.020 -0.020 +0.020 25 Equilibrium 0.080 0 0.120 H 2O Now bring it to equilibrium to find [H3O+]: P+ H3O+ HP + Initial 0.080 0 0.120 Change +X +X -X Equilibrium 0.080+X +X 0.120-X + -5 ([P ][H3O ]/[HP]) = 1.3 x 10 ([0.080 + X][X])/[0.120-X]) = 1.3 x 10-5 X = 1.95 x 10-5 pH = -log(1.95 x 10-5) = 4.71 However, best approach to 27d is simply to bring the reaction to completion, followed by using the Henderson-Hasselbach equation: P+ H3O+ HP + H 2O Initial 0.1 0.020 0.1 Change -0.020 -0.020 +0.020 Equilibrium 0.080 0 0.120 pH = pKa + log([A-]/[HA]) pH = -log (1.3x10-5) + log(0.080/0.120) = 4.71 CHEM162-2003 5TH WEEK RECITATION CHAPTER 14 - ACID AND BASE EQUILIBRIA ACID AND BASE EQUILIBRIA CALCULATIONS 29 Calculate the pH after 0.020 mol NaOH is added to 1.00 L of 0.10 M propanoic acid. (Ka = 1.3 x 10-5) (a) HP + H2O H3O+ + P- 26 Initial Change Equilibrium HP 0.100 -0.020 0.080 H3O+ 0.020 -0.020 0 P0 +0.020 +0.020 HP + H2O H3O+ + PHP Initial 0.080 Change -X Equilibrium 0.080-X H3O+ 0 +X +X P0.020 +X 0.020+X Ka = ([H3O+][P-])/[HP] = 1.3 x 10-5 ([X][0.020+X])/[0.080-X] = 1.3 x 10-5 X = 5.2 x 10-5 pH = -log(5.2 x 10-5) = 4.28 or Henderson-Hasselbach: pH = pKa + log([A-]/[HA]) pH = -log (1.3x10-5) + log(0.020/0.080) = 4.28 27 22. CHEM162-2000 HOURLY EXAM II CHAPTER 14 - ACID AND BASE EQUILIBRIA ACID AND BASE EQUILIBRIA CALCULATIONS What is the pH of an aqueous solution prepared by adding 0.040 moles of NaOH to 1.00 liter of 0.10 M CH3COOH solution? Assume no change in volume. Ka = 1.8 x 10-5 for CH3COOH A. 4.9 B. 4.74 C. 4.57 D. 5.15 E. 4.24 HA + OH- HOH + AHA + OH HOH Initial (1.00 x 0.10)/ 0.040/ (1.00 + 0) (1.00 + 0) = 0.10 M = 0.040 M Change -0.040 -0.040 Equilibrium 0.060 0 Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) pH = (-log(1.8 x 10-5)) + (log([0.040]/[0.060]) = 4.57 + A0 +0.040 +0.040 28 22. This problem is a buffer problem. It belongs in the chapter 15 practice problems, not here. CHEM 162-2000 FINAL EXAM CHAPTER 14 - ACID AND BASE EQUILIBRIA ACID AND BASE EQUILIBRIA CALCULATIONS What is the pH of an aqueous solution prepared by adding 0.10 moles of NaOH to 0.50 moles of HCOOH in 1.00 liter of solution? Ka = 1.8 x 10-4 for HCOOH. A. B. C. D. E. 3.14 3.64 4.14 4.64 5.14 A The Na+ in NaOH is a spectator ion. Get rid of it. HF + OH HOH+F HF Initial Change Equilibrium 0.50 + OH0.10 HOH + F- HF + OH- HOH + F- Initial 0.50 0.10 0 Change -X -X +X Equilibrium 0.50-X 0.10-X +X The K is not equal to Ka, because this is a weak acid reacting with OH-. Note that the right side of the equilibrium equation is the reaction of the conjugate base of HF. Kcb = Kw/Ka. However, we need to reverse that reaction, so we take the inverse of Kcb. Therefore, K = 1/(Kw/Ka) = Ka/Kw. K = (1.8 x 10-4)/ (1 x 10-14) = 1.8 x 1010 Note that if we write the equilibrium expression in the usual manner we will get a quadratic equation, because due to the large K we cannot drop the Xs. To avoid a quadratic equation we will need to bring the reaction to completion, and then use the Henderson-Hasselbalch equation to get the pH. OH- is the limiting reactant. HF Initial Change Equilibrium 0.50 -0.10 0.40 + HOH + F0 +0.10 0.10 OH0.10 -0.10 0 pH = pKa + log([A-]/[HA]) or pH = pKca + log([B]/[BH+]) pH = -log (1.8x10-4) + log(0.10/0.40) = 3.14 3. Chem 162-2005 Exam II 29 Acids and Bases - Chapter 14 Solutions of acids and bases, Ka, Kb, pH, pOH, % dissoc calculations What is the pH of a solution produced by adding 0.20 mol NaOH to 0.60 mol of lactic acid, HC3H5O3. Assume a total volume of 1.0 Liter. Ka = 1.4 x 10-4 for lactic acid. A. B. C. D. 2.95 4.75 4.15 5.35 3.55 HL 0.60 + OH0.20 HOH + L0 E. Initial Change Equilibrium Step 1: Bring to completion to avoid a qudratic equation; Step 2: HendersonHasselbalch equation HL + OH- HOH + LInitial 0.60 0.20 0 Change -0.20 -0.20 +0.20 Equilibrium 0.40 0 +0.20 Henderson-Hasselbalch equation: pH = pKa + log((A-)/(HA)) pH = -log(1.4x10-4) + log((0.20)/(0.40)) = 3.55 CHEM162-2003 5TH WEEK RECITATION CHAPTER 15 - APPLICATIONS OF ACID AND BASE EQUILIBRIA BUFFER CALCULATIONS (INCLUDING HENDERSON-HASSELBALCH EQUATION) 15.23 Calculate the pH of each of the following solutions. d. A mixture containing 0.100 M HC3H5O2 and 0.100 M NaC3H5O2. (HC3H5O2, Ka = 1.3 x 10-5) HP + H2O H3O+ + P30 Initial Change Equilibrium HP 0.100 -X 0.100 - X H3O+ 0 +X +X P0.100 +X 0.100+X (0.100 x X)/(0.100) = 1.3 x 10-5 X = 1.3 x 10-5 pH = -log 1.3 x 10-5 = 4.89 Alternate for (d): Use Henderson-Hasselbach equation pH = pKa + log([A-]/[HA]) pH = -log (1.3x10-5) + log(0.1/0.1) = 4.89 17. CHEM162-2001 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) BUFFER CALCULATIONS How many moles of solid NH4Cl must be added to 1.000 L of a 0.950 M aqueous NH3 solution in order to prepare a buffer solution with pH 9.60? Kb = 1.8 x 10-5 for NH3. Assume no volume change upon addition of the NH4Cl. pH 9.60 pOH = 4.40 -log[OH-] = 4.40 [OH-] = 4.0 x 10-5 NH3 Start Change Equil: + H2O NH4+ Y +X Y+X + OH0 +X 4.0 x 10-5 OH0 +4.0 x 10-5 4.0 x 10-5 0.950 -X 0.950-X NH3 NH4+ Initial 0.950 Y Change -4.0 x 10-5 +4.0 x 10-5 Equilibrium 0.950-(4.0 x 10-5) Y + (4.0 x 10-5) + -5 Kb = ([NH4 ][OH ])/[NH3] = 1.8 x 10 ([Y + 4.0 x 10-5][4.0 x 10-5])/[0.950-(4.0 x 10-5)] = 1.8 x 10-5 31 Y = 0.4275 or Henderson-Hasselbach: pH = -log Ka + log([A-]/[HA]) 9.60 = -log((1x10-14)/(1.8 x 10-5)) + log(0.950/Y) X = 0.4295 A. 0.34 mol B. 0.43 mol C. 0.75 mol D. 1.6 mol E. 2.3 mol 22. CHEM162-2001 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) BUFFER CALCULATIONS Aqueous buffer solutions can be prepared conveniently by partially neutralizing a weak acid with a strong base or a weak base with a strong acid. What is the pH of the buffer solution prepared by adding 0.200 mol of solid NaOH to 1.000 L of 0.500 M HCN solution? (Ka = 6.2 x 10-10 for HCN; assume no change in volume when the NaOH is added.) HCN Start: Change: Equil: 0.5 mol -0.200 0.3 + OH0.200 mol -0.200 0 H2O + CN0 +0.200 0.2 What is K for this reaction? Kw = Ka x Kb (1 x 10-14) = (6.2 x 10-10) x Kb Kb = 0.16 x 10-4 Since the amount of OH- is 0, then clearly some of the CN- will disappear to become OH-. Since we are accustomed to going from left to right in a reaction, lets reverse the reaction. (Since Kb was calculated for the reverse direction, and we are reversing the reverse direction, we dont invert Kb.) H2O + CNHCN + OH Start: 0.2 0.3 0 Change: -X +X +X Equil. 0.2-X 0.3 + X X ((0.3+X) x (X))/(0.2-X) = 0.16 x 10-4 32 X = 1.07 x 10-5 pOH = -log [OH-] = 4.97 pH = 9.03 A. 9.03* B. C. D. E. 9.31 4.69 5.90 8.93 In a previous version I had B as the answer, but that seems to be an error. 24. CHEM162-2001 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) BUFFER CALCULATIONS 25.00 mL of an aqueous nitric acid solution was titrated with 0.250 M aqueous NaOH solution and 32.06 mL of the NaOH solution was required to reach the equivalence point. What was the molarity of the nitric acid solution? This problem is a gift. H3O+ + OH- 0.025L 0.03206 L ?M 0.250 M 2H2O moles of acid = moles of base. Therefore, M1V1 = M2V2 M1 x 0.025 = 0.250 x 0.03206 M1 = 0.3206 A. 0.250 M B. 0.321 M C. 0.403 M D. 0.195 M E. 0.297 M CHEM162-2002 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) BUFFER CALCULATIONS 4. Buffer A consists of 1.00 liter of 0.50 M HNO2 and 0.50 M NaNO2 Buffer B consists of 1.00 liter of 0.050 M HNO2 and 0.050 M NaNO2 33 Which statement is correct? The more buffer components in a solution, the more it can neutralize added acid or base without meaningfully changing the pH, i.e., the higher the buffer capacity. Therefore, the 0.50 M buffer solution has greater buffer capacity. Calculation of pH: HNO2 + Initial: Change: Equil: 0.50 -X 0.50-X H2O H3O+ + 0 +X +X NO20.50 +X +X (X x (0.50 + X))/(0.50 - X) = K Simplify, assuming that K is very small compared to the molarity listed. (X x 0.50)/0.50 = K K = X, i.e., the H3O+ concentration = K, regardless of the value of K. HNO2 + Initial: Change: Equil: 0.05 -Y 0.05-Y H2O H3O+ + 0 +Y +Y NO20.05 +Y +Y (Y x (0.05 + Y))/(0.05 - Y) = K Simplify, assuming that K is very small compared to the molarity listed. (Y x 0.05)/0.05 = K K=Y that is, the H3O+ concentration = the equilibrium constant, regardless of the value of K. Hence, the H3O+ concentrations in both cases are equal to K. Since K is the same in both cases, then the H3O+ concentration in both cases is the same and, correspondingly, the pHs in both cases are the same. A. B. C. Buffer A has a higher capacity than buffer B and a higher pH than buffer B Buffer A has a lower capacity than buffer B and a lower pH than buffer B Buffer A has the higher capacity than buffer B and a lower pH than buffer B Buffer A has a higher capacity than buffer B and the same pH as buffer B E. Buffer A has the same capacity as buffer B and the same pH as buffer B Note: If the equilibrium expressions were calculated to several significant figures, it would show that buffer A has a very slightly lower pH than buffer B; however, answer D, not C is considered as the appropriate answer for the students in this course. D. CHEM162-2002 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) 34 BUFFER CALCULATIONS 8. A buffer consists of 0.50 M NH3 and 0.50 M NH4Cl. If 0.010 mol NaOH is dissolved in 100 mL of this buffer, what is the pH of the resultant solution? A. B. C. D. 11.25 5.26 4.74 10.74 9.43 E. Kb = 1.8x10-5 for NH3 0.010 mol NaOH in 0.1 L = 0.1 M NH4+ + NH4+ Initial Change Equilibrium 0.50 -0.10 0.40 OH NH3 OH0.10 -0.10 0 + H2O NH3 0.50 +0.10 0.60 H2O Henderson-Hasselbach pH = -log(Ka) + log([A-]/[HA]) = -log(1x10-14/1.8 x 10-5) + log (0.60/0.40) = 9.43 CHEM162-2002 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) BUFFER CALCULATIONS 19. What amount of sodium acetate must be dissolved in 400 mL of 0.30 M CH3COOH in order to produce a buffer of pH 5.00? Ka = 1.8 x 10-5 for CH3COOH (acetic acid) CH3COOH pH = 5.00 = -log [H+] [H+] = 1 x 10-5 CH3COOH Initial: Change: 0.30 -X + H2O CH3COOY +X + H3O+ 0 1.0 x 10-5 35 + H2O CH3COO+ H3O+ Equil: 0.30 - X Y+X 1.0 x 10-5 ((Y+X) x (1 x 10-5))/(0.30-X) = 1.8 x 10-5 Y = 0.54 M 0.54 x 0.400 L = 0.216 mol A. B. C. D. 0.52 mol 0.15 mol 0.68 mol 0.34 mol 0.22 mol E. CHEM162-2003 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) BUFFER CALCULATIONS 8. What mass of sodium hypochlorite, NaOCl (molar mass 74.5 g/mol), must be added to 250 mL of 0.60 M HClO (hypochlorous acid, Ka = 3.5 x 10-8), in order to produce a solution having a pH of 7.25? A. B. C. 3.1 g 9.2 g 4.5 g 7.0 g E. 1.3 g moles = M x V = 0.60 mol/L x 0.250 L = 0.15 mol pH = 7.25 [H+] = 10-7.25 = 5.62 x 10-8 HClO + H2O H3O+ + ClOHClO Initial Change Equilibrium 0.60 -X 0.60 - X H2O H3O+ 0 +X 5.62 x 10-8 ClOY Y+X Y+X D. ([H3O+][ClO-])/[HClO] = 3.5 x 10-8 ([5.62 x 10-8][Y + X])/[0.60 - X] = 3.5 x 10-8 ([5.62 x 10-8][Y])/[0.60] = 3.5 x 10-8 Y = 0.3737 M 0.3737 M x 0.250 L = 0.09343 mol g = moles x MW = 0.09343 x 74.5 = 6.96 g or do Henderson-Hasselbalch: 36 pH = -log Ka + log ([A-]/[HA]) 7.25 = -log (3.5 x 10-8) + log (Y/0.60) Y = 0.3734 M 0.3737 M x 0.250 L = 0.09343 mol g = moles x MW = 0.09336 x 74.5 = 6.96 g CHEM162-2003 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) 14. Calculate the pH of a solution formed by adding 0.20 moles of NaOH to 1.00 liter of 0.50 M acetic acid. Ka = 1.8 x 10-5 for acetic acid. A. B. C. 4.24 3.74 4.92 4.57 2.63 D. E. HA + OH- H2O + ABring to completion. HA Initial Change Equilibrium 0.50 -0.20 0.30 OH0.20 -0.20 0 H3O+ 0 +X +X H2O A0 +0.20 0.20 A0.20 +X 0.20+X Determine equilibrium values HA Initial Change Equilibrium 0.30 -X 0.30-X H2O 0 Henderson-Hasselbalch: pH = -logKa + log(A-/HA) pH = -log(1.8x10-5) + log(0.20/0.30) = 4.57 37 49. CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB (BUFFERS & TITRATION) BUFFER CALCULATIONS What is the pH of the buffer solution prepared by dissolving 0.300 mol of NaClO and 0.200 mol of HClO in enough water to make 100.00 mL of solution? Ka = 2.9 x 10-8 for HClO. 0.0200 M HClO and 0.0300 M NaClO HClO + H2O ClO+ H3O+ Initial 2.00 3.00 0 Change -X +X +X Equil 2.00-X 3.00 +X +X ((3.00 + X) x (X))/(2.00 - X) = 2.9 x 10-8 Simplifying: 3.00X/2.00 = 2.9 x 10-8 X = 1.93 x 10-8 pH = - log 1.93 x 10-8 pH = 7.71 A. 6.34 B. 7.54 C. 7.71 D. 8.12 E. 9.43 19. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB (BUFFERS & TITRATION) BUFFER CALCULATIONS The Kb of NH3 is 1.80 x 10-5 . A buffer solution is 0.25 M in NH3 and 0.40 M in NH4Cl. The pH of the buffer is NH3 Initial: Change: Equil: 0.25 -X 0.25-X + H2O NH4+ + OH0 +X +X 0.40 +X 0.40 + X ((0.40 + X)(X))/(0.25-X) = 1.80 x 10-5 Simplify: 0.4X/0.25 = 1.8 x 10-5 X = 1.1 x 10-5 = [OH-] pOH = +4.95 pH = 9.05 38 A. B. C. D. E. 9.05 4.95 4.74 9.26 7.00 162 EXAM III Vlg Chem 162-2003 Final exam + answers Chapter 15A-Applications of Aqueous Equilibria (of Acids and Bases) Buffer calculations (including Henderson-Hasselbalch equation) 10. An acetic acid/acetate buffer solution of pH = 4.80 is prepared. The [CH3COOH] = 0.40 M. What is the [CH3COO-]? Ka(CH3COOH) = 1.8 x l0-5 A. B. C. D. E. 0.45 M 0.40 M 4.47 M 1.6x10-3 M 1.6xl0-5 M CH3COOH + H2O CH3COO- + H3O+ pH = 4.80 [H+] = 10-4.80 = 1.585 x 10-5 Initial Change Equilibrium CH3COOH 0.40 -X 0.40 - X H2O CH3COOY +X Y+X H3O+ 1.585 x 10-5 1.585 x 10-5 ([CH3COO-][H3O+])/[CH3COOH] = Ka ([Y + X][ 1.585 x 10-5])/[0.40 - X] = 1.8 x 10-5 ([Y][ 1.585 x 10-5])/[0.40] = 1.8 x 10-5 Y = 0.454 or Henderson-Hasselbalch pH = pKa + log([A-]/[HA]) 4.80 = -log(1.8 x 10-5) + log(Y/0.4) 39 X = 0.454 Chem 162-2003 Final exam + answers Chapter 15A-Applications of Aqueous Equilibria (of Acids and Bases) Buffer calculations (including Henderson-Hasselbalch equation) 13. Calculate the pH after 10.0 ml of 0.400 M NaOH is added to 20.0 ml of 0.50 M CH3COOH. Ka (CH3COOH) = 1.8 x 10-5 A.7.00 B.4.57 C.0.70 D.2.52 E.13.70 CH3COOH + OH- H2O + CH3COO(0.010 L x 0.400 mol/L) x 1/(0.010 + 0.020) = 0.1333M OH(0.020 L x 0.50 mol/L) x 1/(0.010 + 0.020) = 0.3333 M CH3COOH Since reaction with hydroxide produces a large equilibrium constant, then the conventional approach of a change of -X on the left side and +X on the right side will produce a quadratic equation (the -X and +X cant be dropped). Therefore, bring the reaction to completion, and then come back a little, if necessary. Initial Change Equilibrium HA 0.3333 -0.1333 0.2000 OH0.1333 M -0.1333 0 HOH A0 +0.1333 +0.1333 Henderson-Hasselbalch pH = pKa + log([A-]/[HA]) pH = -log(1.8 x 10-5) + log(0.1333/0.2000) pH = 4.57 Chem 162-2004 Exam II Applications of aqueous equilibria - Chapter 15A Buffers, common ions, Henderson-Hasselbalch equation A buffer contains 0.50 M HC2H3O2 (acetic acid, Ka = 1.8 x 10-5) and 0.50 M NaC2H3O2 (sodium acetate). The pH of the buffer is 4.74. The buffer is diluted by taking 1.00 liter of this buffer and adding 9.00 liters of water, making 10.0 liters total. What is the pH of the diluted buffer? A. 5.74 14. 40 B. C. D. E. 4.74 3.74 2.74 0.474 HA + H2O A- + H3O+ pH = 4.74 [H3O+] = 10-4.74 = 1.820 x 10-5 HA H2O AH3O+ Initial 0.50 0.50 1.820 x 10-5 Change 10 fold dilution 10 fold dilution 10 X dilution Equilibrium 0.050 0.050 X Diluting the concentration of the HA and A- ten fold should have the effect of making their concentrations ten times lower. However, it doesnt seem that the [H3O+] works that way, because if it did then diluting it 1000 fold would make its concentration 1.820 x 10-8, which is a basic solution, which makes no sense. So just dilute the HA and Aand use Henderson-Hasselbalch to determine the pH. Henderson-Hasselbalch: pH = pKa + log([A-]/[HA]) pH = -log(1.8 x 10-5) + log([0.050]/[0.050]) pH = 4.74 The solution to the problem ends here with the pH calculated to be 4.74. However, the solution to this problem raises an interesting question. The solution suggests that diluting the buffer has no effect on the pH. But this is an impossibility because the more the solution is diluted the more the pH should approach 7. Probably the calculations would have to be done without using H-H to get an accurate pH. Providing the pH is unnecessary information because, once knowing the initial concentrations and the Ka, the pH can be calculated. HA H3O+ H2O A Initial 0.50 0.50 0 Change -X +X +X Equilibrium 0.50-X 0.50+X +X To measure the exact pH, without simplifying the equation or using HH: ([A-][H3O+])/[HA] = 1.8 x 10-5 X = 1.79987 x 10-5 pH = -log(1.79987 x 10-5) = 4.74476 Now repeat this exact pH calculation, but with the diluted solution to determine if the prediction of the pH increasing really occurs: 41 HA H2O A- H3O+ Initial 0.05 0.05 0 Change -X +X +X Equilibrium 0.05-X 0.05+X +X ([A-][H3O+])/[HA] = 1.8 x 10-5 X = 1.79871 x 10-5 pH = -log(1.79987 x 10-5) = 4.74504, which shows that the pH does increase very slightly upon dilution, if an exact calculation is performed. 23. Chem 162-2004 Final Exam + Answers Chapter 15 - Applications of aqeous equilibria (of acids and bases) Buffers (including Henderson-Hasselbalch equation) A buffered solution is formed by mixing 1.0L of 0.10 M NaOH and 1.0L of 0.25 M C6H5COOH (benzoic acid). The principal species in this solution responsible for the buffering effect are: A. B. C6H5COOH and H+ C6H5COO- and H+ C6H5COOH and C6H5COOC6H5COOH and OHC6H5COO- and OH- C. D. E. 1.0 L of 0.25 M HB = 0.25 mol OH-/L 1.0 L of 0.10 M NaOH = 0.10 mol OH-/L HB + OH- H2O + BHB OHInitial Change Equilib. 0.25 mol/1.0L 0 -0.10 mol +0.10 mol 0.15 mol/(1.0 + 1.0)L = +0.10 mol/(1.0+1.0)L = 0.075M 0.05 M Hence, the principal species in this solution are benzoic acid and sodium benzoate. However, it wasnt necessary to work this problem out quantitatively. Once we know that a buffer is formed from benzoic acid and NaOH, and that a buffer can be a weak acid and the salt of a weak acid, the only possible answer is benzoic acid and sodium benzoate. 0.10 mol/1.0 L -0.10 mol 0 H2O B- 4. Chem 162-2004 Exam II Applications of aqueous equilibria - Chapter 15A Buffers, common ions, Henderson-Hasselbalch equation 42 A 1.00 liter solution contains 1.00 mol of weak acid HX. After 0.60 mol of OH- is added to this solution, the pH of the solution is 4.00. Calculate Ka for acid HX. A. B. C. D. E. 1.5 x 10-4 8.0 x 10-3 6.7 x 10-5 3.0 x 10-5 3.0 x 10-4 The first ICE table is to bring the reaction to completion. Then, if we set up another ICE table, the equilibrium [H+] would be used. HX H2O X OH Initial 1.00 0.60 0 Change -0.60 -0.60 +0.60 Equilibrium 0.40 0 +0.60 Henderson-Hasselbalch: pH = pKa + log([A-]/[HA]) pH = -log(Ka) + log([A-]/[HA]) 4.00 = -log(Ka) + log([0.60]/[0.40]) Ka = 1.5 x 10-4 9. Chem 162-2004 Exam II Applications of aqueous equilibria - Chapter 15A Buffers, common ions, Henderson-Hasselbalch equation A buffer contains 0.50 M HC2H3O2 (acetic acid, Ka = 1.8 x 10-5) 0.50 and M NaC2H3O2 (sodium acetate). The pH of the buffer is 4.74. What is the pH after 0.10 mol of HCl is added to 1.00 liter of this buffer? A. B. C. D. 4.92 4.74 4.38 5.26 4.57 E. pH = 4.74 [H3O]+ = 10-4.74 = 1.820 x 10-5 43 HA Initial Change Equilibrium 0.50 -1.820 x 10-5 0.50-1.820 x 10-5 H2O H3O+ 0 +1.820 x 10-5 1.820 x 10-5 A0.50 +1.820 x 10-5 0.50+1.820 x 10-5 Simplify: [HA] = 0.50; [A-] = 0.50 2nd step is addition of HCl to the buffer. The equilibrium equation must be A- + H3O+ HA + H2O because HA wont react with HCl. Reactions of strong acids go virtually to completion. AHA H2O H3O+ Initial 0.50 0.10 + (1.820 x 10-5) 0.50 Change -0.10 -0.10 +0.10 Equilibrium 0.40 0 0.60 Henderson-Hasselbalch: pH = pKa + log([A-]/[HA]) pH = -log(1.8 x 10-5) + log([0.40]/[0.60]) = 4.57 17. Chem 162-2004 Exam II Applications of aqueous equilibria - Chapter 15A Buffers, common ions, Henderson-Hasselbalch equation Pyridine, C5H5N, is a weak base with Kb = 1.7 x 10-9. What amount of pyridinium chloride, (C5H5NH)+Cl- must be added to 1.00 L of 0.20 M pyridine in order to produce a buffer having pH 5.00? A. B. C. D. 0.12 mol 0.24 mol 0.54 mol 0.66 mol 0.34 mol E. RNH2 + H2O RNH3+ + OHpH = 5.0 pOH = 14.0 - 5.0 = 9.0 [OH-] = 10-9.0 = 1 x 10-9 44 RNH2 H2O RNH3+ Y +1 x 10-9 Y+(1 x 10-9) OH0 +1 x 10-9 1 x 10-9 Initial 0.20 Change -1 x 10-9 Equilibrium 0.20-(1 x 10-9) Simplifying: [RNH2] = 0.20M [RNH3+] = Y Henderson-Hasselbalch: pH = pKa + log([A-]/[HA]) 5 = -log((1 x 10-14)/(1.7 x 10-9)) + log([0.20]/[Y]) Y = 0.34 mol 19. Chem 162-2004 Final Exam + Answers Chapter 15A - Applications of aqeous equilibria (of acids and bases) Buffers (including Henderson-Hasselbalch equation) Calculate the solubility of Fe(OH)3 in a solution buffered at pH=5.0? Ksp of Fe(OH)3 = 4 x 10-38 A. 1 x 10-4 mol/L 4 x 10-11 mol/L 5 x 10-6 mol/L 8 x 10-5 mol/L 6 x 10-9 mol/L B. C. D. E. Fe(OH)3 Fe3+ + 3OH[H+] = 10-5.0 = 1 x 10-5 [OH-] = (1 x 10-14)/(1 x 10-5) = 1 x 10-9 Fe2+ 3OH Fe(OH)2 Initial Y 0 0 Change -X +X +3X Equilibrium Y-X +X 1 x 10-9 The 3X in the 3OH- column has no meaning, because since the solution is buffered we assume that the addition of OH- does not change the hydroxide ion concentration at equilibrium. [Fe2+][OH-]3 = 4 x 10-38 [X][1 x 10-9]3 = 4 x 10-38 X = 4. x 10-11 45 45. Chem 162-2004 Final Exam + Answers Chapter 15A - Applications of aqeous equilibria (of acids and bases) Buffers (including Henderson-Hasselbalch equation) Calculate the pH after 0.010 mol pure HCl is added to 1 L of a buffered solution of 0.50 M NH3 and 1.50 M NH4Cl (Kb = 1.79 x 10-5) A. B. C. D. E. 6.65 10.32 8.76 9.23 5.34 NH3 + H3O+ NH4+ + H2O NH3 0.50 -0.010 0.49 H3O+ 0.010 -0.010 0 NH4+ 1.50 +0.010 1.510 H2O Initial Change Equilibrium Bring the reaction to completion and then use the Henderson-Hasselbalch equation. pH = pKa + log([A-]/[HA]) Kb = 1.79 x 10-5 Ka = (1 x 10-14)/(1.79 x 10-5) = 5.587 x 10-10 pH = (-log(5.587 x 10-10)) + (log (0.49/1.510)) = 8.76 46 23. Chem 162-2004 Final Exam + Answers Chapter 15A - Applications of aqeous equilibria (of acids and bases) Buffers (including Henderson-Hasselbalch equation) A buffered solution is formed by mixing 1.0L of 0.10 M NaOH and 1.0L of 0.25 M C6H5COOH (benzoic acid). The principal species in this solution responsible for the buffering effect are: A. B. C6H5COOH and H+ C6H5COO- and H+ C6H5COOH and C6H5COOC6H5COOH and OHC6H5COO- and OH- C. D. E. 1.0 L of 0.25 M HB = 0.25 mol OH-/L 1.0 L of 0.10 M NaOH = 0.10 mol OH-/L HB + OH- H2O + BHB OHInitial Change Equilib. 0.25 mol/1.0L 0 -0.10 mol +0.10 mol 0.15 mol/(1.0 + 1.0)L = +0.10 mol/(1.0+1.0)L = 0.075M 0.05 M Hence, the principal species in this solution are benzoic acid and sodium benzoate. However, it wasnt necessary to work this problem out quantitatively. Once we know that a buffer is formed from benzoic acid and NaOH, and that a buffer can be a weak acid and the salt of a weak acid, the only possible answer is benzoic acid and sodium benzoate. 0.10 mol/1.0 L -0.10 mol 0 H2O B- 47 27. Chem 162-2004 Final Exam + Answers Chapter 15A - Applications of aqeous equilibria (of acids and bases) Buffers (including Henderson-Hasselbalch equation) In which titration would you use an indicator having Ka = 1 x 10-5 ? X. Y. Z. A. HCN titrated with NaOH CH3COOH with NaOH NH3 titrated with HCl X, Y and Z Z only X and Y only Y only Y and Z only B. C. D. E. According to the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]) An indicator is a buffer which changes color approximately when [A-] = [HA]. At this point pH = pKa pH = -log(1 x 10-5) pH = 5 This indicator changes color at pH 5 1, i.e., between pH 4 and 6. The equivalence point of a strong acid and a weak base occurs at this pH. 19. Chem 162-2005 Exam II Applications of Aqueous Equlibria - Chapter 15A Buffers, common ions, Henderson-Hasselbalch equation Calculations A buffer consists of 0.50 mol NH3 and 0.50 mol NH4Cl dissolved in 1.0 L of water. If 0.10 mol of NaOH is added to the buffer, what is the pH of the resultant solution? Kb = 1.8 x 10-5 for NH3 A. B. C. D. E. 9.43 9.08 4.57 4.92 10.26 Principal species present: NH3, NH4+, OH Reaction: NH4+ + OH- NH3 + H2O 48 NH4+ + OH- NH3 + H2O Initial 0.50 0.10 0.50 Change Equilibrium Plan: Bring the reaction to completion since it contains a strong base; then use the Henderson-Hasselbalch equation to find the pH. NH4+ + OH NH3 0.50 +0.10 0.60 + H2O Initial 0.50 0.10 Change -0.10 -0.10 Equilibrium 0.40 0 pH = pKa + log(A /HA) Kb = 1.8 x 10-5 Ka = (1 x 10-14)/(1.8 x 10-5) = 5.56 x 10-10 pH = -log(5.56 x 10-10) + log(0.60/0.40) pH = 9.43 49 4. Chem 162-2005 Final Exam + Answers Chapter 15 - Applications of Aqueous Equilibria (of Acids and Bases) Buffers (including Henderson-Hasselbalch equation) Calculations What amount of NaF should be added to 1.0 L of 1.90 M HF solution (Ka = 7.2 x 10-4) to prepare a buffer solution of pH = 3.35? (Assume no volume change) A. B. C. D. E. 3.1 mol NaF 2.3 mol NaF 1.6 mol NaF 1.0 mol NaF 4.9 mol NaF The principal species for the equilibrium reaction are HF and F-. Na+ is a spectator ion. Write equilibrium equation with the uncharged substance, because Ka is given. HF + H2O H3O+ + FpH = 3.35 [H+] = 10-3.35 = 4.47 x 10-4 HF Initial Change Equilibrium 1.90 + H2O H3O+ 0 4.47 x 10-4 H3O+ + + FY FY +4.47 x 10-4 Y + (4.47x10-4) HF + H2O Initial 1.90 0 -4 Change -4.47 x 10 +4.47 x 10-4 -4 Equilibrium 1.90-(4.47x10 ) 4.47 x 10-4 ([H3O+][F-])/[HF] = 7.2 x 10-4 ([4.47 x 10-4][(Y + (4.47 x 10-4))])/[1.90 - (4.47 x 10-4)] = 7.2 x 10-4 Simplify: ([4.47 x 10-4][Y])/[1.90] = 7.2 x 10-4 Y = 3.06 M = NaF concentration or solve this with the Henderson-Hasselbalch equation : pH = pKa + log([A-]/[HA]) 3.35 = -log(7.2 x 10-4) + log([X]/[1.90]) X = 3.06M = [NaF] TITRATIONS and INDICATORS 50 38 Chem 162-2008 Final Exam + Answers Chapter 15B- Acids, Bases and Acid-Base Equilibria Titrations and indicators What is the pH at the stoichiometric point when 0.10 M solution of a weak base (Kb = 1.00x10 is titrated with 0.10 M HCl solution at 298 K? 6) A. B. C. D. E. 7.00 4.65 1.55 8.15 9.35 B RNH2 + H3O+ RNH3+ + H2O Reaction with a strong acid brings the reaction to completion. RNH3+ + RNH2 + H3O+ H2O Initial Change Equilibrium 0.10M -0.10 0 0.10M -0.10 0 0 +0.10 +0.10 0 +0.10 +0.10 Since we want the [H3O+], bring the reaction back to equilibrium. RNH3+ + RNH2 + H3O+ H2O 0.10M 0 0 Initial -X +X +X Change 0.10-X +X +X Equilibrium H2][H3O+])/[RNH3+] = Kca Kw = Kca x Kb 1x10-14 = Kca x 1.00x10-6 Kca = 1x10-8 ([X][X])/[0.10-X] = 1x10-8 Small K rule: ([X][X])/[0.10] = 1x10-8 X = 3.16x10-5 = [H+] pH = -log(3.16x10-5) = 4.50 ([RN 51 16 Chem 162-2008 Final Exam + Answers Chapter 15B- Acids, Bases and Acid-Base Equilibria Titrations and indicators Which indicator would be best to use for a titration of 0.25M NaOH with 0.25M HF (Ka = 6.610-4)? A A. B. C. D. E. m-Nitrophenol (pKa = 7.9) Bromophenol blue (pKa = 2.5) Chlorphenol red (pKa = 5.7) Curcumin (pKa = 10.6) Methyl violet (pKa = 1.8) HF + OH- H2O + FReactions with OH- go to completion. HF + OH F+ H2O Initial 0.125 0.125 0 0 Change -0.125 -0.125 +0.125 +0.125 Equilibrium 0 0 +0.125 +0.125 Bring the reaction back to equilibrium to determine the [OH ], which is needed to determine the pH at the equivalence point. Since students find it more convenient to go from left to right than from right to left, reverse the equation. F- + H2O HF + OHF+ H2O HF + OH0 +X +X Initial 0.125 0 Change -X +X Equilibrium 0.125-X +X [OH ]/[F ] = Kcb Kcb = Kw/Ka = (1 x 10-14)/6.6 x 10-4) = 1.515 x 10-11 (X x X)/(0.125-X) = 1.515 x 10-11 Use the small K rule to avoid a quadratic equation. (X x X)/(0.125) = 1.515 x 10-11 X = 1.38 x 10-6 = [OH-] pOH = -log [OH-] = -log(1.38 x 10-6) = 5.861 pH = 14 5.861 = 8.14 [HF] This problem is treated in the same manner as determining the Ka of the acid, by using the HendersonHasselbalch equation, and realizing that with a perfect buffer ([A-] = [HA]) the pH = pKa. Hence, since the pH at the end point of the titration is 8.14, then the pH of the indicator (which also is a buffer) will be approximately 8.14, and the pKa of that indicator equals approximately 8.14. The closest indicator to 8.14 is meta-nitrophenol. Concern: There should be an option for an indicator of pKa closer to 8.1 than thymol 52 Chem 162-2008 Exam II + Answers Chapter 15B Applic. of Acid & Base Equilibria (Buffers & Titrations) Titrations and Indicators Titration curve 23. Acid HX is a strong acid and acid HY is weak acid. Which of the following statements are true concerning HX and HY? X. A 0.10 M solution of HX has a lower pH than a 0.10 M solution of HY. Y. A 0.10 M solution of HX conducts electricity better than a 0.10 M solution of HY. Z. When titrated with NaOH, the pH of the equivalence point will be greater for acid HX than for acid HY. (a) X only (b) Y only (c) X and Y only (d) Y and Z only (e) X, Y, and Z X. True. A 0.10 M solution of strong acid HX has a pH of approximately 1, whereas a 0.10 M solution of weak acid HY has a pH of approximately 3. Y. True. HX is a strong acid, so completely dissociates, providing a high concentration of ions to complete the circuit. HX is a weak acid, so only partially dissociates, providing a low concentration of ions to complete the circuit, resulting in a weak current. Z. False. For strong acid HX the inflection point will be approximately 7; for weak acid HY the inflection point will be approximately 9. Chem 162-2008 Exam II + Answers Chapter 15B Applic. of Acid & Base Equilibria (Buffers & Titrations) Titrations and Indicators Given the pH at the half-way point in a titration, what is the Ka? 24. 53 In the above graph, NaOH is added to a solution of a weak acid. Based on the graph, what is the approximate Ka value of the acid? (a) 1 x 10-4 (b) 1 x 10-6 (c) 1 x 10-8 (e) 1 x 10-10 (e) 1 x 10-12 Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) At the half-way point [A-] = [HA]. Therefore, log([A-]/[HA]) = 0. Therefore, pH = pKa. The equivalence point is at 50 mL of NaOH. Therefore, the half-equivalence point is at 25 mL NaOH. The pH corresponding to the addition of 25 mL of NaOH = 6.0. Therefore, pKa = 6.0. Ka = 1 x 10-6 54 31 Chem 162-2007 Final exam + answers Chapter 15B Applications of Acid & Base Equilibria Titrations and Indicators The graph below represents the titration of 50 mL of a 0.10 M acid with 0.10 M NaOH 14 12 10 8 pH 6 4 2 0 0 10 20 30 40 50 60 V base added (mL) Which of the following statements are consistent with this titration curve? X. The equivalence point occurs when approximately 25 mL of NaOH are added Y. The curve represents the titration of a weak acid. Z. The pKa of the acid is approximately 9.4. A. X,Y, and Z B. X,Y only C. Y, Z only D. Y only E. Z only X. False. The equivalence point occurs when exactly 50 mL of NaOH are added. Y. True. By the fact that the equivalence point is approximately 9 is consistent with a strong base titrating a weak acid. In addition, by the fact that the initial point is approximately 3 is consistent with the solution in the flask being a weak acid. Z. False. Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) [A-] = [HA] at the half-way point, which is 25 mL of titrant. At the half-way point log([A-]/[HA]) = 0. Therefore, pH = pKa. Since at 25 mL of titrant (the half-way point) the pH = 6.0, then pKa = 6.0. 55 44 Chem 162-2007 Final exam + answers Chapter 15B Applications of Acid & Base Equilibria Titrations and indicators Calculate the pH after 10.0 mL of 0.400 M NaOH is added to 20.0 mL of 0.50 M CH3COOH. (Ka CH3COOH = 1.8 105). A. 2.52 B. 0.70 C. 13.70 D. 4.57 E. 7.00 (0.400M NaOH x 0.0100L)/(0.0100L + 0.0200L) = 0.1333 M NaOH (0.500M HA x 0.020L)/(0.020L + 0.010L) = 0.3333M HA HA + OH- H2O + AReactions of strong bases go to completion. HA + H2O + A OH Initial 0.3333 0.1333 0 Change -0.1333 -0.1333 +0.1333 Equilibrium 0.2000 0 +0.1333 Henderson-Hasselbalch equation: pH = pKa + log([A ]/[HA]) pH = -log(1.8 x 10-5) + log((0.1333)/(0.2000)) = 4.57 CHEM 162-2007 EXAM II + ANSWERS CHAPTER 15B APPLICATIONS OF ACID & BASE EQUILIBRIA (BUFFERS & TITRATIONS) TITRATIONS AND INDICATORS 24. A titration curve is shown below. In this curve, the titrant refers to the substance in the buret, which is added to another substance in the titrating flask. 56 This graph could represent: (a) NaOH in the buret added to HCl in the titrating flask. (b) HCl in the buret added to NaOH in the titrating flask. (c) HCl in the buret added to NH3 in the titrating flask. (d) NaOH in the buret added to CH3COOH in the titrating flask. (e) CH3COOH in the buret added to NaOH in the titrating flask. The graph shows that the pH of the material in the titrating flask is 11, prior to addition of any titrant. This is typical of 0.1M weak base. The pH of excess titrant is 1, which is typical of 0.1 M strong acid. Hence, this is indicative of a weak base in the flask being titrated by a strong acid. Also, the equivalence point of approximately 5 is typical of a weak base being titrated by a strong acid. (a) False. The graph represents a strong acid in the burette being added to a weak base in the titrating flask. (b) False. The graph represents a strong acid in the burette being added to a weak base in the titrating flask. (c) True. The graph represents a strong acid in the burette being added to a weak base in the titrating flask. (d) False. The graph represents a strong acid in the burette being added to a weak base in the titrating flask. (e) False. The graph represents a strong acid in the burette being added to a weak base in the titrating flask. Note: I have recently shown my students two better ways to set up and solve titration problems. This involves (A) using molarity rather than moles in the ICE tables. (B) A dilution factor is used to directly convert molarity to the diluted molarity. These 57 approaches are not shown in most of the problems below, as they were developed only recently. 26 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 15 1ST HALF APPLIC. OF ACID & BASE EQUILIBRIA (BUFFERS & TITRATIONS) TITRATIONS AND INDICATORS Bromothymol blue, an indicator with Ka = 1.010-7, is yellow in its HIn form and blue in its In- form. A few drops of this indicator was added to a solution and then titrated with base. At what pH will the solution turn blue? A. 8 B. 6 C. 4 D. 3 E. 1 Ka = 1.0 x 10-7 pKa = 7 An indicator is a buffer in which the [HIn] = the [In-]. pH = pKa + log([A-]/[HA]) pH = pKa + log(1) pH = pKa The pH of this buffer = 7. Acid = Yellow Base = blue It is yellow below pH 6. At pH6 it begins to change from yellow to green. At high pH 7 and above it is blue. 58 11 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 15 1ST HALF APPLIC. OF ACID & BASE EQUILIBRIA (BUFFERS & TITRATIONS) TITRATIONS AND INDICATORS Consider 50.0 mL of 0.20M solution of a weak base B (Kb = 1.010-5). Addition of which of the following solutions will generate a solution whose pOH = pKb of the weak base? A. B. C. D. 50.0 mL of 0.20M HCl(aq) 50.0 mL of 0.10M NaOH(aq) 60.0 mL of 0.10M HCl 60.0 mL of 0.20M NaOH(aq) 50.0 mL of 0.10M HCl(aq) E. pKb of 1.0 x 10-5M = -log(1.0 x 10-5) = +5 pOH = pKb = +5 pH = 14 5 = 9 A. 50.0 mL of 0.20M HCl(aq) 50.0 mL of 0.20M HCl will completely neutralize 50.0 mL of 0.20M solution of a weak base. The equivalence point of a strong acid and a weak base is approximately pH 5, not pH9. B. 50.0mLof 0.10MNaOH(aq) 0.10 x (50/100) = 0.05M NaOH pOH of 0.05M NaOH = -log(0.05) = 1.30 pH = 14.0 1.30 = 12.70 Since the weak base doesnt react with NaOH, the presence of the weak base will only result in raising the pH above 12.70 C. 60.0mLof0.10MHCl RNH2 Initial + + 0.20x(50/110) = 0.0909M Change -0.0545M Equilibrium 0.0364 Henderson-Hasselbalch equation pH = pKa + log([A-]/[HA]) Kca = Kw/Kb = (1 x 10-14)/(1 x 10-5) = 1 x 10-9 pKca = -log(1x10-9) = +9 pH = (9) + (log([0.0364]/[0.0545])) = 8.82 H3O 0.10x(60/110) = 0.0545M -0.0545M 0 RNH3+ 0 +0.0545M 0.0545M + H2O D. 60.0mLof0.20MNaOH(aq) This will make the pH even higher than 12.70, which is not acceptable. E. 50.0mLof0.10MHCl(aq) RNH2 Initial + H3O+ 0.10x(50/100) = 0.05M -0.05M 0 RNH3+ 0 +0.05M 0.05M + H2O 0.20x(50/100) = 0.10M Change -0.05M Equilibrium 0.05 Henderson-Hasselbalch equation pH = pKa + log([A-]/[HA]) Kca = Kw/Kb = (1 x 10-14)/(1 x 10-5) = 1 x 10-9 pKca = -log(1x10-9) = +9 pH = (9) + (log([0.05]/[0.05])) = 9.0 59 3 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 15 1ST HALF APPLIC. OF ACID & BASE EQUILIBRIA (BUFFERS & TITRATIONS) TITRATIONS AND INDICATORS In a titration, 25.00 mL of barium hydroxide requires 12.35 mL of 0.2315M hydrochloric acid to reach the equivalence point. What is the concentration of the barium hydroxide? A. 0.05718M B. 0.1144M C. 0.2287M D. 0.2342M E. 0.4684M Ba(OH)2 + 2HCl 2H2O + BaCl2 VHCl molHCl molBa(OH)2 MBa(OH)2 0.01235L x 0.2315M x (1 mol Ba(OH)2/2 mol HCl) x (1/0.025L) = 0.0572M 60 10 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA TITRATIONS AND INDICATORS At the equivalence point of a titration of a weak acid with a strong base A. B. pH is less than 7.00 pH is equal to 7.00 pH is greater than 7.00 More data is needed to answer this question. pOH is greater than 7.00 C. D. E. HA+OH HOH + AAt the equivalence point there is only A- present. A- is a weak base. Therefore, the pH must be greater than 7.00. Thereactionofaweakacidwithastrongbasegenerallyhasanequivalence pointofapproximatelypH9.Thiscanbedemonstratedquantitatively.For example,if100mLof0.1Maceticacid(Ka=1.8x105)isreactedwith100mL of0.1MNaOH: Equilibrium Equilibrium 0 0.05-X 0 +X 0.05 +X Kcb=Kw/Ka=(1x1014)/(1.8x105)=5.56x1010 X2/(0.05X)=5.56x1010 X2/(0.05)=5.56x1010 X=5.27x106=[OH] (1x1014)/(5.27x106)=1.90x109=[H+] pH=log(1.06x1011)=8.72 61 17 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA TITRATIONS AND INDICATORS Consider the following indicators and their pH ranges: Indicator pH Range Methyl orange 3.2-4.4 Methyl red 4.8-6.0 Bromothymol blue 6.0-7.6 Phenolphthalein 8.2-10.0 Alizarin yellow 10.1-12.0 For which one of the following titrations would methyl red be the best indicator? A. B. 0.100 M HNO3 + 0.100 M KOH 0.100 M aniline (Kb = 3.8 10-10) + 0.100 M HCl 0.100 M NH3 (Kb = 1.8 10-5) + 0.100 M HCl 0.100 M HF (Ka = 7.2 10-4) + 0.100 M NaOH 0.100 M acetic acid (Ka = 1.8 10-5) + 0.100 M NaOH C. D. E. A. The reaction of a strong acid with a strong base has an equivalence point of ~ pH 7. B. Qualitatively: The reaction of a typical weak base (Kb ~ 10-5) with a strong acid has an equivalence point of ~ pH 5. The equivalence point in this case is probably considerably less than pH 5 because aniline is an extremely weak base (Kb ~ 10-10). Quantitatively: Step 1: Bring to completion. Equilibrium 0 Step 2: Bring back to equilibrium. Equilibrium 0.100-X 0 0.100 +X +X Kca = Kw/Kb = (1 x 10-14)/(3.8 x 10-10) = 2.63 x 10-5 ([H3O+][RNH2])/[RNH3+] = Kca ([X][X])/[0.100-X] = 2.63 x 10-5 Simplifying: X2/0.100 = 2.63 x 10-5 62 X = 1.622 x 10-3 = [H3O+] pH = -log[H+] = -log(1.622 x 10-3) = 2.79 C. Qualitatively: The reaction of a typical weak base (Kb ~ 10-5) with a strong acid has an equivalence point of ~ pH 5. Quantitatively: Step 1: Bring to completion. RNH2 Initial Change Equilibrium 0.100 -0.100 0 + H3O+ 0.100 -0.100 0 RNH3+ 0 +0.100 0.100 + H2O Step 2: Bring back to equilibrium. RNH3+ New initial Change 0.100 -X + H2O H3O+ 0 +X +X + RNH2 0 +X +X Equilibrium 0.100-X Kca = Kw/Kb = (1 x 10-14)/(1.8 x 10-5) = 5.56 x 10-10 ([H3O+][RNH2])/[RNH3+] = Kca ([X][X])/[0.100-X] = 5.56 x 10-10 Simplifying: X2/0.100 = 5.56 x 10-10 X = 7.45 x 10-6 = [H3O+] pH = -log[H+] = -log(7.45 x 10-6) = 5.13 D. Qualitatively: The reaction of a weak acid with a strong base has an equivalence point of ~ pH 9. E. Qualitatively: The reaction of a weak acid with a strong base has an equivalence point of ~ pH9. Methyl red (pH range 4.8 6.0) should be a good indicator for the reaction of NH3 with HCl. 63 25 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA TITRATIONS AND INDICATORS It is found that 25.0 mL of a sodium hydroxide solution completely neutralizes 22.30 mL of 0.253 M sulfuric acid. What is the molarity of the sodium hydroxide? A. B. C. 0.113 M 0.226 M 0.284 M 0.451 M 0.567 M D. E. moles of NaOH = moles of H2SO4 moles H2SO4 = V x M = 0.02230 L x 0.253M = 0.005642 mol H2SO4 H2SO4 + 2NaOH Na2SO4 + 2H2O 0.005642 mol H2SO4 x (2 mol NaOH/mol H2SO4) = 0.01128 mol NaOH M = mol/V = 0.01128 mol/0.025L = 0.4512 M NaOH CHEM162-2003 6TH WEEK RECITATION-2 CHAPTER 15 - APPLICATIONS OF ACID AND BASE EQUILIBRIA (BUFFERS AND TITRATION) TITRATIONS 15.57 Consider the titration of 40.0 mL of 0.200 M HClO4 by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. H+ + OH- HOH c. 40.0 mL Initial Change H+ 0.040 L x 0.200 mol/L = 0.008 mol -0.004 OH0.040 L x 0.100 mol/L = 0.004 mol -0.004 64 Equilibrium 0.004 mol/(0.040 + 0.040 L) = 0.05 M pH = - log(0.05) = 1.30 0 mol CHEM162-2003 6TH WEEK RECITATION-2 CHAPTER 15 - APPLICATIONS OF ACID AND BASE EQUILIBRIA (BUFFERS AND TITRATION) TITRATIONS 59 Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8 x 10-5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. HA + OH- HOH + Ab. 50.0 mL 1st step: Go to completion HA OHAInitial 0.1 L x 0.200 0.050 L x 0.100 0.0 mol mol/L mol/L = 0.005 = 0.020 mol mol Change -0.005 mol -0.005 mol +0.005 mol Equilibrium 0.015 mol/(0.10 0.0 M 0.005 mol/(0.10 + 0.05 L) = 0.1 + 0.05 L) = M 0.0333 M New equation, new equilibrium expression, new calculation HA + H2O H3O+ + A([H3O+][A-])/[HA] = 1.8 x 10-5 HA H3O+ A65 New Initial 0.1 M 0 Change -X +X Equilibrium 0.1 - X +X (X x (0.03333 + X))/(0.1-X) = 1.8 x 10-5 Simplify: (X x 0.03333)/0.1 = 1.8 x 10-5 X = 5.405 x 10-5 pH = - log (5.405 x 10-5) = 4.27 or use the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]) pH = -log(1.8 x 10-5) + log(0.03333/0.1) = 4.27 0.0333 M +X 0.03333 + X CHEM162-2003 6TH WEEK RECITATION-2 CHAPTER 15 - APPLICATIONS OF ACID AND BASE EQUILIBRIA (BUFFERS AND TITRATION) TITRATIONS 65. Calculate the pH at the halfway point for the following titration. b. 100.0 mL of 0.10 M C2H5NH2 (Kb = 5.6 x 10-4) titrated by 0.20 M HNO3 RNH2 + H3O+ Initial Change Equilibrium RNH3+ + H2O RNH2 0.100 L x 0.10 M = 0.010 mol -0.005 mol 0.005 mol/(0.100 + 0.025 L) H3O+ 0.0050 mol = 0.025 L x 0.20 M -0.005 mol 0.000 mol RNH3+ 0 mol +0.005 mol 0.005 mol/ (0.100 + 66 = 0.040 M RNH2 + H2O RNH3+ + OHRNH2 Initial 0.040 M Change -X Equilibrium 0.040 - X 0.025 L) = 0.040 M RNH3+ 0.040 M +X 0.040 + X OH0 +X +X ((0.040 + X) x (X))/(0.040 - X) = 5.6 x 10-4 X = [OH-] = 5.6 x 10-4 [H+] = (1 x 10-14)/(5.6 x 10-4) = 1.79 x 10-11 pH = - log [H+] = - log (1.79 x 10-11) = 10.75 or Use Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]) pH = -log((1 x 10-14)/(5.6 x 10-4)) + log (0.040/0.040) = 10.75 17. CHEM162-2000 HOURLY EXAM II CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB (BUFFERS & TITRATIONS) TITRATIONS In which one of the following titrations, if any, will the pH be 7.00 at the stoichiometric point? A. HNO3 titrated with NaOH B. HCl titrated with NH3 C. HNO3 titrated with NH3 D. HNO2 titrated with NaOH E. None of the other choices will have pH 7.00 at the stoichiometric point. 67 24. CHEM162-2000 HOURLY EXAM II CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) TITRATIONS Calculate the pH at the stoichiometric point when 0.100 mol of HCOOH is neutralized with NaOH in aqueous solution. Assume the final volume of the solution is 1.00 liter. Ka = 1.8 x 10-4 for HCOOH. A. B. C. D. 3.74 10.26 7.00 5.63 E. 8.37 HCOOH + OH- HOH + HCOOHCOOH OH HOH Initial 0.100 0.100 0 Change -0.100 -0.100 Equilibrium 0 0 HCOOH OH HOH New Initial 0 0 Change +X +X Equilibrium X X [HCOO ]/([HCOOH][OH ]) = K K = (1/(Kw/Ka) = Ka/Kw = (1.8 x 10-4)/(1 x 10-14) = 1.8 x 1010 [0.100 - X]/([X][X]) = 1.8 x 1010 X = 2.357 x 10-6 = [OH-] pOH = - log(2.357 x 10-6) = 5.627 pH = 14 - 5.627 = 8.37 The calculation may be simplified so as to avoid a quadratic equation: [0.100 - X]/([X][X]) = 1.8 x 1010 [0.100]/([X][X]) = 1.8 x 1010 X = 2.357 x 10-6 = [OH-] pOH = - log(2.357 x 10-6) = 5.627 pH = 14 - 5.627 = 8.37 HCOO0 +0.100 0.100 HCOO0.100 -X 0.100-X 20. CHEM162-2001 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) TITRATIONS For which of the following acid-base titrations in aqueous solution will the pH at the equivalence point be less than 7.0? Weak bases and acids will tend to have weak effects on raising or lowering the pH, 68 respectively. Strong bases and acids will tend to hav strong effects on raising and lowering the pH, respectively. The only combination of an acid and base that will provide a lowering of the pH is a weak base and a strong acid. A. titration of HCl with NaOH B. titration of CH3COOH with KOH C. titration of NH with HBr D. titration of KOH with HNO3 E. titration of HCOOH with Ba(OH)2 3 CHEM162-2002 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) TITRATIONS 5. In the titration of a weak acid with a strong base, which one of the following statements is true? A. The pH at the equivalence point depends on the indicator used. The pH at the equivalence point is not dependent on the indicator used. The indicator used is dependent on the pH at the equivalence point. The graph of pH vs volume of base added rises gradually at first and then rises steeply near the equivalence point Typical graphs of pH vs volume of titrant, for titration of a weak acid with a strong base or a strong acid with a weak base, rise gradually at first and then rise steeply near the equivalence point. C. The pH = 7.0 at the equivalence point. Titration of a weak acid with a strong base has an equivalence point which is related to the fact that the combination is a weak acid and a strong base, i.e., an equivalence point on the basic side, usually in the pH 8 to 10 range. D. The pH is less than 7.0 at the equivalence point. Titration of a weak acid with a strong base has an equivalence point which is related to the fact that the combination is a weak acid and a strong base, i.e., an equivalence point on the basic side, usually in the pH 8 to 10 range. E. The pH at the equivalence point equals the pKa of the weak acid. The pH at the half-equivalence point (the half-way point) equals the pKa of the weak acid, according to the Henderson-Hasselbalch equation. The pH at the equivalence point is not related to the pKa of the weak acid. CHEM162-2002 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) TITRATION B. 69 11. A 0.750 gram sample of a solid diprotic acid, represented as H2X, is dissolved in water and titrated with NaOH solution, requiring 35.0 mL of 0.275 M NaOH for complete reaction (both protons are neutralized). What is the molar mass of H2X? A. 244 g/mol 156 g/mol 118 g/mol 182 g/mol 105 g/mol B. C. D. E. H2X + 2OH- 2H2O + X2H2X Initial Change Equilibrium 2H2O Ymol/0.035L 0.035Lx0.275M/0.035 L = 0.009625mol/0.035L -1/2(0.009625mol/0.035L -0.009625mol/0.035L 2OH0 0 X20 not relevant not relevant Y - (1/2 x 0.009625) = 0 [The liters are not relevant] Y = 0.0048125 mol H2X MW = 0.750/0.0048125 = 155.8 g/mol CHEM162-2002 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) TITRATION 12. Which one of the following indicators would be most appropriate for determining the equivalence point of a titration of 1.0 M acetic acid (Ka = 1.8 x 10-5) with 1.0 M sodium hydroxide? The equivalence point of the reaction of a weak acid with a strong base is generally in the 8 to 10 pH range. A. Bromphenol blue (changes from yellow to blue over pH range 3 to 5 ) B. Bromcresol green (changes from yellow to blue over pH range of 4 to 6) C. Chlorphenol red (changes from yellow to red over pH range of 5 to 7) D. Alizarin blue (changes from yellow to blue over pH range of 11 to 13) E. Thymol blue (changes from yellow to blue over pH range of 8 to 10) 70 CHEM162-2003 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) TITRATIONS 4. In which titration would you use an indicator having Ka = 1 x 10-5 ? X. HCN titrated with NaOH Y. CH3COOH titrated with NaOH Z. NH3 titrated with HCl A. B. C. D. X and Y only X, Y and Z Y and Z only Y only Z only E. Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) An indicator formula is that of a buffer. The indicator changes color when [A-] = [HA]. At that point the pH = pKa. If Ka = 1 x 10-5, then pKa is equal to 5, and therefore the pH of the color change is equal to 5. In X, a weak acid is titrated with a strong base. The pH at the equivalence point is therefore > 7. In Y, a weak acid is titrated with a strong base. The pH at the equivalence point is therefore > 7. In Z, a weak base is titrated with a strong acid. The pH at the equivalence point is therefore < 7. 162 EXAM IIVl SPRING 2003 2 CHEM162-2003 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) TITRATIONS 13. Consider a titration in which NaOH is being added to a flask which originally contains 1.0 liter of 1.0 M HX. The titration curve for this process is shown below. pH 71 pH at half equivalence point Half equivalence point Volume NaOH added Based on this data, what is the approximate Ka for acid HX? A. B. 1 x 10-10 1 x 10-7 1 x 10-6 1 x 10-5 1x 10-3 C. D. E. Henderson-Hasselbalch: pH = pKa + log (A-/HA) At the half-equivalence point [A-] = [HA] log(1) = 0 Therefore, pH = pKa Since pH = 6 at the half-equivalence point, pKa = 6, and Ka = 10-6. 162 EXAM 11 Vl SPRING 2003 50. CHEM 162-2000 FINAL EXAM CHAPTER 15 - APPLIC OF ACID & BASE EQUIL (BUFFERS & TITRATION) TITRATION 50.0 mL of 0.200 M aqueous HOCl solution is titrated with 0.200 M aqueous KOH solution. Which, if any, of the following statements concerning this titration is not true? (pKa = 7.52 for HOCl) 72 At the stoichiometric point, [ClO-] is approximately 0.100 M. pH > 7 at the stoichiometric point. After 25.0 mL of KOH solution have been added, pH = 7.52. Phenolphthalein (pKIn = 9.4) would be a better indicator than methyl red (pKIn = 5.0) to use in this titration. E. All the other statements are true. A. B. C. D. E 46. CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB (BUFFERS & TITRATION) TITRATION A researcher prepares an approximately 0.5 M solution of HF (Ka = 6.6 x 10-4) and wishes to determine its precise molarity by titrating it with 0.500 M KOH solution. Which of the following would be the best indicator to use in this titration? The equivalence point of a weak acid and a strong base is approximately 9. indicator A. thymolphthalein pH range for color change 9.5 - 10.5 8.0 - 9.5 6.0 - 7.5 4.0 - 6.0 3.0 - 4.0 B. thymol blue C. bromthymol blue D. bromcresol green E. methyl orange 49. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB (BUFFERS & TITRATION) TITRATIONS A 50.0 mL sample of 0.200 M HNO3 solution is titrated with 0.100 M NaOH solution. The pH after the addition of 50.0 mL of the solution of the base is 0.200 M HNO3 diluted from 0.050 L to 0.010 L = 0.100 M HNO3. 73 0.100 M NaOH diluted from 0.050 L to 0.010 L = 0.050 M NaOH. Reactions of strong acids or strong bases go virtually to completion, within stoichiometric limitations. H3O+ + OH- ----> 2 H2O 0.100 0.050 -0.050 -0.050 0.050 0 Initial Change Equil [H3O+] = 0.050 pH = -log 0.050 pH = 1.30 A. B. C. D. 12.70 9.85 4.151 7.000 1.301 E. Chem 162-2003 Final exam + answers Chapter 15A-Applications of Aqueous Equilibria (of Acids and Bases) Reactions between acids and bases (including titrations) TITRATIONS 17. A 100.0 ml sample of 0.2 M (CH3)3N (Kb = 5.3 x l0-5) is titrated with 0.2 M HCl. Calculate the pH at the equivalence point. A. 9.9 B. 3.1 C. 10.3 D. 5.4 E. 7.0 (CH3)3N + H3O+ (CH3)3NH+ + H2O 0.100 L x 0.2 mol/L = 0.02 mol (CH3)3N 0.02 mol (CH3)3N x ((1 mol H3O+)/(1 mol (CH3)3N)) = 0.02 mol H3O+ to reach the equivalence point 74 mol = M x V 0.02 mol H3O+ = 0.2 M x V V = 0.100 L H3O+ 0.02 mol (CH3)3N/(0.100 + 0.100 L) = 0.100 M (CH3)3N 0.02 mol H3O+/(0.100 + 0.100 L) = 0.100 M H3O+ We cannot subtract X and add X in the conventional manner, because since the equilibrium constant of reactions with strong acids is very large, the Xs cannot be dropped, and a quadratic equation will result. Hence, bring the reaction to completion and then slightly back. Initial Change Equilibrium (CH3)3N 0.100 M -0.100 M 0 H3O+ 0.100 M -0.100 M 0 (CH3)3NH+ 0 +0.100 M +0.100 H2O The Henderson-Hasselbalch equation cant be used here because the resulting solution is not a buffer. Hence, bring the reaction slightly back to equilibrium. (CH3)3N H3O+ New initial 0 0 Change +X +X Equilibrium +X +X [(CH3)3NH+]/([(CH3)3N][H3O+]) = 1/(Kw/Kb) [0.100 - X]/([X][X]) = 1/((1x10-14) x (5.3 x 10-5)) [0.100]/([X][X]) = 1/((1x10-14) x (5.3 x 10-5)) X = 4.34 x 10-6 pH = -log (4.34 x 10-6) = 5.36 23. (CH3)3NH+ 0.100 -X 0.100 - X H2O Chem 162-2004 Exam II Applications of aqueous equilibria - Chapter 15A Acid-base titrations; reactions between strong and weak acids and bases Calculate the pH when 30.0 mL of 0.040 M HCl is added to 20.0 mL of 0.030 M NaOH. A. B. C. D. E. 1.92 2.57 3.22 3.07 3.57 H3O+ + OH- 2 H2O Identify the limiting reactant; then bring the reaction to completion. 75 H3O+ 0.030L x 0.040M = 0.0012mol/(0.030L+0.020L) Change -0.0006mol/(0.020L+0.030L) Equilib. 0.0006mol/0.050L = 0.012M + pH = -log[H ] = -log(0.012) = 1.92 M Initial OH- 2 H2O 0.020L x 0.030M = 0.0006mol/(0.020L+0.030L) -0.0006mol/(0.020L+0.030L) 0 Alternatively, an ICE table can be set up to determine the [H3O+] and [OH-] at equilibrium, but this isnt necessary because it would be expected that the change in concentration of H3O+ from 0.012M (i.e., the value of X) would be insignificant. H3O+ + OH- 2 H2O K = 1 x 1014 H3O+ 2 H2O OH Initial 0.012 0 Y Change +X +X -2X Equilib. 0.012 + X +X Y-2X Considering that K is such a huge number, then +X can be dropped, and therefore [H3O+] = 0.012 13. Chem 162-2004 Exam II Applications of aqueous equilibria - Chapter 15A Acid-base titrations; reactions between strong and weak acids and bases Which of the following titrations would have a pH of 7 at the equivalence point? X. Y Z. A. B. C. D. HNO3 and KOH HC2H3O2 and KOH HCl and NH3 Y and Z only X and Y only Z only Y only X only E. Only a strong acid and a strong base (X) will have a pH of 7 at the equivalence point. A weak acid and a strong base (Y) will have an equivalence point above 7, approximately 9; a weak base and a strong acid (Z) will have an equivalence point below 7, approximately 5. 76 CHEM162-2003 5TH WEEK RECITATION CHAPTER 15 - APPLIC. OF ACID & BASE EQUILIBRIA TITRATIONS 27 Calculate the pH after 0.020 mol HCl is added to 1.00 L of 0.1 M sodium propanoate. (HC3H5O2, Ka = 1.3 x 10-5) (b) Approach I. This approach is not recommended, because it requires the use of a quadratic equation; i.e., X is too large to be dropped. P- + H3O+ HP + H2O PInitial 0.100 Change -X Equilibrium 0.100 - X H3O+ 0.020 -X 0.020-X HP 0 +X +X X/((0.100 - X)(0.020 - X)) = 1/(1.3 x 10-5) = 7.69 x 104 X = 0.0199968 H3O+ = 3.2 x 10-6 pH = -log (3 x 10-6) = 5.49 Approach II. This approach is better than approach I, because it doesnt require the use of a quadratic equation, but it is still not the best approach to solving this problem. P- + H3O+ HP + H2O PH3O+ HP Initial 0.100 0.020 0 Change -0.020 -0.020 +0.020 Equilibrium 0.080 0 0.020 P- + H2O HP + OHPNew Initial 0.080 Change -X Equilibrium 0.080 - X HP 0.020 +X 0.020+X OH0 +X +X ((0.020+X) x (X))/(0.080-X) = (1 x 10-14)/(1.3 x 10-5) X = +3.077 x 10-9 = [OH-] H+ = (1 x 10-14)/(3.077 x 10-9) = 3.25 x 10-6 pH = -log(3.25 x 10-6) = 5.49 77 Approach III. Approach dumbest, but by far the fastest, and is therefore the one of choice. Henderson-Hasselbach equation: pH = pKa + log([A-]/[HA]) pH = -log (1.3x10-5) + log(0.080/0.020) = 5.49 27d. Calculate the pH after 0.020 mol HCl is added to 1.00 L of a mixture containing 0.100 M HC3H5O2, 0.100 M NaC3H5O2 and 0.020 M HCl HP + H2O H3O+ + P- Initial Change Equilibrium HP 0.100 +0.020 0.120 H3O+ 0.020 -0.020 0.000 P0.100 -0.020 0.080 HP + H2O H3O+ + P- New Initial Change Equilibrium HP 0.120 -X 0.120 - X H3O+ 0 +X +X P0.080 +X 0.080+X ((0.080+X) x (X))/(0.120-X) = 1.3 x 10-5 X = 1.95 x 10-5 pH = -log (1.95 x 10-5) = 4.71 However, best approach to 27d is Henderson-Hasselbach equation: pH = pKa + log([A-]/[HA]) pH = -log (1.3x10-5) + log(0.080/0.120) = 4.71 78 CHEM162-2003 5TH WEEK RECITATION CHAPTER 15 - APPLIC. OF ACID & BASE EQUILIBRIA TITRATIONS 29 Calculate the pH after 0.020 mol NaOH is added to 1.00 L of 0.10 M propanoic acid. (Ka = 1.3 x 10-5) (a) HP + H2O H3O+ + P- Initial Change Equilibrium HP 0.100 -0.020 0.080 H3O+ 0.020 -0.020 0 P0 +0.020 +0.020 HP + H2O H3O+ + PHP Initial 0.080 Change -X Equilibrium 0.080-X H3O+ 0 +X +X P0.020 +X 0.020+X Ka = ([H3O+][P-])/[HP] = 1.3 x 10-5 ([X][0.020+X])/[0.080-X] = 1.3 x 10-5 X = 5.2 x 10-5 pH = -log(5.2 x 10-5) = 4.28 or Henderson-Hasselbach: pH = pKa + log([A-]/[HA]) pH = -log (1.3x10-5) + log(0.020/0.080) = 4.28 79 18. CHEM162-2000 HOURLY EXAM II CHAPTER 15 - APPLIC. OF ACID & BASE EQUILIBRIA TITRATIONS What is the pH of the aqueous solution formed by mixing 50.0 mL of 0.400 M HCl and 150 mL of 0.100 M NaOH? A. 2.30 B. 1.60 C. 1.90 D. 3.20 E. 2.70 CHEM162-2002 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC. OF ACID & BASE EQUILIBRIA TITRATIONS 16. Calculate the pH of a solution prepared by mixing 20.0 mL of 0.30 M HCl with 15.0 mL of 0.60 M NaOH Initial Change Equilibrium H+ 0.30 x 0.020 = 0.006 mol -0.006 0 OH0.60 x 0.015 = 0.009 mol -0.006 0.003 mol 0.003 mol/(0.020 + 0.015 L) = 0.0857 M pOH = -log0.0857 = 1.067 pH = 14 - 1.067 = 12.93 A. B. 11.48 6.50 12.93 9.62 10.73 C. D. E. 80 CHEM162-2003 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC. OF ACID & BASE EQUILIBRIA TITRATIONS 18. Calculate the pH of a solution prepared by adding 20.0 mL of 0.40 M HCl to 80.0 mL of 0.10 M Sr(OH)2. A. B. C. 11.90 7.00 13.40 12.90 10.50 D. E. This is a strong acid and a strong base. Either one of them warrants bringing the reaction to completion. mol H+ = 0.40 x 0.020 = 0.008 mol mol OH- = 0.10 x 2 x 0.080 = 0.016 mol H3O+ + OH- 2H2O Initial Change Equilibrium H3O+ 0.008 mol/0.020L -0.008mol/0.020L 0 mol/0.020L OH 0.016 mol/0.080L -0.008mol/0.020L 0.008 mol/0.100L 2H2O 0.008 mol/0.100L = 0.08 M [OH-] pOH = -log 0.08 = 1.10 pH = 14.0 - 1.10 = 12.90 81 22. Chem 162-2004 Final Exam + Answers Chapter 15A - Applications of aqeous equilibria (of acids and bases) Reactions between acids and bases (including titrations) Calculate the pH at the halfway point when 100 ml of 0.10 M CH3COOH is titrated by 0.1 M NaOH. Ka(CH3COOH =1.8 x 10-5) A. B. C. D. 5.29 7.00 9.10 8.45 4.74 E. HA + OH- HOH + AHA Initial Change Equilibrium 0.010 mol/0.100L 0 -0.0050 +0.0050 0.0050mol/(0.100 0.0050 mol/(0.100 + 0.050) = + 0.050) = 0.0333M 0.0333M The halfway point is when half of the CH3COOH is titrated with the NaOH. Moles of CH3COOH = 0.100 L x 0.10 M = 0.010 moles CH3COOH Half-way point is when 0.0050 moles CH3COOH is titrated, which means that there are 0.0050 moles CH3COOH remaining in the 0.100 L of the original solution. 0.0050 moles of 0.1 M NaOH were used. 0.0050 mol/0.1 M = 0.050 L NaOH A buffered solution is formed; hence, use the Henderson-Hasselbalch equation. pH = pKa + log([A-]/[HA]) pH = (-log(1.8 x 10-5)) + log ([0.0050]/[[0.0050]) pH = 4.74 This problem could have been solved much more easily. Since the half-way point of a reaction is when the concentration of the weak acid and the conjugate base are equal, then using the Henderson Hasselbalch equation, the term log([A-]/[HA] is equal to zero, and pH = pKa. Hence, pH = -log(1.8 x 10-5). pH = 4.74 0.0050 mol/0.050L -0.0050 0 OHH2O A- 82 49. Chem 162-2004 Final Exam + Answers Chapter 15A - Applications of aqeous equilibria (of acids and bases) Reactions between acids and bases (including titrations) A 100.0 mL sample of 0.10 M C2H5NH2 (Kb = 5.6 x 10-4) is titrated with 0.2 M HNO3. Calculate the pH at the equivalence point. A. B. C. D. 4.87 6.11 6.72 8.67 5.96 E. RNH2 + H3O+ RNH3+ + H2O 0.100L of 0.10M RNH2 Kb = 5.6 x 10-4 0.2M HNO3 At the equivalence point the moles of H3O+ = the moles of RNH2. moles = M x V moles of RNH2+ = 0.10M x 0.100L = 0.01 mol moles = M x V 0.01 mol HNO3 = 0.2M x V V HNO3 at the equivalence point = 0.050L Diluted M of RNH2 = 0.01mol/(0.100L + 0.050L) = 0.0667M Diluted M of H3O+ = 0.01mol/(0.050L + 0.100L) = 0.0667M RNH2 H3O+ RNH3+ H2O Initial 0.0667M 0.0667M 0 0 Change Equilibrium Bring the reaction to completion to avoid a quadratic equation, and since this is a large K reaction (the reaction of a strong acid). RNH2 RNH3+ H2O H3O+ Initial 0.0667M 0.0667M 0 0 Change -0.0667 -0.0667 +0.0667 Equilibrium 0 0 +0.0667M Since there is no RNH2 then this is not a buffer. Hence, calculations cannot be done via Henderson-Hasselbalch. So rewrite the ICE table and determine the equilibrium 83 concentration of H3O+. RNH3+ H2O RNH2 H3O+ 4. Initial 0.06667M 0 0 Change -X +X +X Equilibrium 0.06667-X 0 X X + + K = ([RNH2][H3O ])/[RNH3 ] Kca = (1 x 10-14)/( 5.6 x 10-4) = 1.786 x 10-11 ([X] x [X])/[0.06667-X] = (1.786x10-11) X = 1.091 x 10-6 pH = -log(1.091 x 10-6) = 5.96 Chem 162-2005 Exam II Applications of Aqueous Equlibria - Chapter 15A Acid-base titrations; reactions between strong and weak acids and bases; indicators Concepts The net ionic equation for the reaction that occurs during the titration of nitrous acid, HNO2, with sodium hydroxide solution is: A. HNO2(aq) + Na+(aq) + OH (aq) NaNO2(aq) + H2O(l) HNO2(aq) + OH-(aq) NO2 (aq) + H2O(l) HNO2(aq) + NaOH(aq) Na+(aq) + NO2 (aq) + H2O(l) HNO2(aq) + H2O(aq) H3O+(aq) + NO2 (aq) H+(aq) + OH (aq) H2O(l) - B. C. D. E. HNO2 is a weak acid so it is a main species. NaOH dissociates into Na+ + OH-. Na+ is a spectator ion, so get rid of it. HNO2 + OH- HOH + NO2- 8. Chem 162-2005 Exam II Applications of Aqueous Equlibria - Chapter 15A Acid-base titrations; reactions between strong and weak acids and bases; indicators Concepts Questions 8 and 9 refer to the graph below, which shows the titration curve when 100 mL of 0.025 M acetic acid is titrated with 0.100 M NaOH. 84 Which part of the curve corresponds to the optimum acetic acid/acetate buffer? A. B. C. D. section Y to Z section W to Y point Z point X point V E. A. Section Y to Z is the post-equivalence area. B. Section W to Y is the area surrounding the end-point (the equivalence point). C. Point Z is a point at which there is excess sodium hydroxide. D. Point X is the end-point (the equivalence point). E. Point V is the half-equivalence point, or the point in which half of the acetic acid has been converted into sodium acetate. Since there will be acetic acid and sodium acetate present at this point, it represents the optimum acetic acid-sodium acetate buffer. 9. Chem 162-2005 Exam II Applications of Aqueous Equlibria - Chapter 15A Acid-base titrations; reactions between strong and weak acids and bases; indicators Concepts Questions 8 and 9 refer to the graph below, which shows the titration curve when 100 mL of 0.025 M acetic acid is titrated with 0.100 M NaOH. 85 Which of the following indicators is the best choice for this titration? A. B. Indicator methyl red methyl orange phenolphthalein alizarin yellow bromothymol blue pH Range of color change 4.8-6.0 3.2-4.4 8.2-10.0 11.0-12.4 6.1-7.6 C. D. E. The best choice of indicator is the one which has a pH range at the end point. The end point is 9, so the best indicator is phenolphthalein. 86 14. Chem 162-2005 Final Exam + Answers Chapter 15 - Applications of Aqueous Equilibria (of Acids and Bases) Reactions between acids and bases (including titrations) Concepts Which one(s) of the following statements is (are) true? X. When 50.0 mL of 0.10 M HCN is titrated with 0.10 M NaOH, the equivalence point is reached when the pH = 7 Y. The strength of a weak acid has a significant effect on the shape of its titration curve (pH vs volume of base) Z. Titration can be used to determine values of equilbrium constants A. B. X, Y, and Z X only Y and Z only X and Y only Y only C. D. E. HCN + OH- HOH + CN CN- + H2O HCN + OHThe equivalence point is reached when the HCN has completely reacted with OH- to provide CN-. X. CN- is a weak base which reacts with water to form OH-. The pH > 7.00. Y. The strength of a weak acid does have a significant effect on the shape of its titration curve. If it is a relatively strong weak acid, then the initial pH will be lower than a weak weak acid, and the equivalence point will be lower than that of a weak weak acid. Also, the area of the equivalence point will be markedly more vertical that that of a weak weak acid. Z. Titration can be used to determine values of equilbrium constants. At the half-way point in the titration, substituting the pH into the Henderson-Hasselbalch equation will give the pKa, with which one can find the Ka. 87 23. Chem 162-2005 Final Exam + Answers Chapter 15 - Applications of Aqueous Equilibria (of Acids and Bases) Reactions between acids and bases (including titrations) Concepts Consider the titration of 50.0 mL of 0.10 M HCl with 0.200 M NaOH. Which one(s) of the following statements is (are) true? X. To reach the equivalence point, one should add 25.0 mL of NaOH solution Y. The pH at the equivalence point is 7.00 Z. pH = pKa of the acid at halfway to the equivalence point. A. B. Y only X, Y, and Z X and Y only X only Y and Z only C. D. E. X. mol = M x V MolHCl = 0.10 x 0.050 = 0.0050 mol HCl NaOH: 0.0050 mol = 0.200 x V V = 0.025 L of NaOH needed to neutralize 0.10 L of HCl True. Y. H3O+ + OH- 2H2O H2O is neutral. Hence, the equivalence point would be 7.00. True. Z.False. Henderson-Hasselbalch: pH = pKa + log([A-]/[HA]) This question is basically saying, Can we use the half-way point approach in the Henderson-Hasselbalch equation to determine the pKa of a strong acid. If this were the titration of a weak acid and strong base, or a weak base and a strong acid, then at the halfway point, [A-] = [HA]. Therefore pH = pKa. However, this is the reaction of a strong acid with a strong base. At the half-way point there would only be H3O+ and spectator ions present, not the required [A-] and [HA], so the Henderson-Hasselbalch formula to find the pKa wouldnt apply. Another way to look at this is to calculate the pH of a 0.1M HCl solution at the half-way point. The initial pH of 0.1 M strong acid is (pH = -log(0.1) =)1.0. The half-way point is 50.0 mL 0.1 M HCl + 12.5 mL of 0.200 M NaOH H+ + OH- H2O Initial Ka = 1 x 1014 H+ + (0.2molx0.0125L)/ (0.0125+0.050)= 0.040M -0.040 0M OHH2O (0.1molx0.050L)/ (0.050+0.0125) = 0.080M Change -0.040 Equilibrium 0.040M At the half-way point the [H+] = 0.040. If the pH = pKa, as shown in the Henderson-Hasselbalch equation, then [H+] = Ka. Therefore Ka of HCl = 0.040. This is a ridiculous answer because the Ka of HCl is very 88 89

Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

Rutgers - CHEM 162 - 162
CHAPTERS 22 CHEM 162-2007 PRACTICE PROBLEMSCHAPTER 22 - TRANSITION METALS AND COORDINATION CHEMISTRY COORDINATION COMPOUNDS E. Tavss, PhD1CHAPTER 21 - TRANSITION METALS AND COORDINATION CHEMISTRY49 Chem 162-2008 Final Exam + AnswersChapter 22 Transit
Rutgers - EXPOS - 101
Yogi Patel HumilityOften times people have a sense of ego when they do something or accomplish a difficult goal. The easy way out is to brag about it and boost their accomplishment in other peoples faces for recognition and to build ego. However the hard
Rutgers - BUSINESS W - 132
Yogi PatelAkshay PatelAkshay was born on April 23, 1992 born in Queens, New York. Akshay has 1 brother thats older than him by one year. He moved to Piscataway he was nine where he stayed there the rest of his life and went to Piscataway high school. He
Rutgers - HISTORY - 152
Whigs 17th-18th Century = Parliamentarians/opponents of the Kings prerogatives/Patriots in American Revolution U.S. 19th century = Opponents of King Andrewadvocates of weak executive and strong legislaturesReview of Second American Party SystemWhig org
Rutgers - HISTORY - 152
Patent Office Models of the Good Society Emerson to Carlyle (1840): We are a little mad here with numberless projects of social reform. Not a reading man but has a draft for a New Community in his waistcoat pocket.Sources of Commune Building Romanticis
Rutgers - HISTORY - 152
Entrepreneurs and Inventors: The Culture of EnterpriseUnder My Wing Every Thing Prospers This View of New Orleans Taken from the Plantation of Marigny was painted in 1803. It illustrates two points for us about the role of the state in the development o
Rutgers - HISTORY - 152
Yogi Patel Jared Weigley 2/21/11 Expos: DCKJDJDJVoters always played a giant role in shaping America because it is a democratic nation that firmly believes in democracy. Being a democratic society means that you allow voters to choose they want to place
Georgia State - MATHMATICS - 1101
26Bacteria and Archaea: The Prokaryotic Domains26 Bacteria and Archaea: The Prokaryotic Domains 26.1 How Did the Living World Begin to Diversify? 26.2 What Are Some Keys to the Success of Prokaryotes? 26.3 How Can We Resolve Prokaryote Phylogeny? 26.4
Georgia State - MATHMATICS - 1101
31Animal Origins and the Evolution of Body Plans31 Animal Origins and the Evolution of Body Plans 31.1 What Characteristics Distinguish the Animals? 31.2 What Are the Features of Animal Body Plans? 31.3 How Do Animals Get Their Food? 31.4 How Do Life C
Georgia State - MATHMATICS - 1101
32Protostome Animals32 Protostome Animals 32.1 What Is a Protostome? 32.2 What Features Distinguish the Major Groups of Lophotrochozoans? 32.3 What Features Distinguish the Major Groups of Ecdysozoans? 32.4 Why Are Arthropods So Diverse?32.1 What Is a
Georgia State - MATHMATICS - 1101
33Deuterostome Animals33.1 What Is a Deuterostome?Deuterostomes are triploblastic, coelomate animals with internal skeletons.33.1 What Is a Deuterostome?Living deuterostomes comprise three clades: Echinodermssea stars, sea urchins, and their relative
Georgia State - MATHMATICS - 1101
Study Guide Chapter 42 The Immune SystemWhat materials, tissues, and organs make up the non-specific responses to infection?How is the lympthatic system involved with both specific and non-specific defense?Hematopoetic stem cells produce multiple cell
Georgia State - MATHMATICS - 1101
2211 Fall 2008 Test 1Name_Show all necessary work in your blue book. Please use a different page for each complete problem. Incorrect or non-work will not receive credit. When finished, please place this question paper inside your blue book and turn it
Georgia State - MATHMATICS - 1101
Math 2211Test 2Fall 2009Name_Show all necessary work in your blue book. No credit will be given for incorrect work or answers without work. Use a new page for each problem. Do problems in sequential order. Please use pencil. When finished, place this
Georgia State - MATHMATICS - 1101
2211 Spring 2009Test 1Name_Show all necessary work in your blue book. No credit will be given for incorrect work or answers without work. Please use a new page for each problem. Please use pencil. When finished, place this question paper in your blue b
Georgia State - MATHMATICS - 1101
Georgia State - MATHMATICS - 1101
Saint Louis - NURS - 360
ThefollowingarepossibletopicsfortheComprehensiveFinalTestinNURS360.Muchofthetestwillbea casestudyapproach.Thisisnotanallinclusivelist.Pleasedonotlimityourreviewandstudyingtothe followingtopics.THISISONLYMEANTTOBEAGUIDE!ReviewingtheKeyPointsectionsofyourbo
Saint Louis - NURS - 360
Test 4 Clinical picture for strokes based on location Embolotic-atrial fibulation-dysrhytmia, turbulent blood flow-rupture of rbcs, they stick together, emboli travels to cerebral vessels and blocks it Hemmoragic-due to hypertension or wavering BP Thrombo
Saint Louis - PHILOSOPHY - 100
Study Questions for Platos Apology 1. What is Socratess point in the opening two paragraphs of his defense? (17a-18a)2. What are the two groups of accusers to which Socratess refers? Of which is he more afraid? Why? (18a-e) Old accusers and new accusers,
Saint Louis - PHILOSOPHY - 100
Study Questions for Platos Crito 1. What does Crito want Socrates to do? (44e-45c) What reason does Crito give in 44c, and how does Socrates respond? Crito wants Socrates to save himself, to not go to jail. Crito says that if Scorates does not take the mo
Saint Louis - PHILOSOPHY - 100
Study Questions for Midterm Exam I will pick two of the following essay questions for the midterm exam. 1. From the Maritain readings, explain what philosophy is, how it differs from the special sciences (contrast it with the two errors here), and why thi
Saint Louis - PHILOSOPHY - 100
Study Questions for Book I of Aristotles Nicomachean Ethics Book I 1. What is Aristotles point in chapter 1 of Book I? Why do you think he starts with this point? The good is that at which everything aims. This will be the basis for all other arguments. 2
Saint Louis - PHILOSOPHY - 100
Study Questions for Book II of Aristotles Nicomachean EthicsBook II 1. How, according to Aristotle, do the virtues arise in us? Do they arise by nature? (chptrs 1-2) What are the implications of Aristotles theory of the origin of virtues? What does this
Saint Louis - PHILOSOPHY - 100
Study questions II for Platos Phaedo1. Explain Simmiass objection to Socratess argument. (85e-86d, 91c) Harmony is something, without body. The soul is the first thing to be destroyed when struck by disease and illness. Soul is a harmony of all other hum
Saint Louis - PHILOSOPHY - 100
Study questions I for Platos Phaedo: 1. What is the context for Socratess discussion in this dialogue? It is the day of death for Socrates. 2. What message does Socrates send to Evenus, and why does Socrates think that Evenus will follow his advice? (61b-
Saint Louis - PHILOSOPHY - 100
Study Questions for Platos Republic Book I 1 . Polemarchus initially claims (quoting Simonides) that justice is giving to each what is owed him. He modifies his position as Socrates questions him. a. What does Polemarchus come to think it is that we owe t
Saint Louis - PHILOSOPHY - 100
Study Questions for Platos Republic Book II 1. Explain Glaucons three kinds of goods. What sort of good does Socrates believe justice to be? (357a-e) What sort of good did most people of Socratess time believe justice to be? (358a) 1) things we desire onl
Saint Louis - PHILOSOPHY - 100
Study Questions for Platos Republic Book IV 1. How does Socrates determine what justice is in the city? (427e-435c) 4 cardinal virtues: Prudence (wisdom) - what ought to be done. Primarily in the ruling class. Fortitude (courage) - within the guardians Te
Saint Louis - PHILOSOPHY - 100
Study Questions for Platos Republic Book V 1. How does the philosopher differ from the &quot;lovers of sights and sounds&quot;? What is the difference between knowledge and opinion? (475c-480a) What makes philosophers different from lovers of sights and sounds is t
Saint Louis - THEATER - 100
Michelle Walsh October 5, 2009 Theater 100 Play Performance Worksheet Name of play: Proposals Playwright: Neil Simon Director: Milton Zoth Date of attendance: 09/30/09 Genre: Proposals was a comedy. This play keeps the audience laughing. Plot: 1. A father
Saint Louis - THEATER - 100
Michelle Walsh October 11, 2009 Theater 100 Play Performance Worksheet Name of play: The Bald Soprano Playwright: Eugene Ionesco Director: Tom Martin Date of attendance: October 11, 2009 Genre: Absurd Plot: 1. There is a couple with the name of Smith who
Saint Louis - THEATER - 100
Michelle Walsh October 11, 2009 Theater 100 Play Performance Worksheet Name of play: The Diviners Playwright: Jim Leonard Jr. Publisher: Samuel French Inc. Date of reading: October 24, 2009 Genre: Tragedy Plot: 1. The Layman Family lives in rural Indiana.
Saint Louis - THEATER - 100
Michelle Walsh October 11, 2009 Theater 100 Play Performance Worksheet Name of play: West Side Story (musical) Playwright: Arthur Laurents Date of reading: November 8, 2009 (previously seen performed by my high school) Genre: Tragedy Plot: 1. The musical
Saint Louis - THEATER - 100
Michelle Walsh November 30, 2009 Theater 100 Play Performance Worksheet Name of play: Dancing at Lughnasa Playwright: Brian Friel Date of performance: November 22, 2009 Genre: Drama Plot: 1. The play is told through a narrator, Michael Evans. Michael spea
Saint Louis - THEO - 100
Intro: Many of us spend numerous hours each week studying and socializing in Pius Library here on campus. But how many of us can say anything about whom the library is named after? Before this week I couldnt have told you anything other than he has some r
Saint Louis - THEO - 100
THE100 Study Guide, Fall 2009 Final Exam 1Study Guide For Final ExamThis is NOT everything! I reserve the right to pull PROMINENT items or themes from the chapters you read. Make sure you look at the quizzes and the questions at the back of each chapter
Saint Louis - THEO - 100
Rebecca Austin October 29, 2010 THEO 100 WickmanThe Last Supper and i ts Relation to PassoverThe passage I will be discussing is Luke 22:7-23. I t is Lukes account of the L ast Supper the meal that Jesus shared with his twelve disciples the night before
Saint Louis - THEO - 100
Rebecca Austin December 6, 2010 THEO 100 WickmanCapital Punishment: How Much Authority Should Humans Have Over the Lives of Others?United States courts have long debated the morality of capital punishment as a penalty for certain crimes. Capital punishm
Saint Louis - HIST - 112
Rebecca Austin February 21, 2011 Common SenseThomas Paines Common Sense is a pre-revolutionary pamphlet that effectively combined the emerging ideals of the Enlightenment with new political thoughts. He points out the corruption of monarchies, and emphas
Saint Louis - PHILOSOPHY - 205H
Rebecca Austin January 20, 2011 PHIL 205 Berman1. At 5de, Euthyphro is simply saying that piety is the opposite of its antonym, impiety. Heis not giving an actual definition for piety. If something is pious, it is inherent that it is not impious. Euthyp
Saint Louis - PHILOSOPHY - 205H
Rebecca Austin February 24, 2011 Questions on Kant1. A) One can act in accordance with duty while not be acting from duty. Actingi n accordance with duty means that ones actions are the same as what duty requires. This case is only reliant on the outcom
Saint Louis - PHILOSOPHY - 105
Rebecca Austin November 22, 2010 Clark PHIL 105The Ontological Argument for the Existence of GodThis paper addresses an ontological argument for the existence of God as presented by St. Anselm and Gaulinos counterargument. Anselm holds that God is the g
Saint Louis - PHILOSOPHY - 105
Rebecca Austin December 6, 2010 Clark PHIL 105Free Will: Compatibilism and IncompatibilismThis paper addresses the compatibilist view in support of free will as stated by Harry G. Frankfurt and the opposing incompatibilist argument as presented by Peter
Saint Louis - ENGL - 190
Rebecca Austin November 29, 2010 Cochran ENGL 190Keeping Secrets: How Secrets Affect Familial RelationshipsEveryone has a secret. In fact, most people, in all probability, keep multiple secrets daily. Secrets vary in intensity, importance, and magnitude
Saint Louis - ENGL - 190
Rebecca Austin September 27, 2010 Cochran ENG 190SchizophreniaThere is more to life than meets the eye. For a sizeable portion of the worlds population, there is an entire reality happening in their brain that is completely separate from the real world.
Saint Louis - IB - 200
Rebecca Austin March 30, 2011 Country AssignmentJamaicaJamaica is a small tropical island in the Bahamas. Located south of Cuba, Jamaica is characterized by mountainous, coastal plains and its vast beaches. Jamaica is home to three major cities, with Ki
Piedmont - ACC - 302.1
DueDate:March24,2011ACC 302.1 (2011-1) Practice Quiz 2Student: _1. Listed below are ten terms followed by a list of phrases that describe or characterize five of the terms. Match each phrase with the correct term by placing the letter designating the b
University of Cyprus - ARCH - 100
thThe 14 World Conference on Earthquake Engineering October 12-17, 2008, Beijing, ChinaKINETIC STRUCTURES IN ARCHITECTUREM.C. Phocas and T. Sophocleous-Lemonari112Asst. Professor, Department of Architecture, Faculty of Engineering, University of Cy
Wilfrid Laurier - BUS - 121
[ BU121 Study Notes The Balance Sheet T he balance sheet shows the financial position of a business at a given moment of t ime T he balance sheet simply shows what the fi rm owns, (assets), and what the f i rm owes (liabilities) Assets: Resources owned by
National Taipei University - CE - 208
3 AXIOMS OF PROBABILITY113Axioms of ProbabilityThe question here is: how can we mathematically dene a random experiment? What we have are outcomes (which tell you exactly what happens), events (sets comprising certain outcomes), and probability (which
Purdue - ECE - 301
Name: ECE 301: Quiz 1 Purdue University, Summer 2009 1. Given x(t) as plotted below, plot 4x(2t 3). Make sure to label your axes.1.5 1x(t)0.5 0 0.5 1.510.50t0.511.5Solution:43 4x(2t3)21010 t122. Let the signal y (t) = cos t + j sin t.
Purdue - ECE - 301
Name: ECE 301: Quiz 2 Purdue University, Summer 2009 1. Let the response y (t) to a continuous-time system be related to the input x(t) by the following relationship: y (t) = 2x(t2 ). With yes/no answers (no proofs necessary) state whether the system is:
Purdue - ECE - 301
Solution ECE 301: Quiz 4 Purdue University, Summer 2009 1. Recall that F cfw_x(t) = x(t)ejt dt. Compute the Fourier transform F cfw_et u(t), where &gt; 0 Solution: F cfw_et u(t) = et u(t)ejt dt et(+j) dt0= =1 + j2. Let an LTI system have impulse resp
Purdue - ECE - 301
Name: SOLUTION ECE 301: Quiz 5 Purdue University, Summer 2009 Consider a signal x(t) with a Fourier transform X ( ) = u( + 2) u( 2). Let x(t) be sampled by the system pictured below, with H ( ) = u( + Ts ) u( ) Ts1. What is the largest value of Ts whic
Purdue - ECE - 301
Name: ECE 301: Quiz 6 Purdue University, Summer 2009 1. Let x(t) = et u(t), for some complex number . Find Lcfw_x(t) = x(t)est dt, along with its region of convergence (ROC). Hint: The ROC is the set of s for which the integral converges. Solution: Lcfw
Bahria University - MBA - 1
Principles of Marketing- MBA-1 Bahria University QUIZ-2 Name:1.Marks-25 2. 3.You are directed to study the factors that are larger societal forces that affect your companydemographic, economic, natural, technological, political, and cultural. What are
UDLA - ECON - 101
According to alcoholscreening.org, my alcohol consumption habits are not likely to be harming my health because I dont drink more than the USDA Recommended Guidelines. These results tell me that I am aware of the harm of alcohol and understand how to mana
Liberty - ENGLISH - 101
Eduardo Ortez English 112 March 29, 2011 Poetry Essay Poetry Essay There are many factors that make a poetry piece superb. There is a vast variety of elements that can be present or sometimes even dominant within a poem. Poetic elements like imagery, soun
Simon Fraser - PHYS - 120
Mechanics and Modern PhysicsDr. Bernd StelzerPHYS 120 SFU Fall 2010 Dr. Bernd StelzerAnnouncementsMastering Physics (MP4) online assignment-Due today (9:00 PM) -Recommended browser is: Firefox Last written assignment (PS5) posted on WebCT Extra tuto
Simon Fraser - PHYS - 120
Mechanics and Modern PhysicsDr. Bernd StelzerPHYS 120 SFU Fall 2010 Dr. Bernd StelzerRecap: Simple Harmonic Motion (SHM)Sinusoidal oscillations are the most fundamental and are called simple harmonic oscillations 2t x ( t ) = A cos TWe define an ang