AP Chemistry: Acid-Base Chemistry Practice Problems

Name AP Chemistry: Acid-Base Chemistry Practice Problems Date Due Directions: Write your answers to the following questions in the space provided. For problem solving, show all of your work. Make sure that your answers show proper units, notation, and significant digits. 1. Although HCl and H 2 SO 4 have very different properties as pure substances, their aqueous solutions possess many common properties. List some general properties of these solutions, and explain their common behavior in terms of the species present. Solutions of HCl and H 2 SO 4 conduct electricity, taste sour, turn litmus paper red (are acidic), neutralize solutions of bases, and react with active metals to form H 2 (g). HCl and H 2 SO 4 solutions have these properties in common because both compounds are strong acids. That is, they both dissociate completely in H 2 O to form H + (aq) and an anion. (HSO 4 - is not completely dissociated, but the first dissociation step for H 2 SO 4 is complete.) The presence of ions enables the solutions to conduct electricity; the presence of H + (aq) in excess of 1 10-7 M accounts for all other properties listed. 2. Although pure NaOH and CaO have very different properties, their aqueous solutions possess many common properties. List some general properties of these solutions, and explain their common behavior in terms of the species present. Solutions of NaOH and CaO conduct electricity, taste bitter, turn litmus paper blue (are basic), and neutralize solutions of acids. NaOH and CaO solutions have these properties in common because both compounds form basic solutions in water. That is, they both dissociate in H 2 O to form OH - (aq) and a cation. The presence of ions enables the solutions to conduct electricity; the presence of OH - (aq) in excess of 1 10-7 M accounts for all other properties listed. 3. What is the difference between the Arrhenius and the Brønsted-Lowry definitions of an acid? The Arrhenius definition of an acid is confined to aqueous solution; the Brønsted-Lowry definition applies to any physical state. 4. Give the formulas for the conjugate base of H 2 SO 4 and the conjugate acid of CH 3 NH 2. HSO 4 - is the conjugate base of H 2 SO 4. CH 3 NH 3 + is the conjugate acid of CH 3 NH 2. 5. The hydrogen oxalate ion (HC 2 O - 4 ) is amphoteric. Write a balanced chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base toward water. - 2- Acid: HC 2 O 4 + H 2 O C 2 O 4 + H 3 O + - Base: HC 2 O 4 + H 2 O H 2 C 2 O 4 + OH - 6. The general reaction of an acid dissolving in water may be shown as HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) A conjugate acid base pair for this reaction is A. HA and H 2 O. B. HA and A -. C. H 2 O and A -. D. H 3 O + and A -. E. HA and H 3 O +. Answer: (B) The conjugate acid differs from its conjugate base by a proton, (H + ). The acid form has the proton (HA in this example) and the base form has lost the proton (A - in this case). 7. Write the dissociation reaction for each of the following acids. a. Acetic Acid (HC 2 H 3 O 2 ) HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2- + H 3 O + or HC 2 H 3 O 2 H + + C 2 H 3 O 2- b. The ammonium ion (NH 4 + ) NH 4 + + H 2 O NH 3 + H 3 O + or NH 4 + H + + NH 3 1

8. Identify the Lewis acid and the Lewis base in each of the following reactions. a. 2NH 3 (aq) + Ag + (aq) Ag(NH 3 ) + 2 (aq) base acid b. BF 3 (g) + F - (aq) BF - 4 (aq) acid base 9. Aluminum hydroxide is an amphoteric substance. It can act as either a Brønsted-Lowry base or a Lewis acid. Write a reaction showing Al(OH) 3 acting as a base toward H+ and as an acid toward OH-. Al(OH) 3 (s) + 3H + (aq) Al 3+ (aq) + 3H 2 O(l) (Brønsted-Lowry base, H + acceptor) Al(OH) 3 (s) + OH - (aq) Al(OH) - 4 (aq) (Lewis acid, electron pair acceptor) 10. Would you expect Fe 3+ or Fe 2+ to be the stronger Lewis acid? Explain. Fe 3+ should be the stronger Lewis acid. Fe 3+ is smaller and has a greater positive charge. Because of this, Fe 3+ will be more strongly attracted to lone pairs of electrons as compared to Fe 2+. 11. Differentiate between the terms strength and concentration as they apply to acids and bases. Strength refers to how well an acid or base dissociates in water. Concentration refers to the amount of acid or base dissolved in solution. 12. Strong acids are those which A. have an equilibrium lying far to the left. B. yield a weak conjugate base when reacting with water. C. have a conjugate base which is stronger base than water. D. readily remove the H + ions from water. E. are only slightly dissociated (ionized) at equilibrium. Answer: (B) Strong acids give up their protons (H + ions) easily. If their conjugate base form were strong, then the acid formed would hold strongly to the H +, which would not be a characteristic of a strong acid. 13. Classify each of the following as a strong acid or a weak acid in aqueous solution. a. HNO 2 Weak e. HClO 4 Strong b. HNO 3 Strong f. HClO Weak c. HCl Strong g. H 2 SO 4 Strong d. HF Weak h. HSO 4-14. Arrange the following in order of increasing acid strength. a. H 2 O K a = 1.0 10-14 Weak b. HNO 2 K a = 4.0 10-4 H 2 O < NH + 4 < HOCl < HNO 2 c. HOCl K a = 3.5 10-8 d. NH 4 + K a = 5.6 10-10 15. At the freezing point of water (0 C), K w = 1.2 10-15. Calculate [H + ] and [OH - ] for a neutral solution at this temperature. Since [H + ] = [OH - ] and K w = [H+][OH - ] = 1.2 10-15. 16. Calculate ph and poh for each of the following solutions at 25 C. a. 1.0 10-3 M OH - ph = 11.00; poh = 3.00 b. 1.0 M H + ph = 0.00; poh = 14.00 2

17. The [OH - ] of a certain aqueous solution is 1.0 10-5 M. The ph of this same solution must be A. 1.00 10-14. B. 5.00. C. 7.00. D. 9.00. E. 12.00. Answer: (D) Referring to ph + poh = 14.00 at 25 C, if [OH - ] = 1 10-5 M, poh = 5.0, ph = 14.00 5.0 = 9.0 18. Calculate the ph of 0.10 M HNO 3. This is a strong acid. Therefore the ph = -log[0.10] = 1.00 19. Calculate the ph of 5.0 10-2 M KOH. This is a strong base. Therefore the poh = -log[5.0 10-2 ] = 1.30. poh = 14-1.30 = 12.70. 20. Calculate the ph of a solution made by adding 15.00 g of sodium hydride (NaH) to enough water to make 2.500 L of solution. NaH + H 2 O NaOH + H 2 poh = -log(0.2500) = 0.6021 ph = 14 0.6021 = 13.3980 21. The calculation of concentration and ph for weak acids is more complex than for strong acids due to A. the incomplete ionization of weak acids. B. the low K a value for strong acids. C. the more complex atomic structures of strong acids. D. the low percent ionization of strong acids. E. the inconsistent K b value for strong acids. Answer: (A) Strong acids ionize 100%, therefore the [H+] is equal to the initial concentration of the strong acid. This is not the case for weak acids. Since not all of the acid ionizes, it is necessary to determine both how much of the acid forms H + and how much is left in molecular form. 22. When calculating the poh of a hydrofluoric acid solution (K a = 7.2 10-4 ) from its concentration, the contribution of water ionizing (K w = 1.0 10-14 ) is usually ignored because A. hydrofluoric acid is such a weak acid. B. hydrofluoric acid can dissolve glass. C. the ionization of water provides relatively few H + ions. D. the [OH - ] for pure water is unknown. E. the conjugate base of HF is such a strong base. Answer: (C) By comparing the K a for HF with K w (for water), you can see that even though hydrofluoric acid is a very weak acid it is a much stronger acid than is water, so much so that H + ion contributions of water may be ignored. 23. In many calculations for the ph of a weak acid from the concentration of the acid, an assumption is made that often takes the form [HA] 0 x = [HA] 0. A. is valid because x is very small compared to the initial concentration of the weak acid. B. is valid because the concentration of the acid changes by such large amounts. C. is valid because actual value of x cannot be known. D. is valid because ph is not dependent upon the concentration of the weak acid. E. approximation is always shown to be valid and so need not be checked. Answer: (A) We can expect x to be very small (as indicated by the low value for K a for weak acids), so that the concentration of the weak acid does not significantly change from its initial value. As the final step in such problems, you must determine that the change in the initial [HA] is less than 5% for the approximation to be considered valid. 3

24. Lactic acid (HC 3 H 5 O 3 ) has one acidic hydrogen. A 0.10 M solution of lactic acid has a ph of 2.44. Calculate K a. [H + ] = [C 3 H 5 O 3 - ] = antilog(-2.44) = 10-2.44 = 3.63 10-3 = x R HC 3 H 5 O 3 H + + - C 3 H 5 O 3 I 0.10 M 0 0 C -x + x + x E 0.10 M x 0.10 M 3.63 10-3 x 3.63 10-3 x 3.63 10-3 25. The acid-dissociation constant for benzoic acid (HC 7 H 5 O 2 ) is 6.3 10-5. Calculate the equilibrium concentrations of H +, C 7 H 5 O 2 -, and HC 7 H 5 O 2 in the solution if the initial concentration of HC 7 H 5 O 2 is 0.050 M. 26. Calculate the ph of a solution containing a mixture of 0.10 M HCl and 0.10 M HOCl. (The Ka for HOCl is 3.5 10-8 ). 27. The percent dissociation (percent ionization) for weak acids A. is always the same for a given acid, no matter what the concentration. B. usually increases as the acid becomes more concentrated. C. compares the amount of acid that has dissociated at equilibrium with the initial concentration of the acid. D. may only be used to express the dissociation of weak acids. E. has no meaning for polyprotic acids. Answer: (C) The percent dissociation does change with the concentration. For example, 1.0 M acetic acid does not ionize nearly as much as 0.10 M acetic acid. The term applies to the amount of acid dissociated/the initial concentration, expressed as a percent. 4

28. HA is a weak acid which 4.0% dissociated at 0.100 M. Determine the K a for this acid. A. 0.0040 D. 1.6 B. 0.0016 E. 16.5 C. 0.040 Answer: (B) [HA] 0 = 0.100 M 4.0% of 0.100 = 0.004 [H + ] = [A - ] = 0.004 M Note that although the actual value for [HA] is 0.100 0.004 = 0.096 M, this approximation is within the 5% rule, and you do not get to use your calculator for the multiple choice portion of the test. 29. Saccharin, a sugar substitute, is a weak acid with a pk a = 2.32 at 25 C. It ionizes in aqueous solution as follows: HNC 7 H 4 SO 3 (aq) H + (aq) + NC 7 H 4 SO 3 - (aq) What is the ph of a 0.10 M solution of this substance? K a = 10-2.32 = 4.8 10-3 x 2 = 4.8 10-3 (0.10-x) = 4.8 10-4 - 4.8 10-3 x x 2 + 4.8 10-3 x 4.8 10-4 x = 0.020 and -0.024 (discard) ph = -log(0.020) = 1.70 30. Phosphoric acid, H 3 PO 4, is a triprotic acid. a. Show the three equations involved in the dissociation of this substance. H 3 PO 4 (aq) H + (aq) + H 2 PO 4 - (aq) H 2 PO 4 - (aq) H + (aq) + HPO 4 2- (aq) HPO 4 2- (aq) H + (aq) + PO 4 3- (aq) b. Illustrate how these three equations might be combined to show the complete dissociation of phosphoric acid. H 3 PO 4 (aq) H + (aq) + H 2 PO 4 - (aq) H 2 PO 4 - (aq) H + (aq) + HPO 4 2- (aq) HPO 4 2- (aq) H + (aq) + PO 4 3- (aq) H 3 PO 4 (aq) 3H + (aq) + PO 4 3- (aq) c. If 7.0 M H 3 PO 4 solution dissociated, calculate the ph of the solution. 3 8 13 a1 a2 a3 K = 7. 5 10 K = 6. 2 10 K = 4. 8 10 Note that only the first dissociation forms significant amounts of H +, so use: If x = [H + ]=[H 2 PO 4 - ] Assume [H 3 PO 4 ] 0 = 7.0 x 7.0 M ph = -log[h+]= -log(0.23)=0.64 Note: Check to see that (7.0-x) (7.0-0.23) is about equal to 7.0, (this is within the allowable 5%, as the rule suggests, since 5% of 7.0 is 0.35), therefore the assumption is acceptable. 5

d. Determine the concentration for the ions H 2 PO 4 -, HPO 4 2-, and PO 4 3- (aq) in the dissociated 7.0 M H 3 PO 4 solution. From part (c), [H + ] = 0.23 M = [H 2 PO 4 - ]. H 2 PO 4 - (aq) H + (aq) + HPO 4 2- (aq) K a = 6.2 10-8 [HPO 4 2- ] = 6.2 10-8 M For [PO 4 3- ]: HPO 4 2- (aq) H + (aq) + PO 4 3- (aq) K a = 4.8 10-13 [H+]=0.23 M (from above) [HPO 4 2- ] = 6.2 10-8 M (also from above) e. Determine the poh for this same 7.0 M H 3 PO 4 solution. From part c, ph = 0.64. ph + poh = 14.00; ph = 14.00 0.64 = 13.36 31. Ionic substances known as salts can form acidic, basic, and neutral solutions when dissolved in water. When dissolved in water A. KNO 3 forms a basic solution. B. NaCl forms an acidic solution. C. NaNO 3 forms an acidic solution. D. NaF forms a basic solution. E. KClO 4 forms an acidic solution. Answer: (D) NaF is the salt produced from the reaction of a weak acid with a strong base. 32. Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced chemical equations for the reactions causing the solution to be acidic or basic. The relevant K a and K b values are found in Tables 14.2 and 14.3 of your textbook. a. NaNO 3 neutral b. NaNO 2 basic NO 2- + H 2 O HNO 2 + OH - c. C 5 H 5 NHClO 4 acidic C 5 H 5 NH + + H 2 O C 5 H 5 N + H 3 O + d. NH 4 NO 2 acidic (K a for NH 4 + > K b for NO 2 - ) NH 4 + + H 2 O NH 3 + H 3 O + e. KOCl basic OCl - + H 2 O HOCl + OH - f. NH 4 OCl basic (K b for OCl - > K a for NH 4 + ) OCl - + H 2 O HOCl + OH - 33. An unknown salt is either NaCN, NaC 2 H 3 O 2, NaF, NaCl, or NaOCl. When 0.100 mol of the salt is dissolved in 1.00 L of water, the ph of the solution is 8.07. What is the identity of the salt? 6

34. Calculate the ph of a 0.10 M CH 3 NH 3 Cl solutions. 35. What components must be present in order to have a buffered solution? For what purposes are buffers used? What is meant by the capacity of a buffer? A buffered solution must contain a weak acid or base and its salt. Buffered solutions are useful for controlling the ph of a solution since they resist ph change. The capacity of a buffer is a measure of how much strong acid or strong base the buffer can neutralize. 36. Write reactions to show how a solution containing acetic acid and sodium acetate acts as a buffer when a strong acid or base is added. H 3 O + - + C 2 H 3 O 2 HC 2 H 3 O 2 + H 2 O OH - - + HC 2 H 3 O 2 C 2 H 3 O 2 + H 2 O 37. Calculate the ph of a solution that is 1.00 M HNO 2 and 1.00 M NaNO 2. The K a of nitrous acid = 4.0 10-4. 38. Consider the acids in Table 14.2 of your textbook. Which acid would be the best choice for preparing a ph = 7.00 buffer? Explain how to make 1.0 L of this buffer. The best acid choice for a ph = 7.00 buffer would be the weak acid with a pk a close to 7.0 or a K a 1 10-7. HOCl is the best choice in Table 14.2 (Ka = 3.5 10-8 ; pka = 7.46). To make this buffer, we need to calculate the [base]/[acid] ratio. Any OCl-/HOCl buffer in a concentration ratio of 0.35:1 will have a ph = 7.00. One possibility is [NaOCl] = 0.35 M and [HOCl] = 1.0 M 7

39. Consider the titration of 40.0 ml of 0.200 M HClO 4 by 0.100 M KOH. Calculate the ph of the resulting solution after the following volumes of KOH have been added. a. 0.0 ml Since only the strong acid is present, the ph = -log(0.200) = 0.669 b. 10.0 ml ph = 0.854 c. 40.0 ml ph = 1.301 d. 80.0 ml ph = 7.00 e. 100.0 ml ph = 12.15 f. Which of the indicators in Fig. 15.8 (page 756) could be used for the titration? Bromothymol blue or phenol red 8

40. Consider the titration of 100.0 ml of 0.200 M acetic acid (K a = 1.8 10-5 ) by 0.100 M KOH. Calculate the ph of the resulting solution after the following volumes of KOH have been added. a. 0.0 ml ph = 2.72 b. 50.0 ml ph = 4.26 c. 100.0 ml ph = 4.74 d. 150.0 ml ph = 5.22 e. 200.0 ml ph = 8.79 f. 250.0 ml ph = 12.15 g. Which of the indicators in Fig. 15.8 (page 756) could be used for the titration? o-cresolphthalein or phenolphthalein 9

41. Write balanced net ionic equations for each of the following. a. Ethanol is burned in oxygen. C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 O b. A strip of magnesium is heated strongly in pure nitrogen gas. 3Mg + N 2 Mg 3 N 2 c. Dinitrogen trioxide gas is bubbled into water. N 2 O 3 + H 2 O 2HNO 2 d. A small piece of sodium metal is added to distilled water. 2Na + 2H 2 O 2Na + + 2OH - + H 2 e. A solution of sodium bromide is added to an acidified solution of potassium bromate. This is a redox reaction. Br - Br 2 - BrO 3 Br 2 2Br - Br 2 12H + - + 2BrO 3 Br 2 + 6H 2 O 2Br - Br 2 + 2e - 10e - + 12H + - + 2BrO 3 Br 2 + 6H 2 O 10Br - 5Br 2 + 10e - 10e - + 12H + - + 2BrO 3 Br 2 + 6H 2 O 10Br - + 12H + - + 2BrO 3 5Br 2 + Br 2 + 6H 2 O 10