CHM 152 – Acids and Bases and Salts

Name: __KEY______________________ CHM 152 – Acids and Bases and Salts (Ch. 14) When solving problems identification of the chemicals is very important. Is the chemical a strong acid (ionizes 100%), a weak acid (partial ionization and has a Ka), a strong base (ionizes 100%), a weak base (partial ionization and has a Kb), or a salt? If a salt, is it soluble? If yes, are the product ions neutral, acidic or basic? Neutral ions do not react further but acidic and basic ions react with water. Remember acids and acidic ions produce hydronium ions in water while bases and basic ions produce hydroxide ions in water. Good luck! Acid-Base Concepts 1. List the strong acids: __HCl, HBr, HI, HNO3, H2SO4, HClO4, HClO3___ 2. List the strong bases: __LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2____ 3. Write an equation for the dissociation (ionization) of the following acids in water: (an example has been done for you) a) HClO4 → H+ + ClO4b) H2SO4 → H+ + HSO4c) HC2H3O2 → H+ + C2H3O2d) H2S → H+ + HSe) HCl → H+ + Clf) HNO3 → H+ + NO3- 4. Write the balanced reaction for what happens when nitric acid is put in water. Draw the resulting solution in the beaker. HNO3(aq) + H2O(l)  NO3-(aq) + H3O+(aq) 5. Write the balanced reaction for what happens when acetic acid is put in water. Draw the resulting solution in the beaker. CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq) Page 1 of 11 Name: __KEY______________________ 6. Write an equation for the dissociation (ionization) of the following bases in water: been done for you) a) Mg(OH)2 → Mg2+ + 2OH- (an example has b) NaOH → Na+ + OHc) Ba(OH)2 → Ba2+ + 2OHd) KOH → K+ + OHe) LiOH → Li+ + OH7. Write the balanced reaction for what happens when barium hydroxide is put in water. Draw the resulting solution in the beaker. Ba(OH)2(s)  Ba2+ (aq) + 2 OH- aq) 8. Draw a picture of 3 NaOH units in water in this beaker. 9. What is the conjugate base of HSO4-? __SO42-__________ 10. a. Write the formula and name for the conjugate acid for the following bases: NH3 c. acid = ___NH4+_ name = ___ammonium ion___ b. PO43- c. acid = _HPO42-__name = _monohydrogen phosphate ion_ c. CH3COO- c. acid = __CH3COOH__ name = __acetic acid___ d. F- 11. c. acid = ___HF__ name = ___hydrofluoric acid_ Write the formulas of the conjugate acids for the following bases: a. CN- b. HCO3- c. NH3 d. PO43- conjugate acid is ____HCN______ conjugate acid is _____H2CO3____ conjugate acid is ____NH4+___ conjugate acid is ____HPO42-_____ Page 2 of 11 Name: __KEY______________________ 12. Write a balanced equation for the Bronsted-Lowry acid HPO42- in water. HPO42- + H2O(l)  H3O+(aq) + PO43-(aq) 13. Write and balance the reaction for ammonia in water. NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) 14. What is the strongest base among the following? a. b. c. d. ClOClO2ClO3ClO4- 15. From the following chemical reaction, determine the relative Bronsted-Lowry acid strengths (from strongest to weakest) HClO4 (aq) + H2O (l) → H3O+ (aq) + ClO4- (aq) HNO2 (aq) + H2O (l)  H3O+ (aq) + NO2- (aq) a. b. c. d. H3O+ > HNO2 > HClO4 H3O+ > HClO4 > HNO2 HClO4 > HNO2 > H3O+ HClO4 > H3O+ > HNO2 16. Circle the following acids if they ionize almost 100%: HI HF HNO3 H2CO3 HClO4 17. Which of the following depicts the strongest acid? ____A___ pH 18. What is the equation used for finding pH? _____ -log ([H3O+]) _____________________ 19. What is the equation that relates pH and pOH? ___pH + pOH = 14_______________ Page 3 of 11 Name: __KEY______________________ 20. Are the following solutions acidic, neutral, or basic? a) [H3O+] = 1 x 10-8 M _basic (barely, pH = 8)__ e) [OH-] = 1 x 10-3 M b) [H3O+] = 7.3 x 10-1 M _acidic (pH = 0.137)_ f) [OH-] = 1 x 10-7 M c) [H3O+] = 1.0 M _acidic (pH = 0)__ g) [OH-] = 0.00001 M + -5 d) [H3O ] = 1 x 10 M _acidic (pH = 5)__ h) [OH-] = 4.9 x 10-6 M 21. If the pOH of a solution is 4.52, calculate the pH, the [H+] and the [OH-]. _basic (pH = 11)__ _neutral (pH = 7)__ _basic (pH = 9)__ _basic (pH = 8.69)_ pH = 14 – 4.52 = 9.48 [H+] = 10-pH = 3.3 x 10-10 M [OH-] = 10-pOH = 3.0 x 10-5 M 22. Complete the following table: (first one is done as an example) a b c d e f g h i j k l m n o p [H3O+] [OH-] pH pOH 1.0 x 10-3 1.0 x 10-9 1.0 x 10-10 1.0 x 10-8 0.010 2.19 x 10-9 4.5 x 10-5 1.6 x 10-7 1.0 x 10-7 3.2 x 10-3 0.56 5.98 x 10-11 9.5 x 10-8 0.00453 9.009 x 10-11 1.1 x 10-11 1.0 x 10-11 1.0 x 10-5 1.0 x 10-4 1.0 x 10-6 1.0 x 10-12 0.00000456 2.2 x 10-10 6.2 x 10-8 1.0 x 10-7 3.1 x 10-12 1.8 x 10-14 1.67 x 10-4 1.0 x 10-7 2.21 x 10-12 0.000111 9.3 x 10-4 3.00 9.00 10.00 8.00 2.00 8.660 4.35 6.79 7.00 2.49 0.26 10.223 7.02 2.344 10.0453 10.97 11.00 5.00 4.00 6.00 12.00 5.340 9.65 7.21 7.00 11.51 13.74 3.777 6.98 11.656 3.9547 3.03 23. What would be the pH of each of the following: (DANGER! beware of tricks...) a) 0.0010 M HCl __3.00___ b) 0.0010 M HNO3 __3.00__ c) 0.010 M NaOH __12.00__ d) 0.0035 M HCl __2.46__ e) 1.0 M HBr __0__ f) 0.024 M HCl __1.62__ g) 0.075 M KOH _12.88___ h) 0.000034 M HCl __4.47__ i) 0.000000000001M HCl __7___ j) 12 M HCl _0___ 24. Determine the pH and pOH of the following 4 solutions: a) A 4.5 x 10-3 M HBr solution. Strong acid: pH = -log (H3O+) = 2.35, pOH = 14 – 2.35 = 11.65 Page 4 of 11 Acidic, Basic, or Neutral? Acidic Basic Basic Basic Acidic Basic Acidic Acidic/Neutral Neutral Acidic Acidic Basic Neutral Acidic Basic Basic Name: __KEY______________________ b) A 3.67 x 10-5 M KOH solution. Strong base: pOH = -log (OH-) = 4.435; pH = 14 – 4.435 = 9.565 c) A solution made by diluting 25 mL of 6.0 M HCl until the final volume of the solution is 1.75 L. M1V1 = M2V2 (6.0M)(0.025 L) = (M2)(1.75 L) M2 = 0.0857 M; pH = 1.07; pOH = 12.93 d) 5.0 L of an aqueous solution that contains 1.0 grams of HBr and 1.0 grams of nitric acid. 1.0 g HBr (1 mol / 79.908 g) = 0.0125 mol; 1.0 g HNO3 (1 mol / 63.008) = 0.0159 mol total moles = 0.0284 mol / 5.0 L = 0.00568 M; since both acids are strong, pH = -log (conc.) pH = 2.25; pOH = 11.75 25. What is the pH of a 1.5 x 10-10 HBr solution? If you do the calculation to find –log (1.5 x 10-10), you get an answer of 9.8. However, this is intuitively incorrect – after all, how can a solution that’s made of nothing but pure water (pH = 7) with an acid added be basic overall? The answer, it can’t. The actual pH of the solution is just about 7, with the main acid source being the H+ formed from the autoionization of water. 26. As acid strength increases, the % ionization ____, the [H+] __ __ and the pH __ __. 27. Determine the pH and pOH of a 0.0034 M HNO3 solution. pH = -log[H+] = -log(0.0034) = 2.47 pOH = 14 – 2.47 = 11.53 28. Determine the pH and pOH of a 4.3 x 10-4 M NaOH solution. pOH = -log[OH-] = -log(4.3 x 10-4) = 3.37 pH = 14 – pOH = 14 – 3.37 = 10.63 29. What is the hydroxide concentration and pH if an aqueous solution has a hydronium concentration of 3.57 x 10-9 M. Is this solution acidic or basic? __basic__ [OH-] = Kw / [H+] = 1.0 x 10-14 / 3.57 x 10-9 = 2.80 x 10-6 M pH = -log[H+] = 8.447 30. What is the pH of a solution that contains 25 grams of hydrochloric acid (HCl) dissolved in 1.5 liters of water? 25 g (1 mol / 36.45 g) = 0.68587 moles of HCl /1.5 L = 0.457 M. pH = -log(0.457) = 0.34 31. What is the pH of a solution that contains 1.32 grams of nitric acid (HNO3) dissolved in 750 mL of water? 1.32 g (1 mol / 63.008 g) = 0.02095 mol / 0.750 L = ; pH = -log (0.0279) = 1.55 Page 5 of 11 Name: __KEY______________________ 32. What is the pH of a solution that contains 1.2 moles of nitric acid (HNO3) and 1.7 moles of hydrochloric acid (HCl) dissolved in 1000 liters of water? 1.2 mol + 1.7 mol = 2.9 mol / 1000 L = 0.0029 M; pH = -log(0.0029) = 2.54 33. If a solution has a [H+] concentration of 4.5 x 10-7 M, is this an acidic or basic solution? Explain. The pH of this solution is 6.35, making the solution very slightly acidic. 34. An acidic solution has a pH of 4. If I dilute 10 mL of this solution to a final volume of 1000 mL, what is the pH of the resulting solution? M1V1 = M2V2, where M1 = 10-4 = 1. 0 x 10-4, V1 = 0.010 L, and V2 is 1000 mL; M2 = 1.00 x 10-6 M; pH = 6. 35. What is the pH of a 0.05 M barium hydroxide solution? Ba(OH)2 → Ba2+ + 2OH0.05 M x 2 = 0.10 M; pOH = -log (0.10 M) = 1.00 pH = 14 – 1.00 = 13 36. Calculate the pH of an aqueous solution that contains 2.50 x 10-4 M [H3O+]. pH = -log (2.50 x 10-4) = 3.60 37. Calculate the pH of an aqueous solution that contains 3.50 x 10-3 M [OH-]. pOH = -log (3.50 x 10-3) = 2.4559; pH = 14 – 2.4559 = 11.544 38. What is the pH of a solution made by mixing 100.00 mL of 0.020 M Ca(OH)2 with 50.0 mL of 0.100 M NaOH? Assume that the volumes are additive. (0.020 M * 0.10000 L) = 0.0020 mol Ca(OH)2 x 2 = 0.0040 mol OH-; (0.100 M * 0.0500 L) = 0.005 mol 0.004 + 0.005 mol = 0.009 mol / (0.10000 + 0.0500 L) = 0.06 M pOH = - log (0.06) = 1.222; pH = 12.77 39. Calculate the pH of a 3.56 x 10-4M solution of hydroiodic acid. HI is strong so reacts completely so [H+] = 3.56 x 10-4M thus pH = 3.449 40. Determine the pH of a 4.5 x 10-11 M NaOH solution. Although there is some NaOH present in the solution, the pH isn’t found by taking the –log of anything. The reason for this is that the concentration of base is much, much smaller than the concentration of acid which is naturally found in neutral water. As a result, this base doesn’t really have any affect on the pH, so the pH of the solution is 7.00. 41. Why do we say that a solution with a H+ concentration of 1.00 x 10-7 M is said to be neutral. If it contains acid, shouldn’t it be acidic? It isn’t acidic because while there is some acid in the solution, there is an equal quantity of base. In neutral solutions, the H+ and OH- concentrations are identical, because water breaks up to form them. As a result, the solution is neither acidic nor basic. Page 6 of 11 Name: __KEY______________________ 42. Explain why even a basic solution contains some H+ ions. All aqueous solutions contain H+ ions from the autoionization of water, H2O → H+ + OH-. 43. Explain why even an acidic solution contains some OH- ions. The same answer from #42 applies here. Equilibrium Concentration/Constant of Weak Acids/Bases 44. What is the equilibrium constant expression (Ka) for the acid dissociation of formic acid (HCOOH)? (Hint: Write a balanced equation for the dissociation of formic acid in water first.) HCOOH + H2O H3O+ + COOH- Ka = [H3O+][COOH-] / [HCOOH] 45. What is the pH of a 0.100 M formic acid solution with a Ka = 1.8 x 10-4? HA + H2O  H3O+ + A0.100 0 0 -x +x +x 0.100 – x x x Ka = x2 / (0.100 – x) = 1.8 x 10-4 Use quadratic: x2 + 1.8 x 10-4x – 1.8 x 10-5 = 0 Solve for x: x = 0.0041536; pH = - log [H3O+] = 2.38 46. What is the approximate pH of a 0.50 M solution of nitrous acid? Hint: What is the formula for nitrous acid and its Ka value? HNO2 + H2O  H3O+ + NO2- 0.50 -x 0.50 – x - 0 +x x Ka = x2 / (0.100 – x) = 4.5 x 10-4 Use quadratic: x2 + 4.5 x 10-4x – 2.25 x 10-4 = 0 Solve for x: x = 0.01478; pH = - log [H3O+] = 1.83 0 +x x 47. A tablet containing 500.0 mg of aspirin (acetylsalicylic acid or HC9H7O4) was dissolved in enough water to make 100.0 mL of solution. Given that Ka = 3.0 x 10-4 for aspirin, what is the pH of the solution? 0.5000 g (1 mol / 180.163 g) = 0.0027753 mol / 0.1000 L = 0.027753 M HA + H2O  H3O+ + A- 0.027753 -x 0.027753 – x - 0 +x x 0 +x x Ka = x2 / (0.027753 – x) = 3.0 x 10-4 Use quadratic: x2 + 3.0 x 10-4x – 8.3259 x 10-6 = 0 Solve for x: x = 0.002739; pH = - log [H3O+] = 2.56 Page 7 of 11 Name: __KEY______________________ 48. What is the pH of a 0.30 M pyridine solution that has a Kb = 1.9 x 10-9? C5H5N (aq) + H2O (l)  C5H5NH+ (aq) + OH- (aq) 0.30 0 0 -x +x +x Kb = x2 / (0.30 – x) = 1.9 x 10-9 0.30 – x x x Assume x << 0.30; Kb = x2 / 0.30 x2 = 1.9 x 10-9 * 0.30 = 5.7 x 10-10 x = √5.7 x 10-10 = 2.3875 x 10-5 pOH = - log (OH-) = 4.62; pH = 9.38 49. An amine solution with a concentration of 0.855M has a pH of 10.35. Calculate the equilibrium constant Kb for this amine. (Hint – write a generic reaction) B(aq) + H2O(l)  BH+(aq) + OH-(aq) I .855 0 0 C -x x x E (.855-x) x x pOH = 3.65 so [OH-] = 2.24 x 10-4 = x Kb = (2.24 x 10-4)2 / (0.855 - 2.24 x 10-4) = 5.9 x 10-8 50. Show all your work including the reactions. Calculate the pH if 2.56 grams of sodium fluoride is dissolved in 4.50 liters of water. Ka for hydrofluoric acid is 7.1 x 10-4 2.56 g NaF (1 mole / 41.99g) = .060967 mol / 4.50L = .01355 M NaF(s)  Na+(aq) + F-(aq) reacts all the way so [ions] = 0.01355 M F-(aq) + H2O(l)  HF(aq) + OH-(aq) I 0.01355 0 0 C -x +x +x E (0.01355-x) x x Kb = 1.41 x 10-11 = x2 / 0.01355 pOH = 6.36 so pH = 7.64 base rxn need Kb = 1.41 x 10-11 x = 4.369 x 10-7 = [OH-] check assumption Percent Dissociation 51. What is the percent dissociation of a benzoic acid (C6H5COOH) solution with pH = 2.59? The acid dissociation constant for this monoprotic acid is 6.5 x 10-5. HA + H2O  H3O+ + Ay 0 0 -x +x +x y–x x x (Use y as the initial concentration of benzoic acid; this is what you need to solve for in the equilibrium expression.) Ka = x2 / (y – x) = 6.5 x 10-5 [H3O+]eq = 10-pH = 0.00257 M = x Ka = (0.002572) / (y – 0.00257) = 6.5 x 10-5 y = 0.104 M; % diss. = (0.00257 / 0.104) *100 = 2.47% Page 8 of 11 Name: __KEY______________________ Polyprotic Acids 52. Calculate the concentration of carbonate ion, CO32-, in a 0.010 M H2CO3 solution that has the stepwise dissociation constants Ka1 = 4.3 x 10-7 and Ka2 = 5.6 x 10-11. First dissociation: H2CO3 + 0.010 -x 0.010 – x H2O  HCO3- + H3O+ 0 0 +x +x x x Ka1 = x2 / (0.010 – x) = 4.3 x 10-7 Assume x << 0.010 (and check it!) x2 = 4.3 x 10-9; x = 6.557 x 10-5 = [HCO3-] = [H3O+] Second dissociation: HCO3- + H2O  CO32- + H3O+ Ka2 = [CO32-][H3O+] / [HCO3-] = 5.6 x 10-11 + [H3O ] and [HCO3 ] are known from the first dissociation. Ka2 = ([CO32-] * 6.557 x 10-5) / 6.557 x 10-5 = 5.6 x 10-11 [CO32-] = 5.6 x 10-11 M 53. Write out all the balanced reaction steps for phosphoric acid to become phosphate ion in water. H3PO4 (aq) + H2O(l) H2PO4-(aq) + H3O+(aq) H2PO4-(aq) + H2O(l) HPO42-(aq) + H3O+(aq) HPO42-(aq) + H2O(l) PO43-(aq) + H3O+(aq) Ka and Kb 54. Ammonia has a Kb of 1.8 x 10-5. What is the conjugate acid of ammonia and what is its Ka? Conj. Acid: NH4+; Ka = Kw / Kb = 1.0 x 10-14 / 1.8 x 10-5 = 5.6 x 10-10 Salts 55. Identify the following as strong or weak acids, strong or weak bases, neutral salts, basic salts or acidic salts. LiCl __ neutral salt _______, NH3_______ weak base____, NH4ClO4___ acidic salts _____, HClO4__ strong acid__, Ba(OH)2__ strong base ___, Ba(CH3COO)2__basic salts __, HF__weak acid ______, NaF_____ basic salt_____, KOH_____ strong base ____, AlBr3____ acidic salt __, K2CO3___ basic salt______, Ca(NO3)2___ neutral salts ____. 56. Write a balanced chemical equation to show that sodium cyanide (NaCN) is basic in water solution. NaCN → Na+ + CN- CN- + H2O  HCN + OH- Page 9 of 11 Name: __KEY______________________ 57. What is the Kb expression and value for the equilibrium above? Kb = [HCN][OH-] / [CN-] Kb = Kw / Ka = 1.0 x 10-14 / 4.9 x 10-10 = 2.0 x 10-5 58. Which one of the following salts, when dissolved in water, produces the solution with the highest pH? a. KI b. KBr c. KF d. KCl 59. If an equal number of moles of HCN and KOH are added to water, what is the resulting solution: acidic, basic, or neutral? Explain. KOH is a strong base; HCN is a weak acid. When reacted together, they’ll produce water and KCN. This salt will dissolve 100%, leaving K+ (no further reaction) and CN- (which will react with water). This will produce a basic solution (see eqns from #56). 60. Calculate the pH of a 0.100 M NaCH3COO solution. Ka for acetic acid is 1.8 x 10-5. (Use A- as CH3COO-): A- + H2O  HA + OH0.100 0 0 Kb = x2 / (0.100 – x) = 5.6 x 10-10 -x +x +x Assume x << 0.100 (and check it!) 0.100 – x x x Kb = x2 / 0.100 = 5.6 x 10-10 x2 = 5.6 x 10-11; x = 7.48 x 10-6; pOH = 5.13, pH = 8.87 61. Calculate the pH of a 0.100 M CH3NH3Cl solution. Kb for methylamine, CH3NH2, is 3.7 x 10-4. (Use BH+ as CH3NH3+): BH+ + H2O  H3O+ + B 0.100 0 0 Ka = x2 / (0.100 – x) = 2.7 x 10-11 -x +x +x Assume x << 0.100 (and check it!) 0.100 – x x x Ka = x2 / 0.100 = 2.7 x 10-11 x2 = 2.7 x 10-12; x = 1.64 x 10-6; pH = 5.78 Acid Strength 62. Rank the following acids in order of increasing strength: HCl, HBr, HI, HF, H2S, H2Se Weakest: H2S, H2Se ≈ HF, HCl, HBr, HI Strongest Page 10 of 11 Name: __KEY______________________ Lewis Acids/Bases 63. Which one of the following is not considered a Lewis base? a. H2O b. NH3 c. NH4+ d. Cl- Page 11 of 11