Name____________ANSWER KEY____________ Date____________________ Chapter 14/15 Review Fill in the blanks: Arrhenius defined acids as producing 1 2 in aqueous solution. The 3 and a(n) 5 in aqueous solution and bases as producing definitions are more general: a(n) 4 is a proton acceptor. Water is said to be 6 because it can act as either acid or base. For strong acids and bases, the equilibrium lies far to the the K value is is a proton donor 7 . For weak acids and bases, 8 . An acid containing more than one acidic proton is called concentrations of 10 and/or 11 9 . The ions in solution determines the pH, which is calculated by 12 . 13 can be acidic, basic, or neutral when dissolved in water. dissolved in water. 15 is acidic when dissolved in water, and 16 14 is neutral when is basic. When determining pH of these solutions, a new K value must be calculated using the formula The value of Kw is 17 . 18 . When a weak acid and its conjugate base are dissolved in the same solution, a exists. This solution resists pH changes because the ratio of 20 is almost constant. 1. H+ / hydronium 11. OH– 2. OH– / hydroxide 12. pH = – log [H+] 3. Bronsted-Lowry 13. Salts 4. acid 14. NaCl (various answers) 5. base 15. NH4Cl (various answers) 6. amphoteric 16. NaCH3COO (various answers) 7. right 17. Ka x Kb = Kw 8. small 18. 1.0 x 10–14 9. polyprotic 19. buffer 10. H+ 20. [A–] : [HA] 19__ Acid HClO2 HF HNO2 CH3COOH HOCl HCN Ka 1.2 x 10–2 7.2 x 10–4 4.0 x 10–4 1.8 x 10–5 3.5 x 10–8 6.2 x 10–10 Base NH3 CH3NH2 C6H5NH2 C5H5N (C2H5)2NH Kb 1.8 x 10–5 4.38 x 10–4 3.8 x 10–10 1.7 x 10–9 1.3 x 10–3 1. Calculate the pH for 0.25 M solutions of the following: a. Ba(OH)2 Ba(OH)2 → Ba2+ + 2 OH– [OH–] = 2 x 0.25 M = 0.50 M pH = 13.70 b. LiCN CN– + 0.25 – x H2 O HCN + x x OH– Kb = Kw = 1.0 x 10–14 = x2 Ka 6.2 x 10-10 0.25 – x = [OH ] = 2.0 x 10–3 M pH = 11.30 c. C5H5N C5H5N + H2O C5H5NH+ + OH– 0.25 – x x x Kb = 1.7 x 10–9 = x2 0.25 x = [OH–] = 2.1 x 10–5 M pH = 9.31 d. NH4NO3 NH4+ 0.25 - x NH3 + H+ x x Kw = 1.0 x 10–14 = x2 Kb 1.8 x 10–5 0.25 + x = [H ] = 1.2 x 10-5 M Ka = pH = 4.93 2. Calculate the percent dissociation for a 0.35 M solution of hydrofluoric acid. HF 0.35 – x H+ + F– x x 7.2 x 10–4 = x2 0.35 x = [H+] = 1.6 x 10–2 M 1.6 x 10–2 M x 100 = 4.5% 0.35 M 3. The pH of a 0.063 M solution of HOBr is 4.95. Calculate Ka. [H+] = 10–4.95 = 1.1 x 10–5 M HOBr 0.063 – 1.1 x 10–5 H+ + 1.1 x 10–5 OBr– 1.1 x 10–5 Ka = (1.1 x 10–5)2 = 2.0 x 10–9 0.063 – 1.1 x 10–5 4. Are solutions of the following salts acidic, basic, or neutral? For those which are not neutral, write balanced equations for the reactions causing the solution to be acidic or basic. a. KCl neutral d. KF basic F– + H2O HF + OH– b. NaCH3COO basic CH3COO– + H2O CH3COOH + OH– e. NH4I acidic NH4+ NH3 + H+ c. CH3NH3Cl acidic + CH3NH3 CH3NH2 + H+ f. Mg(CN)2 basic CN– + H2O HCN + OH– 5. a. Calculate the pH of a solution which is 0.50 M CH3NH2 and 0.70 M CH3NH3Cl. CH3NH2 + H2O CH3NH3+ + OH– 0.50 – x 0.70 – x x 4.38 x 10–4 = 0.70 x 0.50 x = [OH–] = 3.1 x 10–4 M pH = 10.50 b. Calculate the pH after 0.10 mol of NaOH is added to one liter of the above solution. OH– + CH3NH3+ → CH3NH2 + H2O 0.10 0.70 0.50 -0.10 -0.10 +0.10 0 0.60 0.60 x = [OH–] = 4.38 x 10–4 M pH = 10.64 c. Calculate the pH after 0.10 mol of HCl is added to one liter of the solution in part a. H+ + 0.10 -0.10 0 CH3NH2 → CH3NH3+ 0.50 0.70 -0.10 +0.10 0.40 0.80 4.38 x 10–4 = 0.80 x 0.40 x = [OH–]= 2.2 x 10–4 M pH = 10.34 6. The molar solubility of silver carbonate is 0.032 M. Calculate Ksp. Ag2CO3(s) 2 Ag+(aq) + CO32–(aq) 0 0 + 2x +x 0.064 0.032 x = 0.032 M Ksp = (0.064)2 (0.032) = 1.3 x 10–4 7. Calculate the molar solubility of Ag2SO4 (Ksp = 1.5 x 10–5) in the following solutions: (a) pure water (b) 0.15 M Al2(SO4)3 (a) Ag2SO4(s) (b) Ag2SO4(s) 2 Ag+(aq) + SO42–(aq) 0 0 +2x +x 2x x 2 Ag+(aq) + SO42–(aq) 0 0.45 + 2x +x 2x 0.45 + x (2x)2 x = 1.5 x 10–5 x = mol sol = 0.016 M (2x)2 (0.45) = 1.5 x 10–5 x = 2.9 x 10–3 M 8. A solution of hydrochloric acid of unknown concentration was titrated with 0.10 M NaOH. If a 100.-mL sample of the HCl solution required exactly 35.0 mL of the NaOH solution to reach the equivalence point, what was the pH of the HCl solution? (M of HCl) (100. mL) = (0.10 M) (35.0 mL) M of HCl = 0.035 M pH = 1.46 9. A 400.0 mL sample of 0.100 M acetic acid is titrated with 0.200 M NaOH. Calculate the pH after the addition of the following volumes of base. a. 30.0 mL c. 200.0 mL b. 100.0 mL d. 300.0 mL a. [CH3COOH] = (0.100 M)(400.0 mL) / (430.0 mL) = 0.0930 M [OH–] = (0.200 M)(30.0 mL) / (430.0 mL) = 0.0140 M CH3COOH + OH– → 0.0930 0.0140 – 0.0140 –0.0140 0.0790 0 CH3COO– + H2O 0 + 0.0140 0.0140 CH3COOH CH3COO– + H+ 0.0790 0.0140 x b. 0.0140 x = 1.8 x 10–5 0.0790 x = [H+] = 1.0 x 10–4 M pH = 3.99 [CH3COOH] = (0.100 M)(400.0 mL) / (500.0 mL) = 0.0800 M [OH–] = (0.200 M)(30.0 mL) / (500.0 mL) = 0.0400 M half-equivalence point: pH = pKa = 4.74 c. [CH3COOH] = (0.100 M)(400.0 mL) / (600.0 mL) = 0.0667 M [OH–] = (0.200 M)(30.0 mL) / (600.0 mL) = 0.0667 M CH3COOH + OH– → 0.0667 0.0667 – 0.0667 –0.0667 0 0 CH3COO– + H2O 0 + 0.0667 0.0667 CH3COO– + H2O CH3COOH + OH– 0.0667 x x x2 = 1.0 x 10–14 0.0667 1.8 x 10–5 x = [OH–] = 6.1 x 10–6 M pH = 8.78 d. [CH3COOH] = (0.100 M)(400.0 mL) / (700.0 mL) = 0.0571 M [OH–] = (0.200 M)(30.0 mL) / (700.0 mL) = 0.0857 M CH3COOH + OH– → 0.0571 0.0857 – 0.0571 –0.0571 0 0.0286 CH3COO– + H2O 0 + 0.0571 0.0571 [OH–] = 0.0286 M pH = 12.457 10. Consider the titration of 100.0 mL of 0.100 M HCl by 0.200 M NaOH. Calculate the pH of the resulting solution after each of the following volumes of NaOH has been added. a. 0.0 mL c. 50.0 mL b. 30.0 mL d. 75.0 mL a. 0.100 M H+ pH = 1.000 b. [H+] = (0.100 M)(100.0 mL) / (130.0 mL) = 0.0769 M [OH–] = (0.200 M)(30.0 mL) / (130.0 mL) = 0.0462 M H+ + 0.0769 –0.0462 0.0307 OH– → H2O 0.0462 –0.0462 0 [H+] = 0.0307 M pH = 1.513 c. [H+] = (0.100 M)(100.0 mL) / (150.0 mL) = 0.0667 M [OH–] = (0.200 M)(30.0 mL) / (150.0 mL) = 0.0667 M d. [H+] = (0.100 M)(100.0 mL) / (175.0 mL) = 0.0571 M [OH–] = (0.200 M)(30.0 mL) / (175.0 mL) = 0.0857 M H+ + 0.0571 –0.0571 0 OH– → H2O 0.0857 –0.0571 0.0286 [OH–] = 0.0286 M pH = 12.457 equivalence point pH = 7.000