pH = -log[H+] pH = -log[H+] pOH = -log[OH-] pH = -log[H+] Acidic/Basic   5.55 9.17 1.12 x10-5 M 6.28 x 10-1 M acidic 2.82 x 10-6 3.55 x 10-9 8.45 acidic 1.48 x 10-5 6.76 x 10-10 4.83 8.93 x 10-10 basic 9.05 4.95 1.59 x 10-14 0.20 13.8 very acidic pH = -log[H+] pH = -log[H+] pOH = -log[OH-] pH = -log[H+] pH = -log[8.93 x 10-10] pH = -log[6.28 x 10-1] 9.17 = -log[OH-] 5.55 = -log[H+] [H+][OH-] = 1 x 10-14 [H+][OH-] = 1 x 10-14 [H+][OH-] = 1 x 10-14 [H+][OH-] = 1 x 10-14 pOH = -log[OH-] pOH = -log[OH-] [OH-] = 10-9.17 [H+] = 10-5.55 [H+][1.12 x 10-5] = 1 x 10-14 [6.28 x 10-1][OH-] = 1 x 10-14 [2.82 x 10-6][OH-] = 1 x 10-14 [H+][6.76 x 10-10] = 1 x 10-14 pH = -log[1.59 x 10-15] pH = -log[1.12 x 10-5] pH + pOH = 14 pH + pOH = 14 [H+] = 8.93 x 10-10 [H+] = 3.55 x 10-9 [H+] = 1.48 x 10-5 [OH-] = 1.59 x 10-14

acid = HC2H3O2 , conjugate base = C2H3O2- ions 2. a) HC2H3O2 + H2O → H3O+ + C2H3O2- acid = HC2H3O2 , conjugate base = C2H3O2- ions base = H2O , conjugate acid = H3O+ ions b) OH- + HCO3- → CO32- + H2O acid = HCO3- ions, conjugate base = CO32- ions base = OH- ions, conjugate acid = H2O c) SO32- + HF → HSO3- + F- acid = HF, conjugate base = F- ions base = SO32- ions, conjugate acid = HSO3- ions

a) @ equivalence point there is 27.4 mL of 0.135 M NaOH . . . . (0.135)(0.0247) = 3.7 x 10-3 mols NaOH since HA is monoprotic, stoichometric equivalent amounts means there will be 3.7 x 10-3 mols of acid HA (aq) + NaOH (aq) → NaA (aq) + H2O (l) b) Molecular weight is old people speak for molar mass molar mass means . . . . . . . . . . grams per mol 184.4 g/mol

when pH = 5.65 . . . . . 10.6 milliliters of 0.135 M NaOH has been added (0.0106L)(0.135M) = 0.00143 mole OH- has been added thus 0.00143 mol acid has been removed 3.7 x 10-3 - 0.00143 = 0.00227 mol acid remaining pH = 5.65 pH = - log[H+] log-1(-5.65) 2.23 x 10-6 M

HC2H3O2 (aq) + K+ (aq) + OH- (aq) → K+ (aq) + C2H3O2- (aq) + H2O (l) 4. a) HC2H3O2 (aq) + KOH (aq) → KC2H3O2 (aq) + H2O (l) HC2H3O2 (aq) + K+ (aq) + OH- (aq) → K+ (aq) + C2H3O2- (aq) + H2O (l) HC2H3O2 (aq) + OH- (aq) → C2H3O2- (aq) + H2O (l) b) NH3 (aq) + HNO3 (aq) → NH4NO3 (aq) NH3 (aq) + H+ (aq) + NO3- (aq) → NH4+ (aq) + NO3- (aq) b) NH3 (aq) + H+ (aq) → NH4+ (aq) c) 2HClO3 (aq) + Sr(OH)2 (aq) → Sr(ClO3)2 (aq) + 2H2O (l) 2H+(aq) + 2ClO3-(aq) + Sr2+(aq) + 2OH-(aq) → Sr2+(aq) + 2ClO3-(aq) + 2H2O(l) H+(aq) + OH-(aq) → H2O(l) d) 2HIO3 (aq) + Cu(OH)2 (aq) → Cu(IO3)2 (aq) + 2H2O (l) 2HIO3 (aq) + Cu(OH)2 (aq) → Cu2+ (aq) + 2IO3- (aq) + 2H2O (l) 2HIO3 (aq) + Cu(OH)2 (aq) → Cu2+ (aq) + 2IO3- (aq) + 2H2O (l)

a) 1.15 g x 0.30 = 0.375 g HCl in 1 mL of solution therefore . . . . . . 375 g HCl in 1000 mL of solution therefore . . . . . . 9.46 mol HCl in 1000 mL of solution 9.46 M 106 mL b) MaVa = MbVb (9.46)Va = (0.2)(5) c) 0.0066 mol OH- therefore . . . . . . 0.0066 mol H+ in 31.67 mL HCl solution 0.2084 M

Lewis acid = BF3 (electron pair acceptor) Lewis base = NH3 (electron pair donor) a) HNO3 (aq) → H+ (aq) + NO3- (aq) b) HF (aq) ⇄ H+ (aq) + F- (aq) c) HBr (aq) → H+ (aq) + Br- (aq) d) HC2H3O2 (aq) ⇄ H+ (aq) + C2H3O2- (aq) e) H2SO4 (aq) → H+ (aq) + HSO4- (aq) f) H2CO3 (aq) ⇆ H+ (aq) + HCO3- (aq) HSO4- (aq) ⇆ H+ (aq) + SO42- (aq) HCO3- (aq) ⇆ H+ (aq) + CO32- (aq) g) H3PO4 (aq) ⇆ H+ (aq) + H2PO4- (aq) H2PO4- (aq) ⇆ H+ (aq) + HPO42- (aq) HPO42- (aq) ⇆ H+ (aq) + PO43- (aq) HSO4- (aq), HCO3- (aq), H2PO4- (aq) and HPO42- (aq)

(3.3 × 10-5)( 0.0133) = 4.39 × 10-7 = [H+] . . . . . pH = 6.36 (1.5 × 10-3)( 0.0045) = 6.75 × 10-6 = [OH-] . . . . . pOH = 5.17 . . . thus pH = 8.83 c) pH = 4.5 . . . . . . [H+] = log-1(-4.5) = 3.162 × 10-5 is 1.33 % of what? . . . . . . 3.162 × 10-5 = (0.0133)(?) 0.0024 M 10) a) 3.00 b) 4.62 c) 11.79 d) 10.88

11. a is neutral KOH and HCl are both strong b is acidic both are weak but the Ka for oxalic acid is larger than the Kb for ammonia c is acidic Cu(OH)2 is weak and HCl is strong d is basic CsOH is strong and H2Cr2O7 is weak e is acidic both are weak but the Ka for acetic acid is larger than the Kb for copper(II) hydroxide HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) 25 mL of 0.10 M HCl . . . . . . 0.0025 mol HCl therefore 0.0025 mol NaOH in 50 mL; 0.05 M NaOH

13. same equivalence point; less NaOH needed to reach equivalence point 14. higher equivalence point; less NaOH needed to reach equivalence point 15. lower equivalence point; more NH3 needed to reach equivalence point