Salts neutralization reactions acids bases strong acid+ strong base non-hydrolyzing salt pH = 7.0 weak acid+ strong base hydrolyzing salt pH 7.0 > strong acid+ weak base hydrolyzing salt pH 7.0 < weak acid+ weak base hydrolyzing salt pH = ? CH3COOH NaOH 1.00 M 1.00 M 500 mL

CH3COOH  CH3COO- + H+ 1.00 M Ka = 1.8 x 10-5 = x2 500 mL 1.00 - x I C 0.00 0.00 C -x +x +x E 1.00 - x x x x = 4.24 x 10-3 = [H+] . pH = 2.37

Equivalence Point CH3COOH NaOH 1.00 M mol acid = mol base 1.00 M 500 mL 1.00 mol L x 0.500 L = 0.500 mol acid 0.500 mol base  1.00 mol L = 0.500 L base CH3COOH  CH3COO- + H+ + OH-  H2O + CH3COO- [CH3COO-] = 0.500 mol = 0.50 M 0.500 + 0.500 L CH3COO- + H2O  CH3COOH + OH- strong conjugate base

Equivalence Point CH3COOH NaOH 1.00 M 1.00 M 500 mL CH3COO- + H2O  CH3COOH + OH- Kb = Kw = 1 x 10-14 = 5.56 x 10-10 1.8 x 10-5 . Ka . x = [OH-] = 1.67 x 10-5 pH = 9.22 5.56 x 10-10 = [CH3COOH] [OH-] = x2 [CH3COO-] 0.500 - x

CH3COOH NaOH 1.00 M 0.100 mol H+ = 0.100 mol OH- 1.00 M 500 mL add 100 mL 0.500 mol - 0.100 mol = .400 mol CH3COOH = 0.667 M . 0.600 L 0.100 mol = 0.167 M CH3COO- 0.600 L . CH3COOH  CH3COO- + H+ [CH3COOH] [CH3COO-] [H+] I 0.667 0.167 0.00 Ka = 1.8 x 10-5 = (0.167 + x) (x) C -x +x +x (0.667 – x) E 0.667 - x 0.167 + x x x = 7.20 x 10-5 pH = 4.14

presence of conjugate base inhibits dissociation Common ion effect presence of conjugate base inhibits dissociation weak acid or base + conjugate base or acid . . . CH3COOH  CH3COO- + H+ . [CH3COOH] [CH3COO-] [H+] I 0.667 0.00 0.00 Ka = 1.8 x 10-5 = (x) (x) C -x +x +x (0.667 – x) E 0.667 - x x x x = 3.46 x 10-3 pH = 2.46

Henderson-Hasselbalch Equation  H+ + A- Ka = [H+] [A-] [H+] = Ka [HA] [A-] [HA] - log [H+] = - log Ka + log [A-] [HA] pH = pKa + log [A-] [HA]

Henderson-Hasselbalch Equation half-way point CH3COOH NaOH 1.00 M 0.250 mol H+ = 0.250 mol OH- 1.00 M 500 mL add 250 mL 0.500 mol - 0.250 mol = .250 mol CH3COOH = 0.333 M . 0.750 L 0.250 mol = 0.333 M CH3COO- 0.750 L . Henderson-Hasselbalch Equation pH = pKa + log [A-] [HA] pH = pKa pH = 4.74 + log .333 = 4.74 .333

Henderson-Hasselbalch Equation CH3COOH NaOH 1.00 M 0.375 mol H+ = 0.375 mol OH- 1.00 M 500 mL add 375 mL 0.500 mol - 0.375 mol = .125 mol CH3COOH = 0.143 M . 0.875 L 0.375 mol = 0.429 M CH3COO- 0.875 L . Henderson-Hasselbalch Equation pH = pKa + log [A-] [HA] pH = 4.74 + log .429 = 5.22 .143