Salts neutralization reactions acids bases strong acid strong
Salts neutralization reactions acids bases strong acid+ strong base non-hydrolyzing salt p. H = 7. 0 weak acid+ strong base hydrolyzing salt strong acid+ weak base hydrolyzing salt p. H > 7. 0 p. H < 7. 0 weak acid+ weak base hydrolyzing salt p. H = ? CH 3 COOH Na. OH 1. 00 M 500 m. L
CH 3 COOH CH 3 COO- + H+ 1. 00 M Ka = 1. 8 x 10 -5 = x 2 500 m. L 1. 00 - x I C E [CH 3 COOH] [CH 3 COO-] [H+] 1. 00 -x 1. 00 - x 0. 00 +x x x = 4. 24 x 10 -3 = [H+] p. H = 2. 37 .
CH 3 COOH 1. 00 M 500 m. L Equivalence Point mol acid = mol base Na. OH 1. 00 M 1. 00 mol x 0. 500 L = 0. 500 mol acid L 0. 500 mol base 1. 00 mol = 0. 500 L base L CH 3 COOH CH 3 COO- + H++ OH- H 2 O+ CH 3 COO[CH 3 COO-] = 0. 500 mol = 0. 50 M 0. 500 + 0. 500 L CH 3 COO- + H 2 O CH 3 COOH + OHstrong conjugate base
CH 3 COOH 1. 00 M Equivalence Point Na. OH 1. 00 M 500 m. L CH 3 COO- + H 2 O CH 3 COOH + OHKb = Kw = 1 x 10 -14 = 5. 56 x 10 -10 -5 1. 8 x 10 Ka x = [OH-] = 1. 67 x 10 -5 p. H = 9. 22 . . 5. 56 x 10 -10 = [CH 3 COOH] [OH-] = x 2 0. 500 - x [CH 3 COO-]
CH 3 COOH 1. 00 M 0. 100 mol H+ = 0. 100 mol OH 500 m. L CH 3 COOH CH 3 I C E 0. 667 -x 0. 667 - x 1. 00 M add 100 m. L 0. 500 mol - 0. 100 mol =. 400 mol CH 3 COOH = 0. 667 M 0. 600 L 0. 100 mol = 0. 167 M CH 3 COO 0. 600 L [CH 3 COOH] Na. OH COO- + [CH 3 COO-] 0. 167 +x 0. 167 + x H+ . . . [H+] 0. 00 Ka = 1. 8 x 10 -5 = (0. 167 + x)(x) +x (0. 667 – x) x x = 7. 20 x 10 -5 p. H = 4. 14
Common ion effect presence of conjugate base inhibits dissociation weak acid or base + conjugate base or acid . CH 3 COOH CH 3 COO- + H+ I C E [CH 3 COOH] [CH 3 COO-] 0. 667 -x 0. 667 - x 0. 00 +x x . . [H+] 0. 00 Ka = 1. 8 x 10 -5 = (x)(x) +x (0. 667 – x) x x = 3. 46 x 10 -3 p. H = 2. 46
Henderson-Hasselbalch Equation HA H+ + AKa = [H+][A-] [HA] [H+] = Ka [HA] [A-] - log [H+] = - log Ka + log [A-] [HA] p. H = p. Ka + log [A-] [HA]
half-way point CH 3 COOH 1. 00 M 0. 250 mol H+ = 0. 250 mol OH 500 m. L 1. 00 M add 250 m. L 0. 500 mol - 0. 250 mol =. 250 mol CH 3 COOH = 0. 333 M 0. 750 L 0. 250 mol = 0. 333 M CH 3 COO 0. 750 L Henderson-Hasselbalch Equation p. H = p. Ka Na. OH . p. H = p. Ka + log [A-] [HA] p. H = 4. 74+ log. 333 = 4. 74. 333 . . .
CH 3 COOH 1. 00 M 0. 375 mol H+ = 0. 375 mol OH 500 m. L Na. OH 1. 00 M add 375 m. L 0. 500 mol - 0. 375 mol =. 125 mol CH 3 COOH = 0. 143 M 0. 875 L 0. 375 mol = 0. 429 M CH 3 COO 0. 875 L Henderson-Hasselbalch Equation . . p. H = p. Ka + log [A-] [HA] p. H = 4. 74+ log. 429 = 5. 22. 143 . . .
- Slides: 9