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Titration Curves

Titration Curves. I. Strong Acid. + Strong Base. 0.1 M HCl. 0.1 M NaOH. 25.0 mL. 25.0 mL. 2.5 x 10 -3 mol. 2.5 x 10 -3 mol. 1. Initial pH. HCl. . H +. + Cl-. 0.1 M. 0.1 M. [H + ] =. 0.1 M. pH =. - log H +. = 1.00. Strong Acid. + Strong Base. 0.1 M HCl.

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Titration Curves

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  1. Titration Curves I. Strong Acid + Strong Base 0.1 M HCl 0.1 M NaOH 25.0 mL 25.0 mL 2.5 x 10-3mol 2.5 x 10-3mol 1. Initial pH HCl  H+ + Cl- 0.1 M 0.1 M [H+] = 0.1 M pH = - log H+ = 1.00 .

  2. Strong Acid + Strong Base 0.1 M HCl 0.1 M NaOH 25.0 mL 10.0 mL 2.5 x 10-3mol - 1.0x 10-3 mol = 1.5 x 10-3 mol 25 + 10 mL V = [H+] = 1.5 x 10-3 mol 35 x 10-3 L 4.28 x 10-2 M [H+] = . . pH = 1.37

  3. Strong Acid + Strong Base 0.1 M HCl 0.1 M NaOH 25.0 mL 20.0 mL 2.5 x 10-3mol - 2.0 x 10-3 mol = 0.5 x 10-3 mol 25 + 20 mL V = [H+] = 0.5 x 10-3 mol 45 x 10-3 L 1.11 x 10-2 M [H+] = . . . pH = 1.95

  4. Strong Acid + Strong Base 0.1 M HCl 0.1 M NaOH 25.0 mL 25.0 mL 2.5 x 10-3mol - 2.5 x 10-3 mol = 0.0 mol [H+] = 1.00 x 10-7 M . pH = 7.00 . . .

  5. Strong Acid + Strong Base 0.1 M HCl 0.1 M NaOH 25.0 mL 35.0 mL 2.5 x 10-3mol 3.5 x 10-3 mol OH- 25 + 35 mL V = [OH-] = 1.0 x 10-3 mol . 60 x 10-3 L . 1.67 x 10-2 M [OH-] = . . . pOH = 1.78 pH = 12.22

  6. Titration Curves Weak Acid + Strong Base 0.1 M CH3COOH 0.1 M NaOH 25.0 mL 25.0 mL Initial weak acid pH = pKa + log [CH3COO-] [CH3COOH] Ka = 1.8 x 10-5 = [H+] [CH3COO-] [CH3COOH] half-way point pH = pKa = 4.74 equivalence point CH3COO- + H2O  CH3COOH + OH- Kb = 5.6 x 10-10 = [OH-] [CH3COOH] [CH3COO-] strong base

  7. Titration Curves Weak Base + Strong Acid 0.1 M NH3 0.1 M HCl 25.0 mL 25.0 mL 1. Initial pH NH3 + H2O NH4+ + OH-  Kb= 1.8 x 10-5 = [OH-] [NH4+] [NH3] [OH-] = 1.34 x 10-3 M pH = 11.12 pOH = 2.87

  8. Titration Curves Weak Base + Strong Acid 0.1 M NH3 0.1 M HCl 25.0 mL 10.0 mL 2.5 x 10-3mol - 1.0 x 10-3 mol = 1.5 x 10-3 mol 25 + 10 mL V = x = 2.67 x 10-5 pOH = 4.57 NH3 + H2O NH4+ + OH-  pH = 9.43 Kb= 1.8 x 10-5 = [NH4+] [OH-] [NH3] . [NH3] [NH4+] [OH-] 0.043 0.029 0.0 0.043 -x 0.029 + x x 1.8 x 10-5 = [x] [0.029 + x] [0.043 - x]

  9. Titration Curves Weak Base + Strong Acid 0.1 M NH3 0.1 M HCl 25.0 mL 20.0 mL 2.5 x 10-3mol - 2.0 x 10-3 mol = 5.0 x 10-4 mol 25 + 20 mL V = x = 4.5 x 10-6 pOH = 5.35 NH3 + H2O NH4+ + OH-  pH = 8.65 Kb= 1.8 x 10-5 = [NH4+] [OH-] [NH3] . . [NH3] [NH4+] [OH-] 0.011 0.044 0.0 0.011 -x 0.044 + x x 1.8 x 10-5 = [x] [0.044 + x] [0.011 - x]

  10. Titration Curves Weak Base + Strong Acid 0.1 M NH3 0.1 M HCl 25.0 mL 25.0 mL 2.5 x 10-3mol - 2.5 x 10-3 mol = 0.00 25 + 25 mL V = x = 5.9 x 10-6 pH = 5.27 NH4+ NH3 +H+  Ka= 5.6 x 10-10 = [NH3] [H+] [NH4+] . . [NH4] [NH3] [H+] 0.05 0.00 0.0 . 0.05 -x x x 5.6 x 10-10 = [x2] [0.05 - x]

  11. pOH = pKb + log [NH4+] [NH3] Titration Curves Weak Base + Strong Acid 0.1 M NH3 0.1 M HCl 25.0 mL 20.0 mL pH = 8.65 2.5 x 10-3mol Ka = 1.8 x 10-5 5.0 x 10-4 mol NH3 2.0 x 10-3 mol NH4+ V = 45 x 10-3 L pOH = 4.74 + log (0.44) (0.11) = 5.34

  12. Polyprotic Acid H2SO3 HSO3- Ka1 = 1.4 x 10-2  + H+ HSO3- SO32- Ka2 = 6.5 x 10-8  + H+ 2 equivalents of base 0.10 M H2SO3 0.10 M NaOH 40 mL 80 mL Initial pH 1.4 x 10-2 = [HSO3-] [H+] = x2 [H2SO3] 0.1 - x x = 0.03 pH = 1.51

  13. Polyprotic Acid H2SO3 HSO3- Ka1 = 1.4 x 10-2  + H+ HSO3- SO32- Ka2 = 6.5 x 10-8  + H+ 2 equivalents of base 0.10 M H2SO3 0.1 M NaOH buffering regions half-way point pH = pKa - log 1.4 x 10-2 = 1.85 - log 6.5 x 10-8 = 7.19 . 1st equivalence point . . 1.84 + 7.19 = 4.52 . 2 2nd equivalence point conjugate base, SO3-

  14. Polyprotic Acid H2SO3  HSO3- Ka1 = 1.4 x 10-2 + H+ HSO3-  SO32- Ka2 = 6.5 x 10-8 + H+ 2 equivalents of base 0.10 M H2SO3 0.1 M NaOH 40 mL 80 mL Initial pH = x2 x = 0.03 1.4 x 10-2 = [HSO3-] [H+] 0.1 - x [H2SO3] pH = 1.51

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