Titration Curves I. Strong Acid + Strong Base 0.1 M HCl 0.1 M NaOH

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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 .

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 V = 25 + 10 mL [H+] = 1.5 x 10-3 mol 35 x 10-3 L [H+] = 4.28 x 10-2 M . pH = 1.37 .

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 V = 25 + 20 mL [H+] = 0.5 x 10-3 mol 45 x 10-3 L [H+] = 1.11 x 10-2 M . . pH = 1.95 .

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 . . .

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- V = 25 + 35 mL [OH-] = 1.0 x 10-3 mol . 60 x 10-3 L . [OH-] = 1.67 x 10-2 M . . pOH = 1.78 . pH = 12.22

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

Titration Curves Weak Base + Strong Acid 0.1 M NH3 0.1 M HCl 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 pOH = 2.87 pH = 11.12

Titration Curves . Weak Base + Strong Acid 0.1 M NH3 0.1 M HCl 25.0 mL 2.5 x 10-3mol - 1.0 x 10-3 mol = 1.5 x 10-3 mol V = 25 + 10 mL 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]

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 V = 25 + 20 mL 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]

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 V = 25 + 25 mL 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]

Titration Curves Weak Base + Strong Acid 0.1 M NH3 0.1 M HCl 25.0 mL pH = 8.65 2.5 x 10-3mol Ka = 1.8 x 10-5 pOH = pKb + log [NH4+] [NH3] 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

Polyprotic Acid H2SO3  HSO3- + H+ Ka1 = 1.4 x 10-2 HSO3-  SO32- + 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

Polyprotic Acid . . . . H2SO3  HSO3- + H+ Ka1 = 1.4 x 10-2 HSO3-  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-

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

Polyprotic Acid H2SO3  HSO3- + H+ Ka1 = 1.4 x 10-2 HSO3-  SO32- + H+ Ka2 = 6.5 x 10-8 2 equivalents of base SO32- + H2O  HSO3- + OH- 0.10 M H2SO3 0.1 M NaOH 40 mL 80 mL Final pH Kb2 = 1 x 10-14/ 6.5 x 10-8 = 1.54 x 10-7 1.54 x 10-7 = [HSO3-] [OH-] = x2 x = 7.16 x 10-5 [SO32-] 0.033 - x pH = 9.86