1. Kinds of Titrations

2. Titration of a Strong Acid With A Base As a strong base is titrated into a strong acid, there is just a slight increase in pH until the equivalence point (point at which the amount of H + and OH - are equal). Near the equivalence point (pH 7) there is a sudden upswing of pH caused by just one drop of the base. If an indicator is chosen that has a transition point close to the equivalence point (pH 7) then a colour change will indicate that the equivalence point has occurred.

3. Features of Strong Acid/Base Curves Since the salts of strong acids and bases are neutral (no hydrolysis effects), the pH of the equivalence point is 7. For this kind of titration, an indicator whose transition point is near 7 should be chosen like phenolpthalein.

4. Titrating A Weak Acid With A Strong Base When a weak acid is titrated with a strong base, there is an initial upswing followed by a gradual rise and then a large upswing around the equivalence point. The pH of the equivalence point is alkaline because the anion salt of the weak acid (CH 3 COO - ) hydrolyzes water, forming OH - ions. CH 3 COO - + H 2 O  CH 3 COOH + OH - An indicator like phenolphthalein should be chosen because it changes colour at pH 9.1, near the equivalence point.

5. Calculating the [Weak Acid] From pH Data For CH 3 COOH + OH -  CH 3 COO - + H 2 O, Consider a.100M solution of CH 3 COOH at the V ½ point in the titration. At this point the [CH 3 COOH] = [CH 3 COO - ] =.050. At this point, the simplified expression for K a becomes: K a = [H 3 O + ][CH 3 COO - ] / [CH 3 COOH] = [H 3 O + ] Thus the pK a = pH ½

6. Calculating the [Weak Acid] From pH Data (Continued) Given the pH start = 2.667, the pH ½V = 4.733, the pH end = 8.033, Then [H 3 O + ] start = 10 -2.667 =.00215278 and K a = 10 -4.733 = 1.849E-5 CH 3 COOH + H 2 O  CH 3 COO - + H 3 O + I [X] i -- 0 0 C -2.153E-3 -2.153E-3 +2.153E-3 +2.153E-3 E [X - 2.153E-3] eq -- +2.153E-3 +2.153E-3 K a = 1.849E-5 = [2.153E-3] 2 /(X – 2.153E-3), X – 2.153E-3 = [2.153E-3] 2 /1.849E-5 =.2506 M X =.2506 M + 2.153E-3 M =.2527999 M =.253 M

7. Sample Problem (Hebden p171) 25.0 mL of a weak acid, furoic acid, is titrated with NaOH. It takes 28.8 mL of NaOH to reach equivalence. The pH of furoic acid is initially 2.021 and the pH ½V is 3.170 when 14.4 mL of NaOH have been added. Find the concentrations of the furoic acid (initial) and NaOH. The initial [H 3 O + ] = 10 -2.021 = 9.52796E-3 The K a = 10 -3.170 = 6.7608E-4 acid cation + H 3 O + IX [furoic]st. __  0 0 C - 9.528E-3 __  +9.528E-3 +9.528E-3 E X - 9.528E-3 __  +9.528E-3 +9.528E-3 From K a = 6.7608E-4 = [H 3 O + ] 2 /[Furoic acid st – 9.528E-3] eq, Furoic acid st - 9.528E-3 = [9.52796E-3] 2 / 6.7608E-4 =.13428 M So [Furoic acid st ] =.13428 M + 9.528E-3 =.14380 M =.144 M The moles Furoic acid = 25.0 mL(.144 mmol/mL) = 3.60 mmol F.A Since the moles NaOH = moles Furoic acid = 3.60 mmol, [NaOH] = 3.60 mmol/28.8 mL = 0.125 M

8. Titration of a Weak Base With A Strong Acid When a weak base (NH 4 OH) is titrated with a strong acid (HCl), there is an initial rapid downswing followed by a gradual lowering of pH. Near the equivalence point there is a rapid downswing with an acidic equivalence point due to the base cation hydrolyzing water. Ex: NH 4 + + H 2 O  NH 3 + H 3 O + An indicator like methyl orange should be chosen because it’s colour transition point is in the acid range close to the equivalence point.

9. Concentrations From Titration pH Readings As with a weak acid and strong base, the initial pH and pH ½ V are useful in determining initial concentrations. The pK b = pOH ½V Convert pH ½V to pOH ½V by 14-pH. To determine the initial weak base concentration, [w. base] EQ = [OH - ] 2 /K b (from pOH ½V ) The [OH - ] is obtained by 10 -(14-pH)

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