Question

Prepare a titration curve (pH vs. equivalents strong base added if you choose a weak acid to be titrated) in which you plot at least 10 data points starting from 0-equiv added to 1 equiv added. Show how you can obtain the pKa for this acid from the plot. You can also choose a weak base to be titrated with strong acid. Use Excel for the calculations and for plotting. Show the calculations for how you obtained the pH at 0, 1 equiv, and any intermediate data points.

Answer 1

The results we get from a weak acid-strong base titration is given .

Here , we assume , the weak acid to be acetic acid equivalent , having concentration to be 1 N . 50 ml of this is taken .

k_a of acetic acid is is 1.8*10^-5 . We put the value here .

Take NaOH equivalent to be the strong base , with a concentration 1 N .

Initial

Here , no base present . The formula to put is ,

0.5 pH log(ka x C)

pH =-log(1.8 x 105 x 1)0.5= 2.37

Middle

Here , we use Henderson-Hasselbalch equation of buffer . The equation is ,

pH= pklogsalt acid)

pka-log(k) = 4.74

Now , by addition of base , [salt] &[acid] changes and pH alters . A pH table is given by addition of base.

NaOH added(ml)	pH
1	3.05
5	3.78
10	4.13
15	4.37
20	4.56
25	4.74
30	4.92
35	5.10
40	5.34
45	5.69
49	6.43
49.5	6.74

Equivalence

At equivalence point , Salt Hydrolysis occurs . The equation is -

pH 7+0.5 x pk, + 0.5 x log salt

pH 7+0.5 x 4.74+ 0.5 x log 50 x 1(50+ 50)] = 9.22

After Equivalence

Here ,

pOH log OH

pH 14 pOH

NaOH = 50.5 ml , pH=11.67

NaOH = 51 ml , pH=11.99

NaOH = 52 ml , pH=12.29

Now , we are going to plot it .

pH-metric plot 14 12 10 8 4 2 0 30 40 10 20 50 60 Volume of NaOH added LD Hd

To determine the pKa , we use the pH corresponding to the half neutralisation point (here , 25 mL) .

At 25 ml , pH=4.74 we get . That is the pKa .