Question

Calculate and plot this titration curve with the following information:

50mL 0.1M acetic acid with 0.1M NaOH from 0-60 mL in 1 mL steps.

Thanks!

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 0.1 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 0.1 N .

Initial

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

0.5 pH log(ka x C)

=-log(1.8 x 105 x 0.1)0.5 = 2.87 pH

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 0.1 (50+ 50)] 8.06

After Equivalence

Here ,

pOH log OH

pH 14 pOH

NaOH = 50.5 ml , pH=10.69

NaOH = 51 ml , pH=10.99

NaOH = 52 ml , pH=11.29

NaOH = 60 ml , pH=11.95 (this point is not included in graph , because , this much high value of pH destroys the glass electrode , thus , practically we don't run experiment up to this much) .

Now , we are going to plot it .

pH-mertic plot 12 10 8 표 6 2 0 10 20 30 40 50 60 Volume of NaOH (m l)

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 .