Complete the following table for the titration of 50.00 ml of 0.100 M propanoic
ID: 523943 • Letter: C
Question
Complete the following table for the titration of 50.00 ml of 0.100 M propanoic acid, CH3CH2COOH, with 0.100 M NaOH. The Ka for propanoic acid is 1.3 x10-5 at 250 C. Show your calculations for each addition . When the table is complete, plot pH versus mL of NaOH using excel program. Draw the curve. What is the pH at equivalence point. Determine the pH by using the half equivalence method and report.
mL NaOH added
Total Volume (mL)
[ H3O]+ Hydronium
Ion concentration
[OH]-
Hydroxide ion
pH of the solution
0.00
5.00
10.00
20.00
25.00
35.00
45.00
49.00
50.00
51.00
60.00
mL NaOH added
Total Volume (mL)
[ H3O]+ Hydronium
Ion concentration
[OH]-
Hydroxide ion
pH of the solution
0.00
5.00
10.00
20.00
25.00
35.00
45.00
49.00
50.00
51.00
60.00
Explanation / Answer
Given Ka = 1.3 x 10-5
The table is as follows
For 0 ml NaOH added
Total volume = 50ml
Ka = [H3O+] x [A-]/[HA]
1.3 x 10-5 = [H3O+] x 0.0130 / 1
[H3O+]=0.169
[OH-] = Kw/[H3O+]
= 1 x 10-14 / 0.169
= 0.59
pKa = - logKa
= -log(1.3 x10-5)
= 4.886
pH=pKa+log[A-/HA]
= 4.886 -1.886
= 3
Similarly
For 5ml
Ka = [H3O+] [A-] / HA
1.3 x 10-5 =[H3O+] [0.1479]/1
[H3O+]= 0.192
[OH-]=0.52
pH = pKa+log[A-/HA]
= 4.8-0.83
=4.05
For 10ml
Ka = [H3O+][A-]/HA
H3O+ = 1.3 x 10-5 / 0.331
[H3O+]=0.43
[OH-]=0.232
pH= Pka+log[A/HA]
= 4.886-0.48
=4.40
For 20ml
Ka = [H3O+][A-] / {HA]
1.3 x 10-5 =[H3O+]x 0.870/1
[H3O+]=0.113
[OH-]= 0.884
pH= pKa+log[A-/HA]
= 4.88-0.06
=4.82
For 25 ml
Ka= [H3O+] x 1.0471/1
[H3O+]=0.136
[OH-]=0.73
pH=Pka+log[A-/HA]
= 4.886-0.02
=4.9
For 35 ml
Ka = [H3O+]x2.57/1
[H3O+] = 0.334
[OH-]=0.29
pH= 4.88+log[A/HA]
= 4.88+0.41
=5.29
For 45ml
Ka = [H3O+]x[4.141]/1
[H3O+]=0.152
[OH-]=0.65
pH=pka+log[4.14]
= 4.88+0.61
=5.95
For 49ml
Ka = [H3O+]x 6.456/1
[H3O+]=0.162
[OH-] = 0.61
pH=pka+log(6.456)
= 4.88+1.81
= 6.69
For 50ml
Ka = [H3O+]x7.943/1
[H3O+]=0.102
[OH-]=0.35
pH=pka+log(7.943)
= 4.88+3.97
= 8.85
For 51ml
Ka = [H3O+] x 11.67/1
[H3O+]=0.104
[OH-]=0.32
pH= 11
For 60ml
Ka= [H3O+]x12.54/1
[H3O+]=0.103
[OH-]=0.31
pH=11.96
The Ph at the equivalence point =pKa
pH=4.886
ml of NaOH added Total volume [H3O]+ [OH]- pH 0 50 0.169 0.59 3 5 55 0.192 0.52 4.05 10 60 0.43 0.232 4.40 20 70 0.113 0.884 4.82 25 75 0.136 0.73 4.9 35 85 0.334 0.29 5.29 45 95 0.152 0.65 5.95 49 99 0.162 0.61 6.69 50 100 0.102 0.35 8.85 51 101 0.104 0.32 11 60 110 0.103 0.31 11.96