Mass (g) of empty beaker: 48.1g Mass (g) of beaker + C6H8O7: 50g Mass (g) of C6H
ID: 570748 • Letter: M
Question
Mass (g) of empty beaker: 48.1g
Mass (g) of beaker + C6H8O7: 50g
Mass (g) of C6H8O7: 1.9g
Mass (g) of beaker + C6H8O7 after cooling: 50g
Molarity (M) of C6H8O7: 0.132M
Trial 1
0.132M C6H8O7
mL pH Derivate
0.00 12.84 -0.29
0.50 12.72 -0.43
1.0 12.54 -0.92
1.5 12.22 -2.03
2.0 11.14 -4.16
2.5 7.08 -4.96
2.75 6.47 -2.90
3.0 6.13 -1.63
3.25 5.93 -1.25
3.5 5.48 -1.11
3.75 5.31 -0.77
4.0 5.15 -0.60
4.25 5.04 -0.57
4.5 4.87 -0.56
4.75 4.75 -0.53
5.0 4.60 -0.48
5.25 4.51 -0.42
5.5 4.40 -0.38
5.75 4.32 -0.31
6.0 4.24 -0.27
6.5 4.11 -0.25
7.0 3.99 -0.23
7.5 3.88 -0.21
8.0 3.78 -0.19
8.5 3.69 -0.17
9.0 3.62 -0.17
9.5 3.53 -0.17
10 3.43 -0.15
10.5 3.38 -0.11
11 3.33 -0.09
11.5 3.29 -0.07
12 3.27 -0.06
12.5 3.23 -0.06
13 3.20 -0.06
13.5 3.17 -0.06
14 3.14 -0.05
14.5 3.12 -0.05
15 3.09 -0.05
Trial 2
0.132M C6H8O7
mL pH Derivate
0.00 12.93 -0.22
0.50 12.84 -0.30
1.0 12.69 -0.74
1.5 12.53 -1.84
2.0 11.84 -4.56
2.5 6.80 -5.73
2.75 6.23 -3.10
3.0 5.93 -1.62
3.25 5.68 -0.98
3.50 5.46 -0.81
3.75 5.28 -0.71
4.0 5.11 -0.63
4.25 4.96 -0.55
4.50 4.84 -0.47
4.75 4.73 -0.43
5.0 4.63 -0.39
5.25 4.53 -0.36
5.5 4.45 -0.32
5.75 4.37 -0.31
6.0 4.29 -0.30
6.25 4.23 -0.32
6.5 4.12 -0.31
6.75 4.06 -0.25
7.0 4.01 -0.22
7.25 3.96 -0.21
7.5 3.91 -0.20
7.75 3.86 -0.21
8.0 3.80 -0.18
8.5 3.72 -0.15
9.0 3.65 -0.12
9.5 3.61 -0.10
10.0 3.55 -0.11
10.5 3.49 -0.11
11 3.45 -0.10
11.5 3.39 -0.10
12 3.35 -0.08
12.5 3.31 -0.07
13 3.27 -0.07
13.5 3.24 -0.06
14 3.21 -0.06
14.5 3.17 -0.06
15 3.15 -0.06
15.5 3.11 -0.05
16 3.09 -0.05
16.5 3.06 -0.05
17 3.05 -0.04
17.5 3.03 -0.04
18 3.01 -0.04
18.5 2.99 -0.04
1. What was the experimental molarity (M) for NaOH in Trial 1? What was the experimental molarity (M) for NaOH in Trial 2? What was the average molarity (M) of the two trials?
2. According to your experimental data, what volume of C6H8O7 represents the half-equivalence (a.k.a. half-neutralization) point in this titration? What is the pH of the solution at the half-neutralization point?
Explanation / Answer
In present case, weak acid, C6H8O7 is titrated against Strong base ,NaOH. It is given that Molarity of C6H8O7 is 0.132M
In this , acid is added in base, pH of solution decreases
Trial-1
Initially vol. Of acid added is 0ml , only NaOH is present, pH=12.84 , pOH =14-12.84 =1.16
[OH-] = 10-pOH
Molarity of NaOH = 10-1.16 = 0.069M
The point at which derivate has maximum value is equivalence point, which is -4.96 corresponds to pH of 7.08 & vol.of acid added is 2.5ml.
pOH is 14-7.08= 6.92
Conc.of NaOH is 10-6.92 = 1.2 *10-7 M
At this point NaOH is neutralised by Acid completely
Half equivalence point is the point at which half neutralisation occurs. So conc. Of base should be 1.2* 10-7/2 =0.6 *10-7M
pOH = -log(0.6*10-7) = 7- log0.6 =7.22
pH at half equivalence point = 14-7.22 =6.78
Vol. Of acid can be obtained by plotting graph between pH vs vol.of acid added.
Trial-2
In a similar way, as above , pH of NaOH = 12.93
So, pOH= 14-12.93 =1.07
Molarity Of NaOH = 10-1.07=0.085M
At equivalence point, max. Value of derivative is -5.73 which corresponds to vol.of acid used is 2.5ml and pH is 6.8
So pOH =14-6.8 =7.2 , conc. Of base= 10-7.2 = 6.3*10-8 M
At half equivalence point half neutralisation occurs
So conc.of base =6.3*10-8/2 =3.15 *10-8M
pOH=-log(3.15*10-8) = 8-log3.15 = 7.5
pH at half equivalence point=14-7.5 =6.5
vol.of acid used at half equivalence point can be obtained from graph.
from trial -1 and trial -2
Trial-1, conc. Of NaOH at equivalence point =1.2*10-7M
Trial-2, conc. Of NaOH at eqv.point= 0.63*10-7M
Molarity of NaOH from trial-1 =0.069M
Molarity of NaOH from trial -2= 0.085M
Average molarity is 0.069+0.085 /2 =0.077M