Table 1: Equilibrium Constants Data Syringe Reading pH After Each 0.5 mL Increme
ID: 578170 • Letter: T
Question
Table 1: Equilibrium Constants Data
Syringe Reading
pH After Each 0.5 mL Increment
Color Observations
0
1.7
Clear yellow
2.5
2
yellow
5
2.5
Light yellow
7.5
3
off yellow
10
3.4
off yellow
12.5
3.8
off yellow
15
4
off orange
17.5
4.2
off orange
20
4.5
off orange
22.5
4.8
off orange
25
7.1
0range
27.5
9
Off orange
30
9.2
Off orange
32.5
9.4
Off orange
35
9.5
pink
37.5
9.7
off pink
40
10
Pink
42.5
10.3
Pink
45
10.5
Dark pnik
47.5
11
Dark pink
49
11.5
Na
50
12.1
na
use graph above to answer questions below, the graph explains everything
(a) According to your experimental data, what volume of 0.10 M NaOH represents the half-equivalence (a.k.a. half-neutralization) point in this titration?
(b) What is the pH of the solution at the half-neutralization point?
(c) Using this info, what is the experimental pKa for acetic acid in this reaction? Explain your answer.
(d) Using the pKa value above, what is your experimental Ka for acetic acid? (You must show all work to receive credit)
(e) Look up the accepted “actual” Ka value for acetic acid. How does your value compare? Calculate the percent error for your experimental value. (You must show all work to receive credit.)
Table 1: Equilibrium Constants Data
Syringe Reading
pH After Each 0.5 mL Increment
Color Observations
0
1.7
Clear yellow
2.5
2
yellow
5
2.5
Light yellow
7.5
3
off yellow
10
3.4
off yellow
12.5
3.8
off yellow
15
4
off orange
17.5
4.2
off orange
20
4.5
off orange
22.5
4.8
off orange
25
7.1
0range
27.5
9
Off orange
30
9.2
Off orange
32.5
9.4
Off orange
35
9.5
pink
37.5
9.7
off pink
40
10
Pink
42.5
10.3
Pink
45
10.5
Dark pnik
47.5
11
Dark pink
49
11.5
Na
50
12.1
na
Explanation / Answer
(a) According to your experimental data, what volume of 0.10 M NaOH represents the half-equivalence (a.k.a. half-neutralization) point in this titration?
the first equivalence point --> "orange" is given at V = 25 mL, then half equivalnece point --> 1/2*25 = 12.5 mL
(b) What is the pH of the solution at the half-neutralization point?
half neutralizaiton point is the V = 25 mL, shown before, the "orange" color shows there is change in pH due to netralization of the 1st proton
(c) Using this info, what is the experimental pKa for acetic acid in this reaction? Explain your answer.
pKa --> acetic acid between 4.5 ant 4.8 --> choose 4.8 since it is nearest
(d) Using the pKa value above, what is your experimental Ka for acetic acid? (You must show all work to receive credit)
Ka = 10^-pKa = 10^-4.8 = 0.0000158
Ka = 1.58*10^-5
(e) Look up the accepted “actual” Ka value for acetic acid. How does your value compare? Calculate the percent error for your experimental value. (You must show all work to receive credit.)
acutal Ka
Ka = 1.8*10^-5
it is very near
%error = (1.8*10^-5 - 1.58*10^-5)/(1.8*10^-5) * 100 = 12.2%
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