Question

# table 1: Equilibrium Constants Data Syringe Reading pH After Each 0.5 mL Increment Color Observations 0...

table 1: Equilibrium Constants Data

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 the graph to answer all questions, we are doing equillibrium constant

(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.)

a) volume of NaOH at half-neutralization point = 12.5 ml

b) pH = 3.75 at half-neutralization point

c)according to Henderson equation

pH = pKa + log { [salt]/[acid]} ( because this is a buffer system)

at half neutralization point half of initial concentration of acid is converted to the salt. sao, at this point [salt] - [acid]

therefore, pH = pKa = 3.75

d)pKa = - log Ka

so, Ka = 1.78*10-4

#### Earn Coins

Coins can be redeemed for fabulous gifts.