Volume of NaOH when solution turns dark pink/orange: 14-10.5= 3.5mL Volume of Na
ID: 883059 • Letter: V
Question
Volume of NaOH when solution turns dark pink/orange: 14-10.5= 3.5mL
Volume of NaOH at the equivalence point (the point on the graph midway between the vertical rise; see Lab 7 [19], Figure 2): 4.5 mL
pH at the equivalence point (should be approximately 8–9): 9
Volume of NaOH at equivalence point: 4.5 mL
Volume at half-equivalence point: 2.25 mL
pH at the half-equivalence point: 4.5 ???????? Not sure if this is correct I divided from my pH at equivalence point.
pH = pKa at this point
Ka (equilibrium constant): ?????????
pKa = –log Ka
Experimental Ka of acetic acid: ?????????????
Theoretical Ka of acetic acid: 1.7 x 105
Compare your experimental value for the equilibrium constant to the theoretical value. Do you notice any variation? If so, why? ?????????????????????
The data I collected are as below:
I my acid was .5ml of 4.5 % acetic acid in 80mL of H20
The base I used was 15mL of .1M sodium hydroxide.
Syringe Reading, mL
pH after each 0.5 mL increment
Color Observations
15
3.6 Initial reading
Clear Yellow
14.5
4.2
Clear Yellow
14
4.4
Clear yellow
13.5
4.5
Clear yellow
13
4.7
Clear yellow
12.5
5.0
Clear yellow
12
5.2
Clear yellow
11.5
5.7
Clear yellow
11
6.4
Dark yellow
10.5
9
Orange
10
11.5
Orange
9.5
12
Orange
9
12
Orange
8.5
12
Orange
8
12
Orange
Syringe Reading, mL
pH after each 0.5 mL increment
Color Observations
15
3.6 Initial reading
Clear Yellow
14.5
4.2
Clear Yellow
14
4.4
Clear yellow
13.5
4.5
Clear yellow
13
4.7
Clear yellow
12.5
5.0
Clear yellow
12
5.2
Clear yellow
11.5
5.7
Clear yellow
11
6.4
Dark yellow
10.5
9
Orange
10
11.5
Orange
9.5
12
Orange
9
12
Orange
8.5
12
Orange
8
12
Orange
Explanation / Answer
pH at the half-equivalence point: 4.5 ???????? Not sure if this is correct I divided from my pH at equivalence point.
Half Equivalence point = 4.5ml/2 = 2.25 ml, that is, between 2 and 2.5 ml
Reading will be 15-2 and 15-2.5 = 13 and 13.5 ml... pH Values are 4.5 and 4.7
I would choose 4.6 or write the range 4.5 < pH < 4.7
pH = pKa at this point
Ka (equilibrium constant): ?????????
pKa = –log Ka
This half equivalence point is a Buffer, a very important one. Let us model it with Henderson Hasselbalch equation:
pH = pKa + log(Acid/Conjugate)
Since Acid = Conjugate (only in the half this is true!)
log(x/x) = 0
pH = pKa
4.6 = pKa
pKa = -logKa = 4.6
Ka = 10^-4.6 = 2.51*10^-5
Which is pretty near to that of the theoretical (1.7*10^-5)
Compare your experimental value for the equilibrium constant to the theoretical value. Do you notice any variation? If so, why? ?????????????????????
There is a slight variation in the coefficient (2.51 vs 1.70) BUT the magnitude is the same (10^5) in both cases!
This is due to the jump between 0.5 ml you have in all your tests. If you want to be precise, you gotta use something smaller such as 0.1 ml or even 0.05 ml if possible. Specially in this range when the pH is arround 4-4.5
You could then calculate a more precise pH, then a pKa and therefore the Ka
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.