Agata Łuszczyńska 13.10.2007

Batory High School I ib 2.5 h

Determination of concentration

of acetic acid in vinegar

DCP		CE
1 asp		1 asp
2 asp		2 asp
3 asp		3 asp

In this experiment we tried to determine the concentration of acetic acid in vinegar.

We did titration by using NaOH with concentration 0.2 M +/- 0.005M.

Our goal was to check if the producent of vinegar didn't cheat and produced the concentration stated on the bottle.

Collecting raw data:

the best way to collect data from this experiment will be a table.

	Initial Reading [± 0.05 ml]	Final reading [± 0.05 ml] the indicator changed color from colourless to pink
Probe 1	0.00 ml	42.35 ml
Probe 2	0.00 ml	42.40 ml
Probe 3	0.00 ml	42.50 ml

Processing raw data:

1.

V₁ = [42.35 ±0.05] ml - [0.00 ± 0.05] ml = 42.35 ml ± 0.10 ml

V₂ = [42.40 ± 0.05] ml - [0.00 ± 0.05] ml = 42.40 ml ± 0.10 ml

V₃ = [42.50 ± 0.05] ml - [0.00 ±0.05] ml = 42. 50 ml ± 0.10 ml
V_average = (42.35+42.40+42.50)/3 = [42.40 ± 0.30] ml

2. Reaction:

CH₃COOH + NaOH --> CH₃COONa + H₂O

3. C_NaOH = [0.200 ± 0.005]M

V_NaOH = [42.40 ± 0.30] ml = [0.04240 ± 0.0003] dm³

n = C x V

So:

n_NaOH = [0.2 ± 0.005] M x [0.04240 ± 0.0003] dm³ = [8.50 x 10^-3 ±0.02]mole

4.

n_NaOH = n_CH3COOH

m = n x M

M_CH3COOH = 60 g/_mole

m_CH3COOH =[8.5 x 10^-3 ± 0.02]mole x 60 g/_mole

m_CH3COOH =[0.51 ± 0.02] g

5.

V_substance = 5 ml = 0.005 dm³

C = n/V

So:

C_s = [8.5 x 10^-3 ± 0.02]mole / 0.005 dm³

C_s = [1.7 ± 0.02] mole/_dm3

6.

C_% = (m_substance/m_solution) x 100%

m_solution = V_solution x d_solution

d_solution = 1.025 g/_cm3 --> from tables

C_% = [m_substance/(d_solution x V_solution)]x 100%

C_% = [0.51/(1.025x5)] x 100%

C_% = 9.95% ± 0.02%

7.

	ad.3.	ad.3.	ad.4.	ad.5.	ad.6.
	VNaOH	nNaOH	mCH3COOH	Csolution	C%
value	0.0424 dm^3	0.0085 mole	0.51 g	1.70 mole	9.95%
error	0.0001 dm^3	0.020 mole	0.02 g	0.02 mole	0.02%

Conclusion and evaluation:

The final percentage concentration was calculated 9.95% ± 0.02%. while the producent stated on the bottle 10%.

Lets look if this method is accurate and so - useful.

[(x_lit - x_exp)/x_lit] x 100% = 0.5 %

x_lit - value on the bottle

x_exp - value calculated

So we see that the result is <2%. That means that this method of indicating the percentage concentration is very useful.

While doing the experiment there could occur some systematic errors like the dilatability of the solution in the burette under the use of temperature. Also there could be some inaccuracy of the automatic pipettes we were using. We can also take into consideration the different volume of a singular drop.

Random errors that could be possible is the burette valve's lack of tightness, through which could leak some of the solution.

Another error that could occur is inaccurate reading of the final voulme.

For improving reading the final volume (which could also be considered a random error) we could put a black piece of paper behind the burette.

This experiment proves that the producent didn't fake the percentage value of the vinegar.

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