# Lab report

Topics: Acid, Sodium hydroxide, Solubility Pages: 5 (946 words) Published: March 26, 2014
﻿Data collection
Quantitative Data
Raw Data
Table 1: Table showing the mass of the amount of unknown acid X measured in grams (±0.001g)

Table 2: Table of reading of the burette initially filled with 25mL of 0.201moldm-3 sodium hydroxide (NaOH) to titrate 25mL (±0.03mL) of unknown acid X in mL (±0.05mL) after each titre.

Reading on the burette initially filled with 25mL of 0.201moldm-3 NaOH (±0.05mL) First titre
21.3
Second titre
18.2
Third titre
15.2
Fourth titre
12.0

Qualitative data
Observations:
When dissolving the acid X in the water, most of it did not dissolve and become whitish foam sitting at the top of the solution. When transferring the acid from the mortar to the flask by washing it with water, the low solubility of the acid made it hard and some of it was still stuck in the mortar and was not transferred into the flask. When phenolphthalein was added to the unknown acid solution, the solution remained clear. At first, when the NaOH base was added to the unknown acid solution, the solution where the base had dropped into changed into a pink colour which disappeared after swirling the flask containing the unknown acid, which returns to being clear. However, after more NaOH base was added to the unknown acid solution, the solution eventually turned pink after swirling the flask.

Data processing:
As we did not refill the burette with NaOH each time after titrating the unknown acid, we need to calculate the difference of the volume before and after each titre. For example, the volume used in the first titre would be calculated by: 25-21.3=3.7 mL

Uncertainty:
(Uncertainty of the previous volume + uncertainty of volume after titration) = (0.05+0.05)
=0.1mL
Calculating the rest of the titres in a similar fashion (previous volume-volume after titration) the following table of results is obtained: Table 3: Table of the volume of 0.201moldm-3 sodium hydroxide (NaOH) used to titrate 25mL (±0.03mL) of unknown acid X in mL (±0.1mL) for each titre.

Amount used to titre 25mL (±0.03mL) of unknown acid X in mL (±0.5mL) First titre
3.7
Second titre
3.2
Third titre
3.0
Fourth titre
3.2

Let HX be the acid X, then the equation of the neutralization of the acid would be: HX + NaOH  H2O + NaX
So the ratio of HX to NaOH is 1:1.
The first value, 3.7mL, is a pilot and is disregarded in the calculations, so we are left with 3 concordant values within 0.2mL of each other. Since the number of moles=concentration*volume, we need to find out the average value of the volume obtained: (Titrate 1+ Titrate 2 + Titrate 3)/3

=(3.1+3.0+3.2)/3
=3.1mL
Uncertainty:
(Largest value-smallest value)/2
=(3.2-3.0)/2
=0.1mL
Average volume of titrate: 3.1mL ± 0.1
Moles=0.201*3.1/1000=6.231*10^-4
Uncertainty:
(% uncertainty of concentration+% uncertainty of volume)
=(0.001/0.201+0.1/3.1)
=3.72…%
Since the ratio is 1:1, then the moles of the unknown acid HX is also 6.231*10^-4. Then there are 6.231*10^-4 moles of HX in 25mL of XH solution, then the concentration is: Number of moles / volume

=6.231*10^-4/0.025=0.024924 moldm-3
Uncertainty:
(% uncertainty of number of moles + % uncertainty of volume) =(3.72…%+0.03/25)
=3.84…%
Then, with the concentration, we can find the number of moles of acid HX originally added to the 100mL of water: Number of moles= concentration * volume
=0.024924*0.1
=2.4924*10^-3 moles.
Uncertainty:
(% uncertainty of concentration+% uncertainty of volumetric flask) =(3.84…%+0.1/100)
=3.94…%

Molar mass=mass/number of moles
=1.503/2.4924*10^-3
=603.03g/mol
Uncertainty:
(% uncertainty of mass + uncertainty of number of moles)
=(0.001/1.503+3.94…%)
=4.01%
=24.2
=20(1sf)
Final obtained molar mass of unknown acid HX is 600g/mol±20 Conclusion:
The obtained value for the monoprotic acid is 600g/mol ±20 so it falls within the range of between 580gmol-1 to 620gmol-1. The unknown acid is benzoic acid, or C6H5COOH with a molar mass of 122.121. My result was...