Chem question

a)
The colour of the indicator changes from colourless to pale pink.
(Indicator is added to the solution in the conical flask. A known volume of solution is added to the conical flask. In the question, a known volume of 25 cm³ of diluted vinegar solution is added to the conical flask. Therefore, the indicator is added to the diluted vinegar soluton.)

b)
Volume used in Run 1 = 22.50 - 0 = 22.50 cm^-3
Volume used in Run 2 = 42.60 - 22.50 = 20.10 cm^-3
Volume used in Run 3 = 22.00 - 2.00 = 20.00 cm^-3
Volume used in Run 4 = 41.90 - 22.00 = 19.90 cm^-3
Volume used in Run 1 is rejected due to great deviation from other volumes.
Reasonable average for the vol. of NaOH = (20.10 + 20.00 + 19.90)/3 = 20.00 cm^-3

c)
CH3COOH + NaOH → CH3COONa + H2O
No. of moles of NaOH used in titration = 0.2 * (20/1000) = 0.004 mol
No. of moles of CH3COOH used in titration = 0.004 mol
Molarity of the diluted vinegar = 0.004/(25/1000) = 0.16 mol dm^-3
Molarity of the commercial vinegar = 0.16 * (250/50) = 0.8 mol dm^-3

d)
Molar mass of CH3COOH = 12*2 + 1*4 + 16*2 = 60 g mol^-1
Concentration of CH3COOH = 0.8 * 60 = 48 g dm^-3

e)
Consider 1 dm^3 (1000 cm^3) of the commercial vinegar.
Total mass of the solution = 1000 * 1 = 1000 g
Mass of CH3COOH = 48 * 1 = 48 g
% by mass of CH3COOH in the commercial vinegar = (48/1000) * 100% = 4.8%

a) For phenolphalein, it is colourless in acidic solution (i.e. vinegar) and red in alkaline solution (NaOH)

Since vinegar is being titrated which is contain in the conical flask, the colour of the solution is colourless.

After being titrated with NaOH, where end point is reached, the colour of the solution changes into red.

Therefore, the colour changes from colourless to red.



b) For calculating reasonable average, the result obtain from the first titration will be discarded.

Then, the average should be calculated as follows:

= volume of NaOH used in 2nd + 3rd+ 4th titration

= [(42.6-22.5)+(22.0-2.0)+(41.9-22.0)] / 3

= 20 cm^3



c) Molarity = no. of mole of ehanoic acid / volume of VINEGAR

First, you should write the equation of the reaction between ethanoic acid and NaOH.

CH3COOH (aq) + NaOH (aq) ---> CH3COONa (aq) + H2O (l)

From (b), we know that volume of NaOH used is 20cm^3, and the concentration of NaOH is 0.2M which is given.

Mole ratio of ethanoic acid to NaOH is 1:1

Therefore,

no. of mole of ethanoic acid in 25cm^3 diluted solution

= no. of mole of NaOH

= 20/1000 * 0.2

= 0.004 mole

no. of mole of ethanoic acid in 250cm^3 diluted solution,

= 0.004 * 10

= 0.04 mole

Before dilution, 50cm^3 of vinegar is used. Hence, molarity of ethanoic acid in the vinegar is calculated as follows:

= 0.04 / 50/1000

= 0.8 mol dm^-3


d) Concentration (g dm^-3) = molarity of ethanoic acid * relative atomic mass

= 0.8 * (12+1*3+12+16*2+1)

= 48 gdm^-3

= 0.048 gcm^-3




e) As density of vinegar is 1gcm^-3, you may assume that there is 1cm^3 of vinegar, that means there is 1g of vinegar.

From (d), concentration of ethanoic acid is 0.048 gcm^-3, that means there is 0.048g of ethanoic acid in 1 cm^3.

Therefore, percentage by mass can be calculated.

= (0.048 / 1) *100%

= 4.8%






Hope my explanations help.

Just a CE level question

2010-04-13 20:35:01 補充：
but it is not copied from CE pp

al定ce????????