[earthnation112]

Hi, just had some work to do been going through it, and so far ive came up with the following answers. I have copied the whole file into the thread and formatted it, incase background information was needed.

My question is have I done the calculation correctly? Are there any big errors present which i have completely missed? All the figures and calculations in "bold" are the calculations i actually did. Thank You

THE STANDARDISATION OF AN APPROXIMATELY 0.1 M SOLUTION OF HYDROCHLORIC ACID


Method

(a)   Preparation of Standard Na₂CO₃ Solution

Remove the anhydrous sodium carbonate from the oven (use tongs) where it has been dried for you and allow it to cool for 30 minutes in a desiccator. Transfer approximately 1.3g into a weigh¬ing bottle and weigh the bottle plus contents plus cap on an accurate balance to 0.0001g.

Tip the contents into a beaker and reweigh the bottle. Dissolve the sodium carbonate in approximately 100 cm³ water by gently heating the solution on a hotplate.

Then mixture was quantitatively transferred when cooled (via a glass funnel) to a clean 250 cm³ volumetric flask.

   Results

   Weight of bottle containing Na₂CO₃: 16.3577g.   
   Weight of empty bottle: 15.0552g.
   Weight of Na₂CO₃: 1.3025g.

 
From the weight of Na₂CO₃ taken, calculate the number of moles of Na₂CO₃ .

(i)   Mr of Na₂CO₃ = 105.99 g mol^-1

(ii)   Number of moles Na₂CO₃ = 0.0123 (3sf)


∴   number of moles in 250 cm³ = 0.0123

∴   number of moles in 1000 cm³ = 0.0123

∴   Concentration of Na₂CO₃ solution = 0.0492 M

(b)   Standardisation of HCl Solution

Take approximately 150 cm³ of the HCl solution from the aspirator in a clean dry 400 or 600 cm³ beaker.

Pipette, using a pipette filler, 25 cm³ of the Na₂CO₃  solution into a clean 250 cm³ conical flask. Add 2-4 drops of methyl red indicator to the flask.

Fill your burette with the acid (using a plastic funnel & then remove the funnel) and read the burette to the nearest 0.05 cm³. It is not necessary to start at 0.00 cm³.

Titrate with the acid until the solution turns brownish red. At this stage, carbon dioxide produced from this reaction must be removed from the solution – this is achieved by boiling. Boil the solution using a hotplate for 30 seconds, during which time the colour of the indicator should revert to yellow. Carefully run the acid into the cooled solution until the red tint of the indicator just appears. Again boil the solution, and if the yellow colour returns, repeat the careful addition of acid until the red tint reappears.

On boiling the solution the red colour should persist and the titration is only then complete. Two boilings are generally necessary. Read the burette again.

Repeat the titration until two titration agree to 0.05 cm³.

Results:

Titre No:                                        I                    II                    III

Final reading (cm³ )                     24.79              24.57                27.09   
First reading (cm³ )                     1.50                1.20                 3.75               
Volume of HCl solution (cm³)        23.29             23.37               23.34

Mean of two titrations (23.37+23.34)/2 = 23.355 = 23.36 cm³

Equation of the reaction:

2HCL(aq) + Na₂CO₃(aq) --------> 2NaCl(aq) + H₂O(l) + O₂(g)

H₂CO₃(aq)$$\rightleftharpoons$$H₂O(l) + CO₂(g)

Calculate the number of moles per dm³ (molarity, M) of the HCl:

25/250*1.3025g = 0.13025g weight of Na₂CO₃ . 
 
0.13025g/105.99 g/mol = 0.00123 moles

HCl 2:1 ratio of moles with Na₂CO₃   
∴ HCL moles 2*0.00123= 0.00246 moles of HCL

0.00246 mol = 0.105 M
0.02336 dm³

(c)   Determination of Relative Molecular Mass of an Unknown Monoacidic Base
   
Titrate 25.00 cm³ of the solution of the base with standardised HCl using phenolphthalein as indicator (pink to colourless).

Note:  The concentration of the unknown monoacidic base, XOH is 10.85 g per dm³.

Burette Results:

Titration No                                 I                       II                   III

2nd reading (cm³)                     25.35                 25.45              25.40

1st reading (cm³)                         0                       0                   0

Volume of HCl solution (cm³)      25.35                 25.45              25.40


Mean of the titrations = 25.4 cm³

Equation of the reaction:

                                     HCl   +  XOH  --------> XCl    +   H₂O
   
Calculation

0.105 M HCl * 0.0254 (L) = 0.00267 moles (3sf)

1:1 ratio of moles of HCL and unknown base. Therefore 0.00267 moles of unknown base.

10.85 g/dm³ * 0.025 dm³ = 0.27125 g (mass of unknown base)

0.27125g / 0.00267 mol = 101.5(RbOH)

[Borek]

0.13025g/105.99 g/mol = 0.00123 moles

Why only 3sf? (and it doesn't apply just here, also in other places).

[mjc123]

You were asked to read the burette to the nearest 0.05 cm³. How do you get numbers like 24.79? Most burettes I have seen are graduated to 0.1 or 0.05 cm³. And in part c, your initial readings are "0". Was that accurately 0.00 each time? If so, say so. "0" means "between -0.5 and 0.5". "0.00" means "between -0.025 and 0.025" (if you can only read to the nearest 0.05).
Apart from that, your actual calculations look OK.

[earthnation112]

Thanks for the input guys, and the reason it was to three significant figures is because the teacher asked for it and this experimental data was already given by the lecturer so your right I've read a burette before and it's not possible to get it as accurate as 24.79. Again thanks for the input.