Practical 2
Finding the Mr of MHCO3
burette
flask
A standard solution is a solution whose concentration is accurately known.
This done by weighing out a sample of solute, transferring it completely to a volumetric
flask, and adding enough solvent to bring the volume up to the mark on the neck of the
flask.
METHYL ORANGE
acid
end-point
alkali
Titrations
• Record your results to the nearest 0.05 cm3
• Calculate your mean titre to two decimal place (does not
have to end in 0 or 5)
• Concordant results to 0.1cm3
Table
Mass /g
Weight of MHCO3 / g (before adding to beaker) (boat + solid)
22.78
Weight of MHCO3 / g (after adding to beaker) (boat + residue)
20.24
Final weight of MHCO3 added to beaker (g)
2.54
Analysis
• Find the Mr of MHCO3
•
•
•
•
•
•
•
•
•
Mol HCl = c x v = 0.1 x 0.02715 (titre dm3) = 0.002715
(1:1 ratio)
Mol MHCO3 in 25cm3 = 0.002715
Mol MHCO3 in 250cm3 = 0.02715
Mr =mass / mol
Mr of MHCO3 = 2.54/0.02715 = 93.63
Mr of M = MHCO3 – HCO3 = 93.63 – 61 = 32.6 (potassium Ar = 39.1)
M = potassium (closest Ar)
K
ERRORS - uncertainties
The margin of error
is the smallest unit
the equipment can
measure by and
then divide by two.
As this can go either
way we put +/before the value
Balance
Volumetric flask
Pipette
Burette
Remember to combine the uncertainties: for a burette its 3 x 0.05 (two readings to determine the
volume and judging end point to one drop)
The balance was used twice as the weighing bottle was reweighed to test for any residue so multiply the
uncertainty by x2
Evaluation
• Titration is reliable if you achieved concordant results
• Mr of KHCO3 = 100.1 (true value for exp error)
The real value will be the
Mr of KHCO3 you have just
calculated from the periodic
table 100.1
Comment: If the exp error is greater than the apparatus error then your results are in
accurate due to human error
Question 5
• Using a balance with such a low accuracy would have a significant
impact on the apparatus error and therefore have a higher likeliness
of inaccurate results
Apparatus error balance(+/- 0.5) : 0.5 x 2/2.54 x 100 = 39.37%
Apparatus error balance(+/- 0.01) : 0.01 x 2/2.54 x 100 = 0.78%
+/- 0.5 is a bigger margin of error than +/- 0.01, so the apparatus
error will be greater
Question 6
• Using a measuring cylinder will have a low accuracy and will therefore
significantly increase the apparatus error. The results are more likely to be
inaccurate due to the larger apparatus error.
• Apparatus error (measuring cylinder): 0.5/25 x 100 = 2.00%
• Apparatus error (pipette): 0.06/25 x 100 = 0.24%
• +/- 0.5 is a bigger margin of error than +/- 0.06 for the pipette, so the
apparatus error for the measuring cylinder will be greater than the pipette