# Calculations reading Burette – 0.05 cm3 per

Calculations for experiment 1Titration Rough 1 2
Final burette reading /cm3 25.10 25.80 31.90
Initial burette reading /cm3 0.00 3.00 7.00
Volume of NaOH used /cm3 25.10 24.80 24.90
Best titre value()
Table 1.1: Titre values for practical 1
Average volume of NaOH used = 24.90+24.802 = 24.85cm3
HCL +NaOHNaCL + H20
Number of moles of NaOH used =24.85 × 0.51000 = 0.012425mol
From equation:
Since reaction between NaOH and HCL is in the ratio 1:1, this implies that 25 cm3 of HCL contains the same number of moles of NaOH in 24.85 cm3.

Hence, 25cm3 of HCL contains 0.012425mol
1000cm3 of HCL contains 0.01242525 ×1000
= 0.497mol
Concentration of HCL = 0.497mol/dm3

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Error analysis for experiment 1:
Every measurement has got an uncertainty which is mainly due:
Inefficiency in the ability of the human senses to detect small changes
Limitations in the measurement scale
To the way the apparatus in being used
Some unexpected source that can increase uncertain are:
Parallax error – eyes are not placed perpendicularly to readings
Thickness of glassware – may not be uniform causing scale lines to vary
Uncertainties in apparatus used:
Pipette – 0.06 cm3 per reading
Burette – 0.05 cm3 per reading
Percentage error in:
25 cm3 pipette
The pipette was used twice since the average titre value depends on the 2 titres used which in turn depends on the two pipette readings.

Percentage error while using pipette for whole experiment = 0.06+0.0625 100%
= 0.48%
50.00 cm3 burette
The burette was used twice since the average of the two titres were used. This implies that, for each titre the error per reading will double (burette was read twice).

Percentage error while using burette for whole experiment =( 20.0524.80 + 20.0524.90) 100%
=0.80%
Calculations for experiment 2
Calculating mass of potassium iodate needed to be dissolved in a 250cm3 standard solution to give a titre of around 25cm3.

I2 + 2S2O32- 2I- + S4O62- ————————-1
IO3- + 6H+ + 5I- 3I2 + 3H2O ———————2
Number of moles of Na2S2O3.5H2O = 0.11000 × 25
=2.50 × 10-3mol
From equation 1:
2 moles of 2S2O32- react with 2 moles of I- causing ratio to be 2:2.

Hence, number of moles of I2 =2.50 10-3mol
=2.50 10-3mol
From equation 2:
5 moles of I- react with 1 mole of IO3- causing ratio to be 5:1.

Hence, number of moles of IO3- used = 2.505 × 10-3
= 5.00 × 10-4m
This implies that 25cm3 contains 5.00 × 10-4mol
250cm3 contains 5.00 × 10-3mol
Hence, mass of KIO3 = Number of moles × Mr(KIO3)
=5.00 × 10-3 × (39.1 + 126.9 + 3(16))
=1.07g
Mass required considering purity = 1.0799 × 100
=1.08g
Preparation of potassium Iodate – KIO3
Mass of beaker and potassium iodate /g 42.200
Mass of empty beaker /g 41.120
Mass of potassium iodate used /g 1.080
Table 2.1: Mass of potassium iodate used
Finding concentration of sodium thiosulfate solution
Titration Rough 1 2
Final burette reading/cm3 30.20 32.60 35.70
Initial burette reading/cm3 0.00 3.00 6.10
Volume of S2O32-used/cm3 30.20 29.60 29.60
Best titre value ()
Table 2.1: Titre values for practical 2
Average volume of Na2S2O3 used = 29.60+29.602 = 29.60 cm3
IO3- + 6H+ + 5I- 3I2 + 3H2O ——————– 1
I2 + 2S2O32- 2I- + S4O62-
3I2 + 6S2O32- 6I- + 3S4O62- ———————- 2
IO3- + 6H+ + 6S2O32- 3H2O + I- + 3S4O62-
Number of moles of IO3- used = 1.07 25214 250

= 5.00 10-4mol
From equation:
6 moles of S2O32- react with I mole of IO3- , causing ratio to be 6:1.

This implies that, number of moles of S2O32- = 6 5.00 × 10-4mol
= 3.00 10-3mol29.60cm3 of Na2S2O3 contains 3.00 10-3mol
1000cm3 of Na2S2O3 contains 3.00 10-3 100029.60=0.101mol
Hence, concentration of sodium thiosulfate = 0.101mol/dm3
Error analysis for experiment 2:
Every measurement has got an uncertainty which is mainly due:
Inefficiency in the ability of the human senses to detect small changes
Limitations in the measurement scale
To the way the apparatus in being used
Some unexpected sources that can increase uncertainty are:
Parallax error – eyes are not placed perpendicular to readings
Thickness of glassware – may not be uniform causing scale lines to vary
Dust on electronic balance
Uncertainties in apparatus used:
Pipette – ±0.06 cm3 per reading
Burette – ±0.05 cm3 per reading
Electronic balance – ±0.0005g per reading
Percentage error in:
25 cm3 pipette
The pipette was used twice since the average titre value depends on the 2 titres used which in turn depends on the two pipette readings.

Percentage error while using pipette for whole experiment = 0.06+0.0625 100%
= 0.48%
50.00 cm3 burette
The burette was used twice since the average of the two titres were used. This implies that, for each titre the error per reading will double (burette was read twice).

Percentage error while using burette for whole experiment =( 20.0529.60 + 20.0529.60) 100%
=0.68%
Electronic balance
The electronic balance was used twice. First to measure the mass of empty beaker and second to the mass of the beaker with the potassium iodate. Causing error to be twice.

Percentage error while using electronic for whole experiment =0.0005 21.08 100%
=0.09%