Equation for titration:
AgNO3 + KX → AgX + KNO3
0.050moldm-3 AgNO3 Solution in the burette
Average Burette reading = 17.40cm3 + 0.01cm3
25.00cm3 + 0.06 cm3 of the unknown potassium halide solution (KX) was transferred into a conical flask using a 25.0cm3 pipette.
The number of moles of AgNO3 used in the titration = concentration x volume
= 0.05 x 17.4
1000
= 0.870 x 10-4 moles AgNO3
1 mole of AgNO3 = 1 mole KX
Therefore the number of moles of KX in 0.025dm3 = 0.870 x 10-3
Thus the number of moles of KX in 0.25dm3 = 10 x 0.870
= 8.70 x 10-3
Therefore the concentration of the KX solution = number of moles
Volume
= 8.70
0.250
= 0.03480moldm-3
1.035g of KX was put into a 250cm3 volumetric flask and made up to a 250cm3 solution.
1.035g of KX was 8.70 x 10-3 moles.
Therefore 1 mole of KX = 1.035 .
8.70 x 10-3
= 118.9gmol-1
One mole of KX has an atomic mass of 118.9gmol-1.
Atomic mass of one mole of KX = 118.9gmol-1
Atomic Mass of one mole of K = 39.10gmol-1
Therefore the atomic mass of one mole of X = 118.9 – 39.10
Atomic mass of X = 79.80gmol-1
Therefore X = Br
The unknown potassium halide was Potassium Bromide.
Percentage Uncertainty
Percentage Uncertainty of Volumetric Flask = 0.3 . x 100 = 0.12%
250.0
Percentage Uncertainty of Pipette: 0.06 x 100 = 0.24%
25.0
Percentage Uncertainty of balance: 0.001 x 100 = 0.10%
1.035
Percentage Uncertainty of Burette: 0.1 x 100 = 0.57%
17.4
Overall Percentage Uncertainty of Equipment = 0.24 + 0.10 + 0.57 + 0.12
= 1.03%
Effect of the total percentage uncertainty on the result [Atomic Mass KX = 118.9gmol-1] = 118.9 x 1.03
100
= 1.22gmol-1 (3sf)
Therefore the atomic mass of KX = 118.9gmol-1 + 1.22gmol-1 (3sf)
Conclusion
Using my observed results I calculated the atomic mass of the unknown potassium halide solution, which I found to be 118. 9gmol-1 , and using this result, I can conclude that the unknown potassium halide used in this titration was Potassium Bromide.
The literature value for the atomic mass of Potassium Bromide is 119.0gmol-1. This value is very close to my observed value, which suggests that the error of my observed result is very small. Using this literature value, I can calculate a quantitative value for the percentage error of my observed result.
Percentage Error = 119.0 – 118.9 x 100
119.0
= 0.0840% (3sf)
Evaluation
It was assumed that the endpoint of the titration occurred when a dark pink precipitate started to form. However, the silver nitrate solution may have been affected by the light in the room, resulting in the precipitate appearing differently each time a titration was carried out depending on how much the silver nitrate solution had been affected by the light in the room. This may have resulted in the end point being overstepped, or not quite reached, which would have affected the accuracy of the results.
Although there was a total percentage error on the equipment of 1.03%, this is negligible, as the greatest error on the results in this titration would be caused by personal judgement, as when the end point is judged to have occurred is dependent on the person carrying out the titration.
The accuracy of the results obtained from this titration may be improved if the silver nitrate solution was held in a container that blocked out light. As a result of this, the silver nitrate would be less affected by the light, and consequently the end points of each titration would be more consistent, therefore improving the accuracy of the results.
