The equivalence point is when the amount of KHP is added is equal to the amount of analyte present in to unknown solution, which in this case is normality. Since the acid is a monoprotic acid, there is only one equivalence point.
At this equivalence point we can then determine the Normality of our unknown with our uncertainties as well.
Experimental:
- Preparation of the KHP (standard solution) (titrant)
Using an Analytical balance 5.3134 g (0.0007 mol) of dried KHP was weighed out. The acid was then accurately using a weighing boat into a 250 mL volumetric flask and diluted to the mark, while swirling throughout, with distilled boiled water. The acid was then put into a 50 mL burette for titration.
- Preparation of the Unknown Base
After acquiring the unknown base, using a 25 mL transfer pipette, 25mL was accurately dispensed to three 250 mL Erlenmeyer Flask, diluted with distilled boiled water to approximately the 50mL mark. Phenolphthalein is then added (2 drops) for titration.
- Titration of the Unknown base
Each Erlenmeyer flasks are placed under the burette for separate titration. After recording the initial volume, the KHP is dispensed into the unknown base until it reaches equivalent point and the unknown will changed into a light pink colour. The Final volume is then recorded.
Results:
- Data
MW KHP = 204.22 g/mol
M KHP Weighed = 5.3134 g
Graph 1: Titration Measurements
- Calculations
Va * Na = Vb * Nb
Normality of unknown
Va: Volume of KHP = 24.92 ± 0.00022
Na: Normality of KHP = (5.3134 ± 0.0001) g KHP * (1 mol KHP / 204.22 g KHP)
= 0.0260 ± 0.0001 mol KHP * (1 g ep KHP/ 1 mol KHP)
= 0.0260 ± 0.0001 / 0.250 ± 0.00012
= 0.104 ± 0.00002 N
Vb: X-bar = (22.36 + 22.40 + 22.37)/3 = 22.38 mL
σ = (22.36 – 22.38)² = 0.0004
(22.40 – 22.38)² = 0.0004
(22.37 – 22.38)² = 0.0001
σ = √ (0.0009 / 2) = √ 0.00045 = 0.02121
Vb = 22.38 ± (0.02121 / √3)
Vb = 22.38 ± 0.01224
Burette correction: 0.06 (found in calibration of burette calibration)
Error = √ (0.01224)² * (0.06)²
Error = 0.0007
Vb = 22.38 ± 0.0007
Nb: ?
Va * Na = Vb * Nb
(24.92 ± 0.00022) mL * (0.104 ± 0.00002) N = (22.38 ± 0.0007) mL * Nb
Nb = {(24. 92 ± 0.00022) mL * (0.104 ± 0.00002) N} / (22.38 ± 0.0007) mL
Nb = 0.1158 ± σ
σ = √ (0.00022/24.92)² + (0.00002/0.104)² + (0.0007/22.38)²
σ = 0.00020
The normality of the unknown #176 is Nb = 0.1158 ± 0.00020 N
95 % Confidence Interval:
= 0.1158 ± (z σ / √n)
= 0.1158 ± ((1.96*0.00020)/√3)
= 0.1158 ± 0.00023
The 95% confidence interval is 0.1158 ± 0.00023 N
Relative standard deviation: (ppt = mg/L)
s / x-bar
s = 0.00023 g eq * 1 mol eq * 204.22 g = 0.04697 g = 0.046971 ppt
1 L 1 g eq 1 mol 1 L
x-bar = 0.1158 g eq * 1 mol eq * 204.22 g = 23.649 g = 23.649 ppt
1 L 1 g eq 1 mol 1 L
Relative Standard Deviation = (0.046971/23.648) = 0.0020 ppt
Discussion
There are nominal values of uncertainties in each step of our experiment. In step 1: Preparation of KHP, there could be a method error, when diluting the KHP in distilled boiled water, the reaction could have not gone to completion, as in the KHP could have not gone to completion. In step 2: Preparation of the unknown base, there could have been another method error, when adding the phenolphthalein we could have added too much then needed. And in step 3: Titration of the unknown base, there could have been personal errors, when reading the burette, there could have been parallax error when reading the calibrated pointer, there also could have been an error in judging when the end-point is achieved (the colour).
The step with the greatest uncertainty in this experiment would have to be in step 3. Personal error would give the greatest uncertainty when achieving the end-point. My estimation on end-point would and could be different then someone else’s judgment on end-point.
With good laboratory practice these uncertainties would be reduced.
Other ways the reduce uncertainties in this lab would be If we were too have longer time to make sure that the step 1, would have gone to completion or in step 2: a different indicator could have been added with a better view of the end-point.
Conclusion
After titration of an unknown with a standard of KHP, I have found that Unknown #176 has a normality of 0.1158 ± 0.00020N, with a 95% confidence interval of 0.1158 ± 0.00023 and a relative standard deviation of 0.0020 ppt.
Reference
