To standardize a sodium hydroxide (NaOH) solution against a primary standard acid [Potassium Hydrogen Phthalate (KHP)] using phenolphthalein as an indicator.
|Independent variables||Mass of KHP (mKHP) |
Volume of KHP solution
|Dependent variables||Volume of NaOH added [since the colour change will not happen at exactly the same volume of NaOH added (VNaOH)]|
|Controlled variables||Concentration of NaOH |
Type of Acid/Base Indicator used – Phenolphthalein
Amount of indicator added – 3 drops
- 2 g of KHP
- Two 100 cm3 Beakers (One for making the KHP solution, one for pouring NaOH into the burette)
- 1 Digital Balance (up to 2 decimal places accuracy)
- 1 Stirring rod
- 1 Funnel
- 100 cm3 volumetric flask
- 100 ml distilled water
- Conical flask
- 100 cm3 burette
- Sodium Hydroxide
Potassium Hydrogen Phthalate ( referred to in the experiment as KHP) was a brittle, white, crystalline substance. The crystals required intense stirring before they could be dissolved in water. The resultant Acidic solution was transparent, with a small amount of undissolved granules of KHP. As the transparent NaOH solution came into contact with transparent phenolphthalein in the KHP solution, it turned pink which on shaking became transparent. However, as NaOH was added further, there came a point when no amount of stirring changed the pink colour.
- % Uncertainty of KHP Mass = (0.01/mKHP) x 100
- % Uncertainty of (aq) KHP in Volumetric Flask = (0.1/100) x 100
- % Uncertainty of (aq) KHP in Pipette = (0.1/10) x 100
- % Uncertainty of NaOH Volume = (0.1/ VNaOH) x 100
Moles (nvf) of KHP in volumetric flask = mKHP/MKHP where MKHP is the Molar Mass of KHP (204.22 g)
[c]KHP in 100 cm3 volumetric flask = n/V
Moles of KHP in 10 cm3 of solution in where V is a given volume of water
conical flask (ncf) = [c]KHP x V
n = m/M mol
[c]1 x V1 = [c]2 x V2
The volume of NaOH added = Final Volume – Initial Volume
n is the number of moles of KHP
m is the mass of KHP in grams
M is the Molar mass in grams
V is the volume in cm3
[c]KHP = (n/V) mol dm-3 = (0.00974/0.1) mol dm-3 = 0.0974 mol dm-3
Where [c]KHP is the concentration of KHP Acid.
Number of moles of KHP in 2.00 grams = (m/M) = (2/204.22) mol = 0.00979 mol
[c]KHP = n/V = (0.00979/0.1) mol dm-3
Number of moles of KHP in 0.01 dm3 of solution in conical flask = [c] x V
= 0.0979 x 0.01 = 9.79 x 10-4 mol.
KHC8H4O4 + NaOH → H2O + NaKC8H4O4
Mole ratio = 1 KHP:1NaOH
From the mole ratio, the number of moles of NaOH = 0.00979 mol.
[c]NaOH = n/V = (0.00979/0.0950) = 0.103 mol dm-3 (cm3 is converted into dm3)
Lab 1: Preparation of KHP Acid
Weight of weighing boat before adding KHP = 2.67 g
Weight of weighing boat with KHP = 4.67 g
Weight of weighing boat after transfer = 2.68 g
Molar Mass (M) of KHP = 204.22 g
Volume (V) of water = 100 cm3 = 0.1 dm3
Mass of KHP Transfer = Weight of weighing boat with KHP – Weight of weighing boat after transfer
= (4.67 − 2.68) g = 1.99 g.
Lab 2: Standardisation of NaOH solution
|Rough Trial||Trial 1||Trial 2|
|Initial Volume of NaOH (±0.1 cm3)||0.0||9.9||0.0|
|Final Volume of NaOH (±0.1 cm3)||9.9||20.9||10.4|
|The volume of NaOH added (±0.1 cm3)||9.9||11.0||10.4|
|Average Volume of NaOH added in 3 trials (±0.1 cm3)||10.4|
*Initial volume is the initial reading of the burette and final volume is the reading after adding NaOH solution
KHC8H4O4 + NaOH → H2O + NaKC8H4O4
Mole ratio = 1 KHP: 1NaOH
From mole ratio, number of moles of NaOH = 0.00974 mol
|Number of moles of KHP in 1.99 grams||m/M||1.99/204.22||0.00974 mol|
|[c]KHP||n/V||0.00974/0.104||0.0937 mol dm-3|
|Number of moles of KHP in 0.01 dm3 of water||[c] x V||0.0937 x 0.01||0.00937 mol|
|[c]NaOH||n/V||0.00974/0.0104||0.0937 mol dm-3|
|Mass of KHP||(0.01/1.99) x 100 = ±0.503%||mKHP = 1.99±0.503% g|
|Moles of KHP in 1.99 g||(1.99 g ± 0.503%/204.22)||n = 0.00974±0.503% mol|
|Volume (aq) KHP in volumetric Flask||(0.1 cm3/100) x 100 = ±0.1%||V = 0.1±0.1% dm3|
|Volume (aq) KHP in pipette||(0.1/10) x 100 = ±1%||V = 0.01±1% dm3|
|[c]KHP||(0.00974 mol ±0.503%/0.1 dm3±0.1%||[c]KHP = 0.0974±0.603% mol dm-3|
|Moles of KHP in 10 cm3 of solution in conical flask||(0.0974 mol dm-3±0.603% x 0.01 dm3±1%)||n = 0.00974±1.603% mol|
|Volume of NaOH||(0.1/10.4) x 100 = ±0.962%||V = 10.4±0.962% cm3 = 0.104±0.962% dm3|
|[c]NaOH||(0.00974±1.603/0.104±0.962%)||[c]NaOH = 0.0937±2.57% mol dm-3|
Data Analysis and Conclusion
One experimental flaw which resulted in readings inconsistent with the literature value was due to human error. This flaw was due to allowing excess sodium hydroxide to flow, causing the KHP solution to become pinker than it should have. This might have caused some deviations because the volume of sodium hydroxide added was excess. The theoretical value of NaOH to be poured was 9.50 cm3, and more or less than 0.1 cm3 of that value. However, the amount I added on an average was 10.4 cm3, which suggests why the solution became unusually dark pink as supposed to light pink.
The theoretical value of the Sodium Hydroxide that was expected to be used was 9.50 cm3. However, there has been a deviation of 0.9 cm3, which is significant, but not high. The resulting percentage error out of this deviation is:
There is almost a 1% deviation. Taking the value of 9.50 cm3 and mass of 2 grams, the concentration of NaOH should have been 0.103 mol, but the value I obtained due to the excessive deviation gave me 0.0937 mol. The percent error that has resulted in:
9.03% is by far a significant error that has resulted from a small error in the volume. Another error was caused by the deviation in the mass of KHP. 2.00 grams was the amount expected to be taken, but the experimental amount was 1.99 grams. However, this, being only 0.01 grams of the expected value, could have only constituted a very small portion of the error. This means that due to systematic error, my accuracy has fallen by 9.03%, which, although not high, is quite a deviation inaccuracy.
The deviation in the volume, however, is not the only indicator of noticeable systematic errors. The volumes of NaOH used up show significant fluctuations. For example, in trial 1, I used 11.0 cm3 of NaOH, which is 1.50 cm3 off 9.50 cm3, and in my rough trial, the volume used was 9.9 cm3 and in trial 2, the amount used up was 10.4 cm3. The difference between these sets of data indicates that the systematic error of allowing the KHP solution to become too pale resulted in strange fluctuations. These fluctuations caused the 0.95% error.
The percentage uncertainty calculated of the concentration of NaOH was 2.57%, which indicates that the level of precision, although not low, could have been better. The expected % uncertainty that was expected was 0.500%, and the uncertainty I obtained was 0.503%. Also, the % uncertainty of the volume of NaOH was ±1.05%, taking the value of 9.50 cm3. Therefore, due to flaws in raw data values taken from systematic errors, there has been a deviation in uncertainty too, indicating the impact of methodical flaws. The uncertainty of 2.57% indicates that my values were accurate up to within ±2.57%.
Overall, the data obtained, although not completely inaccurate, is not as accurate as it could have been. These errors were avoidable.
|Allowing the KHP solution to become too pink.||Due to excessive NaOH, there were accuracy issues in the calculation of the concentration was inaccurate.||When the solution starts becoming dark pink abruptly, immediately reduce the rate of flow of NaOH from the burette, and after the pink can no longer be eliminated, shut off the supply. This way, we avoid excess NaOH from being added|
|Inaccuracy in the mass of KHP||Taking 1.99 grams as supposed to 2.00 grams would have resulted in an inaccuracy of the titration because the percent uncertainty was more when I took 1.99 grams.||Measure KHP accurately.|
|Allowing the KHP to flow along the inner walls of the conical flask||This would have resulted in inaccuracies. The NaOH may not have reacted with the exact amount of KHP expected.||Make sure the conical flask is directly under the pipette, with no contact with the inner walls, so as to get a more accurate measure of the volume.|
This is an awesome source of information, Thank you !
Hello..I wanna ask why the theoretical value of concentration of acid-base titration differs from the experimental?
Due to Uncertainity
What I know is due to various errors which tend to happen when conducting an experiment such as contamination of the sample used (impurities) also external factor like temperature and humidity which results the sample to react with the atmosphere (air)