Quantitative determination of Potassium Acid Phthalate (KHP) by an Acid-Base Tit
ID: 1038366 • Letter: Q
Question
Quantitative determination of Potassium Acid Phthalate (KHP) by an Acid-Base Titration
1. According to this experiment, what reagent meet the criteria of a primary standard? (2pt) 2. What can you do if you do not have a suitable primary standard to carry out your analysis? (2 pts) 3. Identify three possible error sources in this experiment. (2 pts) 4. Using the following data, obtained after the experiment for the analysis of an unknown KHP sample, calculate: (4 pts) a. The concentration of the KOH solution. b. The average percent of KHP in the unknown. C. The relative deviation of HP (%) in parts per thousand (ppth). d. If the accepted value for the KHP unknown sample is 26.75%, determine the relative error in ppth (parts per thousand) Titration Data KHP PRIMARY STANDARD MASS (G) KOH VOLUME (mL) (99.97%) 0.7123 0.7131 0.7325 26.41 26.89 26.52 MASS (G) 0.8424 0.8534 0.8429 KOH VOLUME (mL) 7.80 8.20 8.50 KHP UNKNOWN SAMPLEExplanation / Answer
Hello,
This question involves the laboratory standardization of KOH (aq) by Potassium Hydrogen Pthalate which is commonly abbreviated as KHP or Quantitative determination of KHP. If you’ve ever seen pellets of KOH, one very distinctive feature of theirs is being hygroscopic i.e. they absorb moisture from air and the surface becomes slippery. So, suppose you intend to find the mass of KOH in anhydrous form using the balance, you’d end up getting incorrect results since KOH pellets upon absorbing moisture would gain weight & thus the further calculations would get messed up. So, it is imperative to standardize the sample of KOH for precise measurement of its concentration.
For this purpose, we use an organic acid, which is non-hygroscopic, with which an acid base reaction would ensure that the concentration of KOH is being identified correctly. The organic acid we use in this case is Potassium Hydrogen Pthalate (KHP) or KHC8H4O4. KHP behaves as a mono basic acid which is non-hygroscopic as well as a crystalline solid. KHP can be obtained as a solid of extremely high purity so it is utilized as a primary standard. Since KOH is a monoacidic base and KHP is a monobasic acid, so in aqueous medium, the reaction between the two turns out to be a 1:1 reaction i.e. for 1 mole of KOH, 1 mole of KHP is consumed. Phenolpthalein is used as an indicator for this titration.
In this titration, KOH is taken in the burette and KHP is taken in the titration flask. Few drops of Phenolpthalein indicator is added to the titration flask, where it remains colourless. KOH from the burette is added very slowly and just when the colour of the contents of titration flask turns faint pink, the burette reading is noted. The titration is repeated 2-3 times to reduce the chances of error.
The following neutralization reactions take place in this titration:
KHC8H4O4 (aq) + KOH(aq) à K2C8H4O4 (aq) + H2O(l)
Now coming to your questions:
1) According to this experiment, what reagent meets the criteria of a primary standard?
As earlier stated, we require an acid which is non-hygroscopic, little or no reactivity with air, can be obtained in high purity & reasonably large molar mass so that the relative error with weighing the standard is as low as possible. As seen from the data provided we are using KHP at the purity level of 99.97% so, Potassium Hydrogen Pthalate or KHP meets the criteria of the primary standard.
2) What can you do if you do not have suitable primary standard to carry out your analysis?
In such a scenario, there are some other compounds which also fulfil the criterions of a primary standard. These are again weak organic acids. The ones which can be used are Oxalic Acid (H2C2O4), Benzoic Acid (C6H5COOH), Succinic Acid (HOOC-(CH2)2-COOH) or Potassium Hydrogen Tartarate (KHC4H4O6).
3) Identify three possible error sources in this experiment.
The three possible sources of error that might creep in this experiment are:
i) During the titration, accidental addition of more KOH than required to reach the end point would result in an incorrect estimation of its concentration. Ideally we should stop adding KOH just when a persistent very faint shade of pink colour of Phenolpthalein indicator is observed, but a dark pink colour is not desired, which is obtained due to addition of more than optimum amount of KOH.
ii) The distilled water used for preparing the aqueous solution of KOH if not boiled for a few minutes would result in an error. Boiling the distilled water should be done for a few minutes to removed the dissolved CO2. This is done so as to ensure that a reaction between KOH and CO2 doesn’t take place and we estimate the correct concentration of KOH.
iii) KHP if not heated to 110oC would result in an another error as heating would help in removing any moisture adsorbed by the sample and thus a correct weight of the sample can be measured.
4. Using the following data, obtained after the experiment for the analysis of unknown KHP calculate:
KHP Primary standard (99.97%)
Mass (grams)
KOH volume (ml
1
0.7123
26.41
2
0.7131
26.89
3
0.7325
26.52
KHP Unknown Sample
Mass (grams)
KOH volume (ml
1
0.8424
7.80
2
0.8534
8.20
3
0.8429
8.50
i) The concentration of KOH solution
Using primary standard KHP (99.97%)
For Sample 1 –
Mass of sample = 0.7123 grams
Moles of KHP in the sample = 0.7123 grams/204.2 grams = 0.0034 moles
Moles of KOH used in the titration = 0.0034 moles (since the reaction takes place with 1:1 stoichiometry so the moles of KHP in the sample should correspond to the amount of KOH utilized)
Volume of KOH = 26.41 ml
Molarity of KOH based on Sample 1 = 0.0034 moles/0.02641 litres = 0.128 M
For Sample 2 –
Mass of sample = 0.7131 grams
Moles of KHP in the sample = 0.7131 grams/204.2 grams = 0.0035 moles
Moles of KOH used in the titration = 0.0035 moles (since the reaction takes place with 1:1 stoichiometry so the moles of KHP in the sample should correspond to the amount of KOH utilized)
Volume of KOH = 26.89 ml
Molarity of KOH based on Sample 2 = 0.0034 moles/0.02641 litres = 0.129 M
For Sample 3 –
Mass of sample = 0.7325 grams
Moles of KHP in the sample = 0.7325 grams/204.2 grams = 0.00358 moles
Moles of KOH used in the titration = 0.00358 moles (since the reaction takes place with 1:1 stoichiometry so the moles of KHP in the sample should correspond to the amount of KOH utilized)
Volume of KOH = 26.52 ml
Molarity of KOH based on Sample 1 = 0.0034 moles/0.02641 litres = 0.135 M
Average molarity of KOH using sample 1, 2 & 3 = 0.130 M
ii) Average percentage of KHP in the unknown
For Unknown Sample 1 –
Volume of KOH = 7.8 ml
Moles of KOH used in the titration = 0.130 M x 7.8 ml = 0.001 moles
Now the number of moles of KOH reacting with unknown sample should be equal to number of moles of H+ derived from KHP and by extension should be equal to the number of moles of KHP in the unknown sample.
So, the number of moles of KHP in the unknown sample 1 should be = 0.001 moles
Mass of KHP in unknown sample 1 = 0.001 moles x 204.2 grams = 0.2042 grams
Mass of unknown sample 1 = 0.8424 grams
% of KHP in the unknown sample 1 = (0.2042/0.8424) x 100 = 24.24 %
For Unknown Sample 2 –
Volume of KOH = 8.2 ml
Moles of KOH used in the titration = 0.130 M x 8.2 ml = 0.001 moles
Now the number of moles of KOH reacting with unknown sample should be equal to number of moles of H+ derived from KHP and by extension should be equal to the number of moles of KHP in the unknown sample.
So, the number of moles of KHP in the unknown sample 1 should be = 0.001066 moles
Mass of KHP in unknown sample 1 = 0.001066 moles x 204.2 grams = 0.2176 grams
Mass of unknown sample 1 = 0.8534 grams
% of KHP in the unknown sample 1 = (0.2042/0.8534) x 100 = 25.50 %
For Unknown Sample 3 –
Volume of KOH = 8.5 ml
Moles of KOH used in the titration = 0.130 M x 8.5 ml = 0.001105 moles
Now the number of moles of KOH reacting with unknown sample should be equal to number of moles of H+ derived from KHP and by extension should be equal to the number of moles of KHP in the unknown sample.
So, the number of moles of KHP in the unknown sample 1 should be = 0.00105 moles
Mass of KHP in unknown sample 1 = 0.00105 moles x 204.2 grams = 0.2256 grams
Mass of unknown sample 1 = 0.8429 grams
% of KHP in the unknown sample 1 = (0.2042/0.8429) x 100 = 26.76 %
The average percentage of KHP in unknown sample 1, 2 & 3 = (24.24 + 25.5 + 26.76)/3 = 25.5 %
iii) Relative standard deviation pf KHP (%) in parts per thousand:
Average percentage of KHP in unknown sample = 25.5 %
Deviation for unknown sample 1 = |25.5 – 24.24|= 1.26
Deviation for unknown sample 2 = |25.5 – 25.5| = 0.0
Deviation for unknown sample 3 = |25.5 – 26.76| = 1.26
Average deviation = (1.26 + 0 + 1.26)/3 = 0.84
Deviation in Parts Per Thousand = (0.84/25.5) x 100 = 3.29 (this is unitless)
iv) If the accepted value of KHP unknown sample is 26.75%, determine the relative error in parts per thousand.
Accepted percentage of KHP in unknown sample = 26.75 %
True percentage of KHP in unknown sample = 25.5%
Relative error in parts per thousand = [(26.75 – 25.5)/25.5] x 100 = 4.9 %
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KHP Primary standard (99.97%)
Mass (grams)
KOH volume (ml
1
0.7123
26.41
2
0.7131
26.89
3
0.7325
26.52