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The following is a typical problem in acid-base titration analysis of a real sample. The detailed resolution of the problem is presented with an in-depth explanation of the most important steps involved and that can be easily extrapolated to the resolution of any other acid-base titration problem, as well as other types of titrations such as precipitometries or titrations. redox.
There are several different ways to solve this type of problem, but we will emphasize the use of normality and the number of titling and titling equivalents at the end point of the titration. This allows any such problem to be solved by following exactly the same procedure, but changing the number of equivalents per mole of the titrant and the titrant depending on the type of reaction involved in the titration.
The problem in question consists of two acid-base titrations: one to carry out the standardization of the titrating agent using a primary standard, and the second to carry out the analysis of a real sample, so it is a very good approach to the kind of problems that must be solved in a real analysis laboratory. For simplicity, neither experimental errors nor the statistical analysis of the results will be considered.
Problem: Analysis of a toilet bowl cleaner by acid base titration
Statement:
It is desired to analyze a sample of commercial scale remover frequently used for cleaning toilets. The active principle of this product is hydrochloric acid (HCl) at 6.75% m/V, and it is analyzed by means of an acid-base titration with potassium hydroxide.
The potassium hydroxide solution was standardized by titrating a sample containing 0.4956 g of potassium acid phthalate, KHC 8 H 4 O 4 or KHP (MM=204.221g/mol). During the titration, 25.15 mL of KOH were consumed to reach the end point.
To analyze the sample, 10.00 mL of the cleaner was first taken and diluted to 250 mL with distilled water. Then, a 25.00 mL aliquot of this solution was taken and titrated with the previously standardized potassium hydroxide solution using phenolphthalein as indicator. The end point is reached after adding 17.50 mL of titrant. What is the actual concentration of HCl in the toilet bowl cleaner?
Solution:
As can be seen, the main objective of the problem is to determine the actual concentration of HCl in the toilet bowl cleaner, which should be around 6.75% m/V. Since the sample is too concentrated to be analyzed directly, it is diluted before being titrated. This means that the titration of the sample will not give us the concentration that we are looking for directly, but we must first find the concentration of the dilution and then, with this concentration, we will calculate the actual initial concentration of the cleaner.
To be able to calculate the concentration of the diluted solution by means of the titration, it is mandatory to know the concentration of the titrant, in this case, potassium hydroxide. However, this concentration is not provided by the exercise directly, but instead provides us with information from another titration that was carried out with the same titrating agent, but on a known sample of potassium hydrogen phthalate or KHP.
After this analysis, it is clear that, in order to solve the problem, we must first calculate the concentration of potassium hydroxide using the data from the first titration (the standardization), then this must be used to determine the concentration of the diluted sample and , finally, the concentration of the original concentrated solution, that is, of the sample, is determined.
Data:
- 1st titration (KOH standardization)
| Titrant: KOH (W KOH = 1 eq/mol) | Titrated = KHP (W KHP = 1 eq/mol) |
| VKOH = 25.15 mL | m KHP = 0.4956 g |
| NKOH = ? | MM KHP = 204.221 g/mol |
The number of equivalents per mole (W) of KOH is 1 since it is a base that only has a hydroxide ion, while potassium acid phthalate is an amphoteric salt that in this case acts as an acid (because it is reacting with a base) monoprotic (because it has only one proton), so it also has 1 equivalent per mole.
The titration reaction is:
- 2nd degree (sample analysis)
| Titrant: KOH (W KOH = 1 eq/mol) | Titrated = HCl (W HCl = 1 eq/mol) |
| VKOH = 17.50 mL | V aliquot = 25.00 mL |
| NKOH = ? | N aliquot = ? |
Like phthalate, hydrochloric acid is also a monoprotic acid, so the number of equivalents per mole of this acid is also 1.
In this case, the titration reaction is:
- Dilution
| V concentrate = 10.00 mL | Diluted V = 250.0 mL |
| N concentrated = ? | N diluted = ? |
calculations
The purpose of using normality instead of another unit of concentration when solving titration problems is that, at the end point of the titration, which is assumed to be equal to the equivalence point, it is true that the equivalents of the titrant are equal to the degree equivalents. That is:
where the number of equivalents can be obtained either from the mass of the substance and its molecular weight, or from its normal concentration as follows:
where m is the mass, W is the number of equivalents per mole, MM is the molar mass, N is the normal concentration, and V sol is the volume of the solution.
These three equations are usually enough to solve any titration problem.
KOH solution standardization
The above three equations can be combined into one to find the normal concentration of the potassium hydroxide solution, that is, of the titrant. At the end point of the standardization, it holds that
Titration of the diluted sample aliquot
Now that we have the concentration of the titrant, we can use it to determine the concentration of HCl in the aliquot. Again combining the relation of the equivalents at the end point with the normality formula, we can write:
Dilution
We already found the concentration of the aliquot that was titrated, which corresponds to the concentration of the diluted solution of the original sample. Now we only need to use the dilution equation to determine the concentration of the original stock solution.
This is the agreement we were looking for. The only thing left is to convert it to percentage m/V in order to compare it with the value reported on the label. For this, we consider that the solution contains 1,689 equivalents of HCl in 1 L = 1000 mL of solution. This, together with the molar mass of the HCl and the number of equivalents per mole will allow the percentage m/V to be calculated:
The actual concentration of HCl in the scaler tested is 6.158% m/V and that is only slightly different from ours. If we compare this value with the values most likely to be complete
References
Ahumada Forigua, DA, Morales Erazo, LV, Abella Gamba, JP, & Gonzalez Cardenas, IA (2019). Acid-base titration techniques: metrological considerations. Colombian Journal of Chemistry , 48 (1), 26–34. Retrieved from https://www.redalyc.org/jatsRepo/3090/309058491010/309058491010.pdf
Wisdom (nd). ACID-BASE titration and REDOX titration exercises. Retrieved from https://sapiencia-web.blogspot.com/p/itulacion.html
Skoog, DA, West, DM, Holler, J., & Crouch, SR (2021). Fundamentals of Analytical Chemistry (9th edition). Boston, Massachusetts: Cengage Learning.
TP Chemical Laboratory. (2015, November 15). Acid Base Titrations . Retrieved from https://www.tplaboratorioquimico.com/quimica-general/acidos-y-bases/titulaciones-acido-base.html
Jiménez, AG, and Hernández, AR (sf). Standard substances for the standardization of acids and bases Retrieved from http://depa.fquim.unam.mx/amyd/archivero/DOCUMENTOPATRONESPRIMARIOSACIDOBASE_34249.pdf
