Numbers of Atoms: Real chemists still do titrations
In our last Numbers of Atoms post, we considered volumetric analysis – a simple approach to quantitative analysis that allows for relatively quick determination of the concentration of an analyte in a solution. In a titration, a quantity of reactant (the titrant) is added to a known quantity of a sample in solution, until all of the analyte has reacted. That “right amount” of titrant is detected by an indicator, and the concentration of the analyte in the sample is derived from the quantity of titrant that was used.
In that last post, we walked through the classic introductory example: the determination of the concentration of a strong acid by titration with a strong base.
I’m sure all of that keeps students busy for a few hours, but I imagine real analytical chemists don’t do titrations. They must have fancy machines to do all of that work.
There are machines that can automate titrations, and some shiny new instruments have replaced the need for them, but a resourceful chemist will know how to do an analysis by titration. Not only does it allow for analysis without an expensive instrument, but it also provides a convenient way to validate that an instrument is working properly.
Besides, we mustn’t trivialize the chemistry behind the titration. If the sample is properly treated to extract only the analytes of interest – consider our articles on separating analytes in samples – such that the analyte is the only thing in the solution that will react with the titrant, then the titration can be a very efficient approach for finding its concentration. The limitations of the method will be the accuracy at which one can measure the masses and volumes used in the analysis.
Alright then, but seriously – how often does a chemist need to know the exact concentration of a strong acid?
That isn’t the only use of a titration. Many compounds can be analyzed by a titration of some form, as long as the reaction involved is fast enough and works to completion. It is possible to titrate a solution of a weak acid such as acetic acid with NaOH, even it is known that only a fraction of the acetic acid is dissociated at any one time – this is why it is a weak acid. (A strong acid such as HCl is completely dissociated at all times.) The NaOH that is added does react completely and rapidly with the acetic acid, and the stoechiometry of one mole of NaOH reacting with one mole of acetic acid holds. This is, in fact, a fairly common titration.
Now, let us consider a titration that is more involved, even though it remains an acid/base reaction: the determination of protein content in food. There are modern instruments that can do this with reasonable precision, such as near-infrared spectroscopy. But the gold standard method for quantifying protein content in food remains the Kjeldahl Method, which has existed for 130 years.
The Kjeldahl method involves four steps. The first three steps will get the nitrogen within the proteins into solution:
- The sample of food is first mixed with concentrated sulfuric acid, which causes the proteins to decompose and the nitrogen within it to turn in ammonium sulfate ((NH4)2SO4).
- The ammonium sulfate is then treated with concentrated sodium hydroxide, releasing ammonia (NH3) gas. Of course, this gas must be trapped – any loss of ammonia and the final result we will calculate will be incorrrect.
- The ammonia gas is then reacted with an acid – hydrochloric acid (HCl) or boric acid (B(OH)3) is often used. The ammonia gas turns into the ammonium ion (NH4+), which is now in solution.
An ingenious trick is usually played in that third step. The acid solution has a known concentration, and an exact volume is added – a volume that is more than what will be necessary to react with all the ammonia gas.
Why all this ingenuity? Wouldn’t it be easier to just titrate the gas with the acid, like we did with the strong base in the last example?
Not in this case, since the ammonia gas is, well, a gas, and the reaction to turn it into ammonium ions is not a rapid reaction. In this case, the best approach is to add an excess of acid to be certain that all ammonia has reacted. In the final step, we titrate the unreacted acid – a process known as a back titration.
A back titration? As in a titration in reverse?
More of an indirect titration. We are, of course, interested in how much NH4+ is in solution, since we can assume that every atom of nitrogen from the protein in the original food sample corresponds to one molecule of ammonia gas, and after that, one NH4+ ion.
The back titration will tell us how much excess acid was added. The quantity of acid that was not in excess is what reacted with the ammonia gas.
This looks like alot of steps – show me an example.
No problem. Let us consider a sample of 1.035 g of milk, for which we want to determine its protein content. The sample has undergone the treatment described above. The trapped ammonia gas is treated with 10.00 mL of 0.05188 mol/L HCl. The solution was titrated with 0.009274 mol/L NaOH, of which 11.88 mL was needed to reach the equivalence point. Note that we can convert the units of the concentrations into mmol/mL without the number changing, and it saves us from converting the units of the volumes. We find that we had added 0.5188 mmol of HCl, of which 0.1102 mmol was in excess (because that was the quantity of NaOH necessary to neutralize the excess acid). The difference, 0.4086 mmol, represents the quantity of HCl consumed by the ammonia gas. It also represents the quantity of nitrogen that was in the original milk sample.
Our sample contained 0.4086 millimoles of nitrogen – and we can find that the mass of nitrogen (based on 14.0067 mg N/mmol) was 5.723 mg.
Wait a minute – can we assume that all of the nitrogen in the milk came from protein?
That’s a good point – and in the case of milk, we assume that on average, 6% of the nitrogen comes from sources other than proteins. (This number varies by the source – this article says 5%.) So we can take 94% of that mass as relevant to protein determination – leaving us with 5.380 mg N from protein.
Each protein consists of a certain percentage of nitrogen atoms. Proteins from milk have, on average, 15.65% nitrogen by mass – in other words, 1 mg of protein will contain 0.1565 mg N. So our 5.380 mg of N would have come from 34.38 mg of protein. That is the mass of protein in our original milk sample, which weighed 1.035 g (1035 mg). Therefore, the protein content in this milk sample was 3.32%.
OK, I think I get that. The equipment doesn’t seem too complicated, and preparing the solutions would be easy.
Preparing the solutions is easy, but there is a catch that I’ve avoided until now. Of course, we always need to be certain of the concentration of the titrant. Titrants are usually prepared by either diluting concentrated solutions, or weighing them in solid form and dissolving. But for some titrants, including NaOH, it is not as simple as weighing and dissolving. In the case of NaOH, its solid form is in pellets that can adsorb water from the air, even while sitting on the scale being weighed.
But aren’t we dissolving the sodium hydroxide in water? What difference does that make?
Consider this – a technician want to prepare one litre of 0.1 M NaOH, by weighing 4 g of NaOH pellets and dissolving it in water. (One mole of NaOH weighs 39.997 g, but this technician knows she can be a touch over or under that number as long as she records the exact mass.) The mass that appears on the scale will include any water that is adsorbed into the pellets. Even if the scale read exactly 3.9997 g, there will be slightly less than that in actual NaOH.
How much less? That’s the problem – we need to know the concentration of the titrant with as much precision as possible, and “close to 0.1M” just won’t do.
OK then. That sounds like a problem, but I take it there is a solution.
There is – and you have to prepare it.
Yes, did you miss cringing at my bad jokes?
To find the actual concentration of NaOH in the prepared solution, we would titrate it with a solution of a primary standard. That’s right – in this preliminary step, the NaOH would be the solution being analyzed in the flask, and the primary standard solution would be the titrant in the buret.
Only a select few compounds make reliable primary standards – these compounds will be solids that are easy to weigh and dissolve into solution. Primary standards are available at high purity (at least 99%), they are stable compounds with low hygroscopicity (in other words, they do not decompose when heated, or adsorb water when exposed to air), and they have a high molar mass.
Why does the molar mass matter? What is the lowest molar mass that is allowed?
There is no hard-and-fast rule, but the idea is that solutions are prepared with an approximate concentration (moles per litre) in mind. The higher the molar mass of the substance, the more that will be dissolved, resulting in a smaller relative error when weighing its mass before preparing the sample.
All of these characteristics speak to the preparation of the solution, since we need to trust that the mass of the primary standard on the scale is the mass of that substance that is in solution when it is prepared, and we must also trust that the concentration will be the same after being in solution for a few hours. Generally, the primary standard is heated to just over 100°C to evaporate all water, then left to cool in a desiccator, a sealed apparatus with a desiccant – a salt that easily adsorbs any moisture within the dome, so that the primary standard does not do so.
Back to our NaOH solution – before conducting the titration of an acid solution, it will be titrated with a primary standard solution. The best primary standard is potassium hydrogen phthalate, or KHP. (A common mistake is to think of “P” as phosphate or phosphorous, particularly when calculating its molar mass.) Phenolphthalein is used as an indicator for this titration, and since we are adding an acid to a base, the equivalence point will be found when the indicator changes from pink to clear. The stoichiometry of the NaOH/KHP reaction is 1:1, so the number of moles of KHP needed to reach the equivalence point is equal to the actual number of moles of NaOH in the aliquot being analyzed. From that, we can calculate the actual concentration of NaOH that we will use as a titrant in analyzing the acid solution.
Wow, that’s alot to think about. You mentioned that there are other types of titrations…
Yes I did, and they all operate on the same principle: finding the right amount of titrant to react exactly with all of the reagent in a solution. In the next post we will look at complexometric titrations of metals and redox titrations.