Analytical Chemistry Technique: Titration

For non-science people, the terms "analytical" and "chemistry" may incite bad memories in high school AP chemistry class or just another moment that makes you go, "Ugh, science = math!" But, set aside your apathy for couple of minutes and read through this article. I guarantee that at the end, you will be glad that you chose to read through this article.

Now, imagine this. Let's say that one day around your house, you find this mysterious blue liquid. You have narrowed it down to either a blue fruit punch or a blue liquid for windshield washer fluid. You think that it is the fruit punch, but you are not confident enough to actually taste it (smell failed for you this time, hypothetically). What would you do to determine the liquid?

If you were fortunate to have a friend who was a chemist, he or she will likely perform several tests, one of the options being titration. Titration is defined as a process in which "a solution of accurately known concentration, called a standard solution, is added gradually to another solution of unknown concentration, until the chemical reaction between the two solutions is complete."¹

In English, this means that you add a known liquid (standard solution, or sometimes known as titrant) to an unknown liquid that you are analyzing called analyte. Usually, this applies to an analysis for adding bases, such as sodium hydroxide (NaOH), to determine the pH of a liquid. The determination can take places in two ways: by eyes or by a device called pH meter.

Going back to our storyline of blue fruit punch versus blue windshield washer fluid, let's talk about these two liquids. Blue fruit punch, like other fruit punches, will likely to be acidic. So, if we add sodium hydroxide and phenolphthalein, a chemical that will show the color change, there should be a change in the color of blue fruit punch.

On the other hand, let's assume that blue windshield washer fluid is a basic solution. Then, adding bases to a basic solution will not show any color change. So, if we add sodium hydroxide and see no change, then we can feel confident that we should not drink that liquid to quench our thirst.

Think of this process as a detective work because in essence, that is what many real-time chemistry laboratory works are - trying to determine what is the compound sitting in front of you. Usually, titration is experimented in undergraduate chemistry courses by observing changes to ascorbic acid (also known as Vitamin C) when triiode ion (I₃^-) is added. Another possibility for experiment that I personally did was to add sodium hydroxide to potassium hydrogen phthalate (KHP). You can find the first experiment here, and the second one here, and probably other ones in the Internet. Please note that none of the institutions is the one I attend; I just found them on the internet.

In all titrations, the experimenter performs stoichiometry, a fancy term for predicting the amount of products that will form from given reactants, to see how much of titrant will add to change the pH of an analyte. This amount determined mathematically is known as equivalence point. In the lab, the experimenters use phenolphthalein to observe with their eyes, or a pH meter, to estimate end point, which is a point that will come pretty close to the equivalence point on amount of titrant used to change the analyte.

To put it simple, the chemist will predict how much titrant will be needed through calculations and compare the experimentally observed value to the prediction. Because nothing in science is ever perfectly done as expected, due to factors known as systematic error/random error, the equivalence point and end point will never be same.

So, a brave soul that you are to still keep up reading this article will have a chance to see titration in action! I can assure you that you will not understand everything (if you do, that's great!) on what you are about to see. The point that I want you to see is the application of titration beyond just hypothetically made-up story. I believe that by seeing the mathematics behind titration, you will be able to gain more then say, just reading a Wikipedia article on titration.

Caution: I will try my best to hyperlink the key concepts to Wikipedia entries so you can pause and look them up as you go, if you would like. But, the overall level of understanding is really up to you and your experiences with chemistry.

Following problem is from Additional Exercises that are publicly and freely available from a website of Quantitative Chemical Analysis by Daniel C. Harris. The link is provided in the footnote.²

"A mixture weighing 27.73 mg containing only FeCl₂ (FM 126.75) and KCl (FM 74.55) required 18.49 mL of 0.02237 M AgNO₃ for complete titration of the chloride. Find the mass of FeCl₂ and the weight percent of Fe in the mixture."

Okay, first of all: breathe. All these information probably makes no sense to you now, but it will be better at the end. Now, as a rule for SI units, let's convert everything to the proper units. Then, the mixture has a weight of 0.02773 g, and AgNO₃ (known as silver nitrate) has a volume of 0.01849 L.

The next step is to find the moles of silver nitrate used to calculate the moles of FeCl₂ (iron (II) chloride) and KCl (potassium chloride). The concept of mole is not easy for people with no background in chemistry. To oversimplify the idea, in titration, whatever the amount of titrant reacted is equal to the amount of analyte in the reaction. This "amount" in chemistry refers to the moles (not mass), so that's the reasoning here. Capiche? I hope so!

Doing so will give 4.1362 x 10^-4 mol of silver nitrate in the reaction. Now, this number must be equal to the moles of iron (II) chloride and potassium chloride. So, let the mass of iron (II) chloride be x g, and the mass of potassium chloride be 0.02773 - x g. The latter expression came from the fact that the sum of two compounds equals to the mass of a mixture.

Here, it will gets a little tricky. Based on dimensional analysis, we can see that dividing the mass of a compound by its formula mass (abbreviated as FM and has units of g/mol) yield mole as the final unit. Mathematically, this is:

2x/(126.75) + (0.02773-x)/(74.55) = 4.1362x10^-4

Notice the 2 as the coefficient of x in the first fraction. This number came from stoichiometry, so I won't go in details, but if you are curious, write down the full reaction for potassium chloride with silver nitrate.

Anyway, solving for x yields x = 0.01761 g for iron (II) chloride and 0.01012 g for potassium chloride. Now, to find the weight percent, we need to calculate the mass of just iron in the mixture.

The idea here is again, moles. We can take the mass of iron (II) chloride, divide by its formula mass, and understand that the moles of iron (II) chloride will be same for moles of iron because there is no subscript attached to iron. Then, multiplying by the atomic weight of iron, we get 0.007759 g. Dividing this number by 0.02773 g, or the mass of the entire sample, and multiplying by 100, we get the weight percent, which is 27.98 percents.

Whew! Take time to breathe. This problem was particularly different because of stoichiometry, but I hope that you now know what titration is. The idea is relatively simple, but it's just the math that makes the concept rather difficult to do.

¹ Brian B. Laird. University Chemistry (McGraw-Hill: New York, 2009): 622.

²http://bcs.whfreeman.com/qca7e/content/cat_050/ch05-ch08_supp.pdf