How to solve this?

Question

Vitamin C (ascorbic acid, C₆H₈O₆, 176.12 g/mol) can be measured by redox titration with iodine solution. Iodine is reduced to iodide ion, while ascorbic acid is oxidized to dehydroascorbic acid (C₆H₆O₆). The iodine solution is usually made in the presence of iodide ion, forming the more stable triiodide ion, I₃⁻, giving the following overall reaction C₆H₈O₆(aq) + I₃⁻(aq) → C₆H₆O₆(aq) + 3I⁻(aq) + 2H⁺(aq) The titration is carried out in the presence of starch, which forms a dark blue complex with the excess iodine when the endpoint is reached. You are testing a new multivitamin pill for its vitamin C content. One pill weighing 0.763 g is dissolved in dilute acid and titrated with 0.0236 M I₃⁻ solution, requiring 28.3 mL to reach the endpoint. Calculate the percent by mass of vitamin C in the pill.

Answer 1

Although this looks like a more complicated titration problem because it's a redox titration, we can solve it the same way we would solve any other titration problem.

Titrations are helpful analytical methods that allow us to determine the amount of an unknown. In this case, we're interested in determining the concentration of ascorbic acid in our vitamin C pill. We can do this by slowly adding our titrant, I₃^-, until all of the ascorbic acid is reacted. In this particular titration, once all of the ascorbic acid has reacted, we have reached our "equivalence point" or "endpoint" which we will physically be able to see because I₃^- forms a blue complex with starch so that the solution in our Erlenmeyer flask turns blue!

We know, based on our chemical equation, that we can react every 1 molecule of ascorbic acid with 1 molecule of triiodide (I₃^-). This is because the stoichiometric coefficients (the numbers in front of the chemicals in the equation) are both 1. Sometimes the problem will give you an unbalanced equation, so it's always best practice to make sure you are working with a balanced equation!

The first thing we want to solve for is the amount of titrant (I₃^-)at the equivalence point. At this point, the mol of titrant = the mol of analyte!

If we look at the word problem, we know that the titration required 28.3 mL of titrant to reach our equivalence point. We also know that our titrant's concentration is 0.0236 M. It's important here to recognize the definition of molarity as M =(mol/L). To solve for mols, we will first convert our volume in mL to L (1), and then use molarity to convert to mol (2).

(1) 28.3mL titrant = 0.0283 L titrant (by multiplying 28.3 mL by (1L/1000mL))

(2) 0.0283 L * (0.0236 mol I₃^-/L titrant) = 0.00066788 mol

Now this is a long number and not very convenient to carry through. At this point, we can just keep the significant figures. If we look at our volume, we have 3 sig figs. If we look at our concentration, we ALSO have 3 sig figs. That means we only need to report 3 sig figs for our mols of I₃^-.

This means we have 0.000668 mol or 0.668 mmol of I₃^- at the equivalence point

Next we will convert mol of titrant at the equivalence point to mol of analyte at the equivalence point

Again, looking at our equation, we see 1:1 stoichiometric coefficients so in this case, we also have 0.668 mmol of ascorbic acid at our equivalence point. If we had a non 1:1 molar equivalency, we would just multiply by the mol ratio found in the equation (see below for how that would be set up).

0.668 mmol of I₃^- * (1 mmol of ascorbic acid/ 1 mmol of I₃^-) = 0.668 mmol of ascorbic acid

Now we have mol of ascorbic acid but the question is asking us for percent by mass of ascorbic acid in the vitamin C tablet.

First we must think about the definition of "mass %".

Let's look at the "%" part. We know a percent is just a part of a whole. In the case, we want to know what part of the whole pill is ascorbic acid.

The "mass" in "mass%" tells us how we will measure our part.

Typically in chemistry you'll run into volume% and mass% and it essentially just tells you what units you need to be in! In our case, when we see mass, we should think grams! The word problem can help us too, if we note that the mass of the tablet is already given in grams. This means we just need to solve for the mass of the analyte (in grams).

To convert mol of analyte to grams of analyte, we just need the molecular weight of ascorbic acid which is given as 176.12 (g/mol).

0.000668 mol I₃^- (176.12 g ascorbic acid / mol I₃^-) = 0.118 g ascorbic acid

Finally to find our % mass (our part of a whole in grams) we take

0.118g ascorbic acid / 0.763 g pill = 15.5 wt% (ascorbic acid/pill)