Science Buddies: "Ask an Expert"

Hi there,
I am a grade 11 IB student in SL Chemistry. I will be starting my II (Internal Investigation) soon and I was wondering if this would be a good topic: determining the content of sugar in different fruits using redox titration. Specifically, glucose and fructose, as I have read that the safest fruits are those that are high in glucose and low in fructose. Do you think this would be feasible? Any suggestions for improvement or a completely different idea?

I would like to do my project on titration specifically as our two options are titration or calorimetry and I am not very confident in the latter.
Thanks!

Hello,

Determining the glucose and fructose in various fruits is a great topic. Analysis of both sugars in the presence of others in fruits would require you to identify separate redox titrations specific for glucose and for fructose. Feasibility of the project depends on finding sugar-specific titrations and availability of reagents and equipment at your lab or work area. Researching titration procedures for fructose and glucose would be a good first step for the project.

I hope this helps. Please ask again if you have more questions.

A. Norman

Thank you for your response! After doing some research, I read about determining glucose content through iodometric titration. This is word-for-word a paper I read by A.L. Galant, R.C. Kaufman & J.D. Wilson.

"Early methods for the quantification of glucose relied on its ability to act as a reductant in solution. Although an assortment of metal ions could oxidize the glucose carbonyl group, copper (Cu2+) was the most popular due to the potential for formation of stable precipitates or colorimetric end-products. Benedict’s solu- tion was one of the first widely-used reagents to take advantage of the relevant chemistry, employing sodium carbonate in the presence of copper citrate or tartrate to precipitate Cu2O (Benedict, 1908). This chemistry was co-opted by Otto Folin and Hsien Wu, who coupled the reduction of copper to the oxidation of phosphomolybdic acid (the Folin–Wu method). The result was a blue end-product, which was stable for several days, and which could be compared directly against solutions of known starting sugar content (Folin & Wu, 1919). One short-coming of Benedict’s solution and the Folin–Wu method was the inability to differentiate glucose from other reducing sugars in a complex solution. To address this issue, Cajori devised a method by which excess iodine was first reduced by glucose, and subsequently titrated against thiosulfate to determine the quantity of glucose present. Any fructose and sucrose present would not react with iodine, but fructose that remained after the glucose had been con- verted to gluconic acid could be reacted with copper as previously described by Folin and Wu."

My concern is that it seems like this titration method is not very accurate considering it cannot find glucose and fructose separately. I am confused about the part where it states "To address this issue...fructose remained after the glucose had been converted..." Does this mean that fructose and glucose can be found separately from iodometric titration? Also, I'm unsure if this method is applicable to fruits. Sorry for the long post, chemistry is not my strong suit.

Hi kierracov,

Any fruits that you might test contain glucose, fructose and other sugars. Successful analysis of such mixtures requires that you have test procedures that are specific for glucose and fructose. In other words, the test procedure must be able to differentiate glucose and fructose.

The passage you posted briefly describes tests that can differentiate glucose and fructose. The glucose-specific test involves reaction with iodine and titration against thiosulfate and the fructose-specific test involves reaction with copper. These two tests work for sugar mixtures because glucose reacts with iodine but fructose does not.

I hope this helps. Please ask again if you have more questions.

A. Norman

Hmm...I'm worried that doing glucose and fructose titration might be a little too complicated for my skill level because the two need to be separated. My teacher recommended determining vitamin C content in different fruits using titration, so I would like to do something along those lines. I don't want to do vitamin C particularly since it's very common. However, when I researched titration with fruits, the only thing I could find was vitamin C. Do you know of any other compounds that are present in fruits that I could find with titration? Thank you for your help once again!

Hi kierracov,

You could test juices for titratable acidity. In this kind of experiment, the titration doesn't differentiate the different acids that may be present in juice. Rather, the measurement indicates the total amount of acid in the juice.

I hope this helps. Please ask again if you have more questions.

A. Norman

Interesting! To put a twist on that, would the following make for a good experiment:
Determining and comparing the effect of fermentation on either vitamin C content OR titrable acidity in various fruits (e.g. orange, apple, grapes, banana, apricot)

Essentially, I would be just be doing redox titration to determine the vitamin C content for each fruit, but would do the same for the fermented version of each fruit to see it provides higher vitamin C content and thus, more health benefits.

Hi kierracov,

Investigation of the effect of fermentation on vitamin C content of fruit juices could be an interesting project. But there are many kinds of fermentation processes. You might want to find out if there is a fermentation process that can be used to increase or decrease the vitamin C in juice.

I hope this helps. Please ask again if you have more questions.

A. Norman

norman40,

Thank you for your continuous help. My teacher thought the fermentation idea was creative, but too complex and would get into too much biology. I'm now looking at different conditions that affect vitamin c content in fruits or fruit juices (e.g. temperature, storage, cooking etc.) that I could test for my lab. Many of these ideas are common and overdone, so any other ideas would be greatly appreciated. Once again, thank you so much!

Hi kierracov,

One experiment that you might try is a comparison of vitamin C in fresh and commercially prepared juices.

There is a Science Buddies project that involves comparing vitamin C contents of fresh and prepared orange juices. You may be able to modify the project to fit with your interests and the background information provided may be useful to you.

https://www.sciencebuddies.org/science- ... -vitamin-c

I hope this helps. Please ask again if you have more questions.

A. Norman

Hi norman40,

I decided to investigate the effect of heat time on vitamin C content in vegetables (red bell peppers & broccoli). I am using the potassium iodate titration method. So my flask will contain: a vegetable sample, potassium iodide, hydrochloric acid and starch indicator. And the burette will contain the potassium iodate solution.

However, I’m wondering how much I need of each solution. The potassium iodate and potassium iodide are provided by my school, but they are in solid form and don’t have a given concentration. The hydrochloric acid is 0.5M, and soluble starch is being used. I’m using 20mL samples of vegetable juice. How do I figure out how much of each solution I need? There’s obviously experiments online with different measurements, but how do I make such measurements proportionate for my lab? And do I need to know the concentration of potassium iodide and potassium iodate to do the experiment? Please let me know ASAP.

Thanks.

Hi kierracov,

Vitamin C titration is a common test and there are many procedures available online that include appropriate reagent concentrations. One example is at the following link:

https://www.thoughtco.com/vitamin-c-det ... ion-606322

If you follow the procedure at the above link (or the Science Buddies project from my previous post) you don't need to know the iodide concentration. Instead, you titrate a known vitamin C tablet and use that result to calculate the amount of vitamin C in a juice sample. My suggestion is to make the solution quantities specified in the above link. You should have plenty for your experiments.

I hope this helps. Please ask again if you have more questions.

A. Norman

The link provided seems to be for iodine titration, not iodate which is what I'm doing. I've figured out how to determine how many g of potassium iodate and potassium iodide I need for the lab, but I'm wondering what an appropriate amount is? I will be using the vitamin C tablet and use it to find the content of vitamin C in my vegetable samples. But does the concentration and grams I use of the potassium solutions affect the titration anyway (i.e. does it make it go faster/slower, does it help with colour change)?

Hi kierracov,

The concentrations of iodate and vitamin C will affect the volume of iodate solution you use. Higher iodate concentration means you'll add less iodate solution to your juice sample to complete the titration. Lower vitamin C concentration also means you'll add less iodate solution to the juice sample. But you'll want to titrate enough iodate solution to get a good volume reading from your burette at the endpoint – maybe 10-20 mL. Assuming equal accuracies of the standard and unknown sample titrations, the solution concentrations won't affect the final vitamin C result.

I hope this helps. Please ask again if you have more questions.

A. Norman

Hi there.

I changed my procedure from potassium iodate titration to iodine titration as I found the latter to be much easier. The problem is, my school ran out of iodine solution (Iodine concentrated, Lugol), so I need to use J Crows’ 2% Lugol’s iodine solution to perform my vitamin C titration on red peppers and broccoli instead. However, my teacher wants us to write the report as if we used the school’s iodine solution, so the measurements, concentrations and any calculations must correspond with their solution and the values I would have used given that solution.

I’m unsure how to figure out the amount of 2% solution I need so that it is proportional to the amount I would have used given the pure iodine solution. My original plan was to create a 0.03 M iodine solution. Therefore, I was going to:

Take 20mL of iodine solution
Add 646mL of water to it to dilute to 666mL

I figured this out doing the following dilution calculation:
(1M) 0.02L= 0.03M(?L)

How much of the 2% J Crows’ solution do I need to create a 0.03M iodine solution? How much do I need to dilute it to?

Any help ASAP would be greatly appreciated!

Hi kierracov,

Lugol's solution is 2% by weight I3K in water. The approximate molar concentration is 0.049M.

Let's say that you want 0.1 L (100 mL) of solution at a concentration of 0.03M. The calculation for this is as follows.

(0.049 M)(V)/0.1L = 0.03M

The original concentration (0.049 M) multiplied by V gives the moles in that volume. The number of moles divided by the desired final volume of solution is the desired concentration.

Rearranging the equation to solve for V gives the following.

V = (0.03M)(0.1L)/0.049M

And the result is that you need to dilute 0.061 L (61 mL) of 2% Lugol's solution to 0.1 L (100 mL) to get a 0.03M concentration.

I hope this helps. Please ask again if you have more questions.

A. Norman

Thanks Norman! By any chance, do you have a reference stating that the molarity of Lugol's 2% solution is 0.049 M? I tried looking it up but wasn't able to find anything

Hi kierracov,

My apologies for a mistake in my previous post. The math for getting to the volume (V) of Lugol's solution you need (and the 0.061 L result) is correct. But I wrote 6.1 mL instead of the correct value of 61 mL. I have edited my post to show the correct volume in mL.

You can find details about Lugol's solution here:

https://en.wikipedia.org/wiki/Lugol%27s_iodine

You can convert the concentration in weight percent to molarity given the molecular weight of I3K (419.8 g/mol) and the definitions of the solution types.

A 2% (by weight) solution contains 2 g of solute and 98 g of water. The number of moles of solute (I3K) is 2 g/419.8 g/mol or 0.0048 mol.

The molar concentration is the number of moles divided by the volume of the solution. For this conversion we have to assume that the solution volume is 98 mL. So the approximate molarity is 0.0048 mol/0.098 L or 0.049 M.

I hope this helps. Please ask again if you have more questions.

A. Norman

Thanks Norman! Completely separate question - I'm currently learning about redox reactions and I'm wondering if there's a difference between oxidation state and oxidation number and what that difference is? It seems like it's used interchangeably throughout my textbook, but I read that oxidation number is in roman numerals while oxidation state has the sign and then the number.

Hi kierracov,

“Oxidation state” and “oxidation number” refer to the same thing. More information is available here:

https://en.wikipedia.org/wiki/Oxidation_state

Roman numerals are commonly used to indicate the oxidation state of a metal in a compound's name. For example iron (III) chloride contains iron in the +3 oxidation state while iron (II) chloride has iron in the +2 oxidation state.

I hope this helps. Please ask again if you have more questions.

A. Norman

Hi Norman,

So how much water would I need to add for a 0.015 M lugol's iodine solution of 1 M?
I did the following calculation:

M1V1 = M2V2
(0.02)(1) = (0.015)(?)
? = 1333.33 mL

But 1333.33 mL seems like way too much. Did I do something wrong? Is there a way to decrease the amount of water needed so that it's still a 0.015 M solution using the 1 M Lugol's iodine? Considering I'm using a 1L volumetric flask to hold the iodine, it doesn't make sense to go over 1000mL.

Please reply ASAP. Thanks!

Hi kierracov,

As you've written the equation, M1 is 0.02 M, V1 is 1 L, and M2 is 0.015 M. The unknown that you solved for (V2) is the final volume (1333.3 mL) needed to dilute 1 L of a 0.02 M solution to a concentration of 0.015 M. In other words you could prepare the 0.015 M solution by adding water to 1 L of the 0.02 M solution to get a final volume of 1333.3 mL.

A much more common lab practice is to make a specific volume of solution using a volumetric flask. To do this, the desired final volume (the flask volume) becomes V2 in your equation. And the unknown becomes V1 (the volume of the concentrated solution needed). I worked through an example of this calculation in my reply to you on January 30 with different values for the concentrations and the desired volume. You should be able to work through the math in the previous post with any concentration and volume values you choose.

You mentioned in a previous post that you were using 2% Lugol's solution for your experiments. If that is still the case, the molar concentration is about 0.049 M as explained in my post from January 31.

I hope this helps. Please ask again if you have more questions.

A. Norman

Hi Norman,

Thanks again. The teacher wants us to write our report as if we used 1M Lugol's iodine solution, which is why I have to write my calculations differently. I made a typo in the previous post, I meant to write that I want a 0.005M iodine solution to titrate. I believe this is the correct calculation if I want 1L of the iodine solution:

M1V1 = M2V2
(1M)(V) = (0.005M)(1L)
V = 0.005L or 5mL

Just to clarify, this would mean that I need to take 5mL of the Lugol's iodine solution, and add 995mL of water to create my desired iodine solution? Would this be too dilute considering I'm only taking 5mL of the solution, which doesn't seem like much at all.

Thanks for all your help once again.

Hi kierracov,

Your calculation is correct. If you want to make 1 L of 0.005 M solution you need to dilute 5 mL of the 1 M solution.

To make this solution in the lab, you would transfer 5 mL of the 1 M solution to a 1 L volumetric flask. Then you'd fill the flask to the mark. In effect, you'd be adding 995 mL of water to the 5 mL of 1 M solution.

A 0.005 M solution is reduced in concentration by a factor of 200 relative to the 1 M solution. So only 5 mL of the 1 M is required for the dilution you want.

I hope this helps. Please ask again if you have more questions.

A. Norman

Great, thanks so much again! I assumed the concentration of the Lugol's iodine was 1M because that's what a classmate was saying, but it turns out they weren't sure. Our teacher told us to look it up, so I believe this is the one the school purchased:

https://www.boreal.com/store/product/11 ... oncentrate

It doesn't say the concentration though, would you happen to know why or where I can find it?

Hi kierracov,

Sorry, but I don't know the concentration of the Lugol's iodine that your have. My suggestion is to contact the supplier or manufacturer of the solution to get the concentration.

I hope this helps. Please ask again if you have more questions.

A. Norman

Hi kierracov,

After some additional reading about how Lugol's iodine solutions are prepared, I realized that I made a mistake in calculating the molar concentration in my posts on January 30 and 31, 2019. I assumed that a 2% Lugol's iodine solution contained 2 g of I3K per 100 g of solution. And I used the molecular mass of I3K to calculate the molarity. This is incorrect because the solution is not made this way.

A 2% Lugol's iodine solution contains 2 g of iodine (I2) per 100 mL of water. An excess amount (4 g) of potassium iodide (KI) is included to help dissolve the iodine. The added iodide is not included in the 2% concentration designation for the solution. In other words the “2%” refers only to the iodine (I2).

The molecular mass of iodine (I2) is 253.8 g/mol so the number of moles in 2 g is 0.0079 mol (2 g/253.8 g/mol). And the molar concentration of iodine is 0.0079 mol/0.1 L or 0.079 M.

It appears that Luogl's iodine is commonly made on a weight per volume basis as described above. But sometimes it is prepared on a weight percent basis. For a 2% solution that means that 2 g I2, 4 g KI and 94 g water are combined. Preparing the solution this way would result in a higher molar concentration of iodine due to the smaller volume of water. In this case the I2 molarity would be 0.0079 mol/0.094 L or 0.084 M.

Sorry for any confusion my mistake may have caused.

I hope this helps. Please ask again if you have more questions.

A. Norman