Chemistry Project Topic
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Re: Chemistry Project Topic
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
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
Last edited by norman40 on Thu Jan 31, 2019 4:05 am, edited 1 time in total.
Re: Chemistry Project Topic
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
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Re: Chemistry Project Topic
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
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
Re: Chemistry Project Topic
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.
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Re: Chemistry Project Topic
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
“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
Re: Chemistry Project Topic
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!
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!
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Re: Chemistry Project Topic
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
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
Re: Chemistry Project Topic
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.
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.
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Re: Chemistry Project Topic
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
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
Re: Chemistry Project Topic
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?
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?
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Re: Chemistry Project Topic
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
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
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Re: Chemistry Project Topic
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
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