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Offline ben809

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Simple Stoichiometry Question
« on: December 09, 2014, 02:53:49 PM »
I'm making a batch with a 46% KOH solution.

I just want to be sure I did this right when making my KOH (doing w/v):

For a 500ml total solution I multiplied = 500ml (2.044g/ml) = ~1022 g (2.044g/ml is density of my KOH)

..then 1022g (46% KOH) = 470.12 g of KOH need to be added to that solution.

Just want to get a second thumbs up (or down) if I'm right.

Offline Borek

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Re: Simple Stoichiometry Question
« Reply #1 on: December 09, 2014, 03:16:53 PM »
While there are things that suggest you are on the right track, you are wrong about 2.044 g/mL being a density of the 46% w/v solution. As far as I can tell it is much lower, below 1.5 g/mL.
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Offline ben809

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Re: Simple Stoichiometry Question
« Reply #2 on: December 09, 2014, 04:12:04 PM »
While there are things that suggest you are on the right track, you are wrong about 2.044 g/mL being a density of the 46% w/v solution. As far as I can tell it is much lower, below 1.5 g/mL.

I'm using a solid KOH that we have, on our SDS it has a density of 2.044 g/ml...so I'm guessing if that's the case then what I made up should be correct!?!?

Thank you for your input!

Offline Arkcon

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Re: Simple Stoichiometry Question
« Reply #3 on: December 09, 2014, 04:19:17 PM »
Like Borek:, I find your procedure hard to follow.  You have to make a w/v solution.  So you should use a certain weight, and add it to a certain volume.  Why do you need the density of a solid, to know its weight?  You use the density of a liquid, to get the weight, of a specific volume.  But you're not doing that.  So I'm at a loss.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline ben809

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Re: Simple Stoichiometry Question
« Reply #4 on: December 10, 2014, 08:33:41 AM »
Like Borek:, I find your procedure hard to follow.  You have to make a w/v solution.  So you should use a certain weight, and add it to a certain volume.  Why do you need the density of a solid, to know its weight?  You use the density of a liquid, to get the weight, of a specific volume.  But you're not doing that.  So I'm at a loss.

Got yah, I thought you also had to take into consideration the density for solids when making a % solution. If its not close to waters or the solvent being used.

So I was using the density to figure out how much in grams I need to add to my total solution of 500ml of water.

Offline mjc123

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Re: Simple Stoichiometry Question
« Reply #5 on: December 10, 2014, 08:57:02 AM »
Be careful - you don't add anything TO 500 ml of water. 46% w/v means 46g KOH per 100 ml of total solution. (Your "total solution of 500 ml of water" is seriously ambiguous.) If you add the required mass to 500 ml water, you will have more than 500 ml of less than 46% solution. You add the solid to a lesser amount of water, then add water to make the total volume up to 500 ml.
This is where your solid density may come in useful. Your required mass of KOH (have you worked it out yet?) would occupy ca. 130 ml. So assuming, to a first approximation, no volume change on mixing, you would need ca. 370 ml water. I would try dissolving your KOH in ca. 300 ml water, then when it is dissolved (add a little more water if necessary), add more water in small portions with good mixing until the volume is 500 ml.

Offline DrCMS

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Re: Simple Stoichiometry Question
« Reply #6 on: December 10, 2014, 10:44:40 AM »
The KOH solid ben809 has is most likely only ~85% KOH. 

If you want to use 30-50% KOH just buy it rather than waste a load of time making it up from KOH flake and then having to titrate and standardise it.

46% w/v equates to ~34%w/w


Offline ben809

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Re: Simple Stoichiometry Question
« Reply #7 on: December 12, 2014, 08:29:40 AM »
Be careful - you don't add anything TO 500 ml of water. 46% w/v means 46g KOH per 100 ml of total solution. (Your "total solution of 500 ml of water" is seriously ambiguous.) If you add the required mass to 500 ml water, you will have more than 500 ml of less than 46% solution. You add the solid to a lesser amount of water, then add water to make the total volume up to 500 ml.
This is where your solid density may come in useful. Your required mass of KOH (have you worked it out yet?) would occupy ca. 130 ml. So assuming, to a first approximation, no volume change on mixing, you would need ca. 370 ml water. I would try dissolving your KOH in ca. 300 ml water, then when it is dissolved (add a little more water if necessary), add more water in small portions with good mixing until the volume is 500 ml.


Well I don't really have a mass, I just needed a 46% solution to do a quick test and we ran out of our stock solution right when I needed it.

what does "ca." mean in your statement!?

Offline mjc123

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Re: Simple Stoichiometry Question
« Reply #8 on: December 12, 2014, 11:09:06 AM »
Quote
Well I don't really have a mass
Isn't that what you were looking for?
"ca." = "circa" = "about"

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