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Author Topic: Quantitative analysis HELP  (Read 339 times)

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Fatpants15

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Quantitative analysis HELP
« on: January 15, 2017, 07:27:06 AM »

So I have some homework problems that I think I have but am not sure. They are:

1. a) A series of sulfate samples is to be analyzed by precipitation as BaSO4. If it is known that the sulfate content of the samples range between 20.0% and 55.0%, what minimum sample mass should be taken to ensure that a precipitate mass no smaller than 0.300 g of barium sulfate is produced?

b) What is the maximum barium sulfate mass to be expected if this minimum quantity of sample mass taken?

2. The density of 49.5% (m/m) aqueous hydrofluoric acid is 1.15g/mL. How would you prepaare 500.0 mL if 50.0 ppth (m/m) HF solution from the concentrated stock?

For part A I did simple stoichiometry to go from .300 g BaSo4 to grams of sulfate and then divided that by the minimum 20% sulfate content to get a sample mass of 0.618 g as the answer. For part B I multiplied 0.618 g by 55% sulfate content and then used that to get from grams sulfate to grams BaSO4. I got 0.826 g BaS04
For #2 I'm not sure what to do
« Last Edit: January 15, 2017, 08:18:55 AM by Fatpants15 »
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AWK

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Re: Quantitative analysis HELP
« Reply #1 on: January 15, 2017, 07:36:32 AM »

Calculate mass of HF in 50 ml of 50 ppm solution (use density of water for this solution, ie 50 ml=50 g).
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Fatpants15

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Re: Quantitative analysis HELP
« Reply #2 on: January 15, 2017, 08:20:27 AM »

Sorry there was a typo it's actually 500.0 mL. So I would just take 500.0 mL multiplied by 50/1000 and then multiply that by the 1.15 g/mL density of aqueous HF?
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AWK

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Re: Quantitative analysis HELP
« Reply #3 on: January 15, 2017, 09:10:37 AM »

First, calculate mass of HF in 500 ml of 50 ppm solution (use density of water for this solution, ie 500 ml=500 g) - you need calculate mas of fluoride ion, then mass of HF.
Then calculate volume of concentrated HF that contains this mass of HF (this will be very small volume).
A serial dilution will be useful.
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Fatpants15

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Re: Quantitative analysis HELP
« Reply #4 on: January 15, 2017, 10:41:44 AM »

Ok. So I got that you would need 28.75 grams of HF and using the 49.5% stock solution with the density of 1.15 you would need 50.50 mL of the stock solution.  Does this sound right?
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Borek

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Re: Quantitative analysis HELP
« Reply #5 on: January 15, 2017, 10:53:36 AM »

Ok. So I got that you would need 28.75 grams of HF

Close, but wrong. Please show how you got it.
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Fatpants15

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Re: Quantitative analysis HELP
« Reply #6 on: January 15, 2017, 11:07:26 AM »

Ok I redid it and now I got something a little different. So I took the 500 mL and changed it to 500 g because of water density. Then I set it up like this: 500g*50/1000=25.0 g of HF. Then using the density I got that you would need 25.0g HF * 1.15g/mL= 28.75 mL of HF (this is where I think I got the grams and mL mixed up). Then I just set up 28.75/x=.495/1 to find the total amount of stock solution needed and I got 58.08 mL as the final answer. Is that better?
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Borek

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Re: Quantitative analysis HELP
« Reply #7 on: January 15, 2017, 11:19:26 AM »

Then I set it up like this: 500g*50/1000=25.0 g of HF.

OK

Quote
Then using the density I got that you would need 25.0g HF * 1.15g/mL= 28.75 mL of HF

No. Units of that multiplication are g2/mL (whatever it means), not mL.

You know you need 25 g of HF. What mass of the stock solution contains 25 g of HF?
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Fatpants15

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Re: Quantitative analysis HELP
« Reply #8 on: January 15, 2017, 01:00:25 PM »

Ok so I got that for 25.0 g HF * 1g sol'n/0.495 g HF=50.50 g stock sol'n. Then I can just do 50.50 g sol'n*1 mL sol'n/1.15g sol'n. Right?
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Re: Quantitative analysis HELP
« Reply #9 on: January 15, 2017, 01:18:52 PM »

Sorry, I was misdirected by your abreviation ppth - this is just 0.5 % solution.
Result of your division with 3 significant digits is correct.
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Borek

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Re: Quantitative analysis HELP
« Reply #10 on: January 15, 2017, 08:58:42 PM »

Sorry, I was misdirected by your abreviation ppth - this is just 0.5 % solution.

50/1000 = 5%
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Borek

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Re: Quantitative analysis HELP
« Reply #11 on: January 15, 2017, 09:00:54 PM »

Ok so I got that for 25.0 g HF * 1g sol'n/0.495 g HF=50.50 g stock sol'n. Then I can just do 50.50 g sol'n*1 mL sol'n/1.15g sol'n. Right?

Yes.

Actually you need a bit more acid, as 5.00% HF solution density is not 1 g/mL, but 1.0165 g/mL.
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