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Offline Roddy

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Preparation of a Solution
« on: November 21, 2009, 11:16:02 PM »
I need help with this problem.
Calculate the following to prepare a solution:

a) From solid iron (III) sulfate nonahydrate, calculate the grams required to prepare 500.0 mL of 2.117 x 10-4M Fe(III).

b) The Fe(III) will hydrolyze and precipitate as Fe(OH)3 unless the acid concentration is greater than 5.2 x 10 -3 M H+.  A stock solution of 1.0 M HCl is available.  How many mL are required to add to the solution in part a) above?

In part a) I calculated the g of Fe(III) needed as being 0.0297g, but I am not sure if it is correct.  Part b) I don't know where to start. 

Offline Borek

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Re: Preparation of a Solution
« Reply #1 on: November 22, 2009, 05:46:01 AM »
0.0297g seems OK.

How many moles of acid will you need in the final solution? Assume volume (500 mL) will not change - you will be able to check it later comparing initial volume with the volume acid added (if the latter is much smaller, it can be ignored).
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Offline Roddy

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Re: Preparation of a Solution
« Reply #2 on: November 22, 2009, 10:32:31 AM »
What about part B?  That is the part that I really need the help on.

Offline Borek

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Re: Preparation of a Solution
« Reply #3 on: November 22, 2009, 11:56:30 AM »
You have either not read the question or my answer.
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Offline Roddy

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Re: Preparation of a Solution
« Reply #4 on: November 22, 2009, 12:04:50 PM »
I read both, but neither made sense...

What I tried is the dilution formula C1V1=C2V2
C1=2.117x10-4 mol/L
V1=500.0mL
C2=5.2x10-3 mol/L
V2= unkown

I solved for V2 and got 20.4mL.  Is that the correct way to set it up?

Offline sjb

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Re: Preparation of a Solution
« Reply #5 on: November 22, 2009, 12:48:21 PM »
How many ml of a 1M solution of acid do you need to take to make 500 ml of a 5.2 mM solution?

Offline Borek

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Re: Preparation of a Solution
« Reply #6 on: November 22, 2009, 12:58:42 PM »
I read both, but neither made sense...

Seems like you have not understood the question:

Quote
C1=2.117x10-4 mol/L

Concentration of what?

Quote
C2=5.2x10-3 mol/L

Concentration of what?
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Offline Roddy

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Re: Preparation of a Solution
« Reply #7 on: November 22, 2009, 01:02:43 PM »
OK, I think I see what you mean.  I was using the correct formula, just not the correct initial concentration and initial volume is the unknown, correct?

Redoing it I got:
C1V1=C2V2
C1=1.0 M
V1= unkown
C2=5.2x10-3 mol/L
V2= 500 mL

I solved for V1 and got 2.6 mL.
That seems to be a more reasonable answer.

Offline Borek

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Re: Preparation of a Solution
« Reply #8 on: November 22, 2009, 01:32:03 PM »
Now, as I signalled earlier - final volume is 502.6 mL, that means final concentration is not exactly what you wanted. But you are off by less than 2.6/500 - about half %. In the case of final concentration given with two significant digits half % doesn't matter.
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Offline Roddy

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Re: Preparation of a Solution
« Reply #9 on: November 22, 2009, 01:59:48 PM »
OK, great thanks for your help on this Borek.  ;D  I still am working on that second problem I posted in the other thread.  I am not really sure where to go.  

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