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Topic: EDTA - Please, i still need *delete me*  (Read 19486 times)

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dpascali

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EDTA - Please, i still need *delete me*
« on: January 23, 2006, 06:13:12 PM »
Ill set up this problem...

A 0.5215g sample of CaCO3 is dissolved in 12M HCl and the resulting solutiom is diluted to 250.0mL in a volumetric flask.

What would be the molarity of Ca2+ ? I put down .84M by doing moles of Ca2+ divided by .25L Ca2+. Would this be correct?

Also, there is a second part of asking how many mL of EDTA are needed to titrate the Ca2+  ion in the aliquot? This problem is from a 25.00mL aliquot from the solution in part 1, that is titrated with EDTA. The amount of blank contains a small amount of Mg 2+ that requires 2.60mL to reach the end point. And that same amount of Mg 2+  is added to the required 28.55mL of the EDTA to reach an end point.  I'm so confused on how to set this question up and in order to get to the rest of my review questions i have to get pass this one...ekkk. Any help?
Thanks!
« Last Edit: January 23, 2006, 09:48:27 PM by dpascali »
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Offline Borek

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Re:EDTA - Determination of the hardness of water.
« Reply #1 on: January 23, 2006, 06:46:19 PM »
Quote from: dpascali on January 23, 2006, 06:13:12 PM
What would be the molarity of Ca2+ ? I put down .84M by doing moles of Ca2+ divided by .25L Ca2+. Would this be correct?

Approach sounds correct, result is off by factor of 40. Check your math.

How many moles of Ca2+ have you found?
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dpascali

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Re:EDTA - Determination of the hardness of water.
« Reply #2 on: January 23, 2006, 06:48:51 PM »
the moles i found for Ca2+ is 2.08808 X 10 -1.
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Offline plu

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Re:EDTA - Determination of the hardness of water.
« Reply #3 on: January 23, 2006, 07:07:10 PM »
Quote from: dpascali on January 23, 2006, 06:13:12 PM
A 0.5215g sample of CaCO3 is dissolved in 12M HCl and the resulting solutiom is diluted to 250.0mL in a volumetric flask.

You don't have the initial volume of HCl.  How are you supposed to dilute it?  :-\
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dpascali

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Re:EDTA - Determination of the hardness of water.
« Reply #4 on: January 23, 2006, 07:07:22 PM »
I got that number from

.5215g CaCO3 x(1mol CaCO3 /100.1g CaCO3 ) x (1 mol Ca/ 1 mol CaCO3 ) x (40.08g Ca/ 1 mol Ca) = .208808 mol Ca2+
Is that right or did i miss something?
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Offline plu

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Re:EDTA - Determination of the hardness of water.
« Reply #5 on: January 23, 2006, 07:08:47 PM »
Quote from: dpascali on January 23, 2006, 07:07:22 PM
I got that number from

.5215g CaCO3 x(1mol CaCO3 /100.1g CaCO3 ) x (1 mol Ca/ 1 mol CaCO3 ) x (40.08g Ca/ 1 mol Ca) = .208808 mol Ca2+
Is that right or did i miss something?


That last conversion brought you over to grams of Ca2+.  Too many steps  :P
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Offline Borek

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Re:EDTA - Determination of the hardness of water.
« Reply #6 on: January 23, 2006, 07:10:30 PM »
Quote from: plu on January 23, 2006, 07:07:10 PM
You don't have the initial volume of HCl.  How are you supposed to dilute it?  :-\

No need for that - there was enough HCl to dissolve carbonate, then solution was filled up to 250.
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dpascali

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Re:EDTA - Determination of the hardness of water.
« Reply #7 on: January 23, 2006, 07:12:51 PM »
Quote from: plu on January 23, 2006, 07:07:10 PM
You don't have the initial volume of HCl.  How are you supposed to dilute it?  :-\

Well the set up is this-
A 0.5215-g sample of CaCO3 is dissolved in 12 M HCl and the resulting solution is diluted to 250.0mL in a volumetric flask.
 
Then there are 3 questions-
a. how many moles of CaCO3 are used (formula mass = 100.1) i got 5.210 moles
b. what is the molarity of the Ca 2+ in the 250 mL of solution. I got .84M
c. how many moles of Ca 2+ are in the 25.0 mL aliquot of the solution in 1b. I got 2.08x10 -2 moles.

Did i answer those correctly (showed work in previous post)?

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Offline plu

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Re:EDTA - Determination of the hardness of water.
« Reply #8 on: January 23, 2006, 07:13:06 PM »
Quote from: Borek on January 23, 2006, 07:10:30 PM
No need for that - there was enough HCl to dissolve carbonate, then solution was filled up to 250.

Apologies.  I overlooked the fact that HCl is not involved in the question  :D
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dpascali

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Re:EDTA - Determination of the hardness of water.
« Reply #9 on: January 23, 2006, 07:15:54 PM »
I just can't seem to figure out the steps that i wewnt too far on >:(...can i get alil more help? thanks!
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dpascali

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Re:
« Reply #10 on: January 23, 2006, 07:17:32 PM »
See that is were im lost i found moles of just Ca 2+ (or so i thought) i dont know how to deal with just Ca 2+ from CaCO3. I just use the same moles as CaCO3?
« Last Edit: January 23, 2006, 07:21:20 PM by Borek »
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Offline plu

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Re:EDTA - Determination of the hardness of water.
« Reply #11 on: January 23, 2006, 07:19:32 PM »
mol CaCO3 = mol Ca2+
.5215 g CaCO3 x (1 mol / 100.1 g) = 5.210 x 10-3 mol Ca2+
« Last Edit: January 23, 2006, 07:20:48 PM by plu »
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dpascali

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Re:EDTA - Determination of the hardness of water.
« Reply #12 on: January 23, 2006, 07:26:01 PM »
Thank you! Its funny when you work into it so much you start over thinking things... I had that 2 hours ago. I dont know why i do this....

So for c.) it would just be (.00521 M x .025L )= 5.21 x 10 -4?
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Offline Borek

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Re:EDTA - Determination of the hardness of water.
« Reply #13 on: January 23, 2006, 07:28:28 PM »
That's why I think american way of doing calculations is wrong ;) It is too easy to get automatic and loose control over what you are doing.

The same applies to automatic vs. manual gear box ;)

I am doing all such calculations with ratios and cross-multiplication.

Did I mentioned I am using manual gear box?
« Last Edit: January 23, 2006, 07:29:45 PM by Borek »
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Offline Borek

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Re:EDTA - Determination of the hardness of water.
« Reply #14 on: January 23, 2006, 07:31:26 PM »
Quote from: dpascali on January 23, 2006, 07:26:01 PM
So for c.) it would just be (.00521 M x .025L )= 5.21 x 10 -4?

No. 0.00521 is not a concentration yet. It is number of moles.
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