Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: dpascali 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!
-
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?
-
the moles i found for Ca2+ is 2.08808 X 10 -1.
-
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? :-\
-
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?
-
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
-
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.
-
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)?
-
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
-
I just can't seem to figure out the steps that i wewnt too far on >:(...can i get alil more help? thanks!
-
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?
-
mol CaCO3 = mol Ca2+
.5215 g CaCO3 x (1 mol / 100.1 g) = 5.210 x 10-3 mol Ca2+
-
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?
-
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?
-
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.
-
There is a second part to this problem that has also taken me many hours to figure out how to set up.
the set up is:
25.00 mL aliquots of the solution from problem 1 are titrated with EDTA to the Eriochrome Black T end point. A blank containing a small measured amount of Mg 2+ requires 2.60 mL of the EDTA to reach the end point. An alquot to which the same amount of Mg 2+ is added requires 28.55mL of the EDTA to reach the end point.
the questions are:
a.) how many mL of EDTA are needed to titrate the Ca 2+ ion in the aliquot?
would you just add the 25.00mL and 28.55mL then subtract the 2.60 mL from that answer to get the total?
b.) how many moles of EDTA are there in the volume obtained from Part a?
and obviously i need to figure out a. to get to b.
-
No. 0.00521 is not a concentration yet. It is number of moles.
I see...
so it would be... .02083 M x .025L= 5.20979x 10-4
-
Has my luck run out for help haha ;D ???
-
There is a second part to this problem that has also taken me many hours to figure out how to set up.
the set up is:
25.00 mL aliquots of the solution from problem 1 are titrated with EDTA to the Eriochrome Black T end point. A blank containing a small measured amount of Mg 2+ requires 2.60 mL of the EDTA to reach the end point. An alquot to which the same amount of Mg 2+ is added requires 28.55mL of the EDTA to reach the end point.
the questions are:
a.) how many mL of EDTA are needed to titrate the Ca 2+ ion in the aliquot?
would you just add the 25.00mL and 28.55mL then subtract the 2.60 mL from that answer to get the total?
b.) how many moles of EDTA are there in the volume obtained from Part a?
and obviously i need to figure out a. to get to b.
-
You are thinking too complicated again! You need 2.60 mL to titrate the Mg2+. Now, you need 28.55 mL to titrate the Mg2+ and the Ca2+. 28.55 - 2.60 = 25.95 mL to titrate the Ca2+.
-
so for b) how would you figure out the molarity of the EDTA to get the # of moles of EDTA? (jw could this be the same as the molarity of Ca2+ from 1b?)