# Calculate Molarity of Cl- in Solution

by nautica
Tags: molarity, solution
 P: n/a A Cl solution with a volume of 13.53 mL was treated with 30.0 mL of Cr3+ solution (containing excess Cr3+) to convert to hexachlorochromateIII: 6Cl- + Cr ----- Cr(Cl)6 The excess Cr3+ was then tirated with 11.85 mL of 0.01280 M (en) according to equation: Cr + 3en ------------- Cr(en)3 Cr(Cl)6 does not react with en. If 41.35 ML of en were required to react with 20.17 mL of the original Cr3+ solution, calculate the molarity of Cl in the 13.53 mL Chloride sample. THIS IS WHAT I DID 0.04135 L * 0.01280 M / 3 moles / 0.02017 L * 0.03 L * 6 mol Cl / 0.01353 L with an answer of 0.1164 M Cl- in solution my problem is - no where did I include the 11.85 mL of the en that was calculated. Do I need to or does this look correct??? Thanks Nautica
P: 173
You can find Cl- molarity by equating no: moles.
You can find molarity of Cr3+ by:
 41.35 ML of en were required to react with 20.17 mL of the original Cr3+ solution
Then you can find no moles of Cr3+ excess in reaction b/w Cl- by:

 The excess Cr3+ was then tirated with 11.85 mL of 0.01280 M (en)
Therefore you can find no moles of Cr3+ required to react with CL- and thus no moles of Cl- by:
 13.53 mL was treated with 30.0 mL of Cr3+ solution
The dividing by volume you get Molarity.

Central idea is that there is excess Cr3+.
Sorry for long post
regards
 P: n/a That sounds better. Let me try it and I will post my results. Thanks Nautica
 P: n/a Calculate Molarity of Cl- in Solution I am still doing something wrong 0.04135 L * .01280 M / 3 moles * .02017 L * .01185 L / .03 L * .0135 L * 6 moles gives me 1.14 x 10 ^-7 then I divided that by the 13.35 mL to get 8.4 x 10^-6 M of Cl- but that does not sound right. thanks
 P: 173 I will try to do it in steps so that we can find out mistakes if any easily. molarity of cr3+=41.35 *0.01280 /3*(20.17) -->(A) (note:both V are in mL so conversion factor cancels out) Total no moles in Cr3+soln = (A)*30.0 /1000 -->(B) No: moles unreacted=11.85 *0.01280 /3 -->(C) Therefore total no moles unreacted=(B)-(C) Therefore total no moles of CL- == [(B)-(C)]*6 Molarity of Cl- = [(B)-(C)]*6/.01353 Too lazy to do numericals now .Try this.
 P: n/a That answer looks better. I am getting 0.022 M of Cl-
PF Gold
P: 613
I will not contribute to the answering effort, as poolwin2001 did it quite well. I wondered some other thing:

 Quote by nautica Cr(Cl)6 does not react with en.
It is interesting, since ethylenediamine is a chelator, strong enough to make hexachlorochromate(III) ion to convert [Cr(en)3]3+, at least this is what I thought when I read this thread.

The theoretical reaction should be like this:

$[CrCl_6]^{3-}~+~3~en \rightarrow [Cr(en)_3]^{3+}~+~6~Cl^-$

I have some suggestions though. It is possible that ethylenediamine does not react readily with the labile chromium complex, so a rapid treatment would only chelate the uncomplexed chromium.

Is there anyone to clear this?

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