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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: plu on January 07, 2006, 10:54:07 PM

Title: iron thiocyanate calculations
Post by: plu on January 07, 2006, 10:54:07 PM
Hello there!  I was recently approached with a question that I could not answer.  :-[  Here it is:

A student was to determine the equilibrium constant for the reaction between iron(III) and thiocyanate ions.  A standard reference solution was prepared by mixing 18.0 mL of a 0.20 mol/L Fe3+ solution with 2.0 mL of a 0.0020 mol/L SCN-.  The absorbance of this solution was found to be 0.520.  A second solution was prepared with the same 0.0020 mol/L SCN- and dilute 0.0020 Fe3+ solutions.  Solution #2 contained the following: 5.0 mL Fe(NO3)3, 2.0 mL KSCN, and 3.0 mL H2O.  The absorbance of the second solution was found to be 0.138.  Assuming FeSCN2+ follows Beer's Law, find the equilibrium constant for the reaction.  Note: The molar absorptivity of FeSCN2+ is 7.00 x 103 L/cm.mol.

I can work the question out if I assume that the pathlength used is a standard 1.00-cm cell.  However, it is apparently possible to solve the question without knowing the pathlength   :-\  Assistance requested!
Title: Re:iron thiocyanate calculations
Post by: Mitch on January 07, 2006, 11:56:19 PM
Its a common assumption to assume the path length is 1cm.
Title: Re:iron thiocyanate calculations
Post by: kkrizka on January 08, 2006, 02:07:37 AM
We are doing this lab in class right now, so I might be some help. According to my teacher, we can use the first testube, our reference test tube to figure out the molarities of others throug cross multiplication. This means we don't need to use Beer's Law or at least not directly.

So it would be setup like this:
0.002M     ?M?
--------- ------
0.520       0.260

Does anyone here see a problem with this method?

Btw, we had to cancel the lab because the concentration of KSCN was too high so the resulting solution was too dark and we couldn't get the reading. Doing it next class!
Title: Re:iron thiocyanate calculations
Post by: sdekivit on January 08, 2006, 05:14:24 AM
if law of Lambert-Beer is applicable to the experiment (that is: when extinction and concentration are proportional to each other), than there's no problem using cross multiplication

--> remember that when E and c are proportional, when E get's 2 times higher c does too. Now cross multiplication is no problem.
Title: Re:iron thiocyanate calculations
Post by: plu on January 08, 2006, 11:26:51 AM
So it would be setup like this:
0.002M     ?M?
--------- ------
0.520       0.260

How are you getting the 0.260?
Title: Re:iron thiocyanate calculations
Post by: kkrizka on January 08, 2006, 11:36:35 AM
How are you getting the 0.260?

Random number I made up for my example.
Title: Re:iron thiocyanate calculations
Post by: plu on January 08, 2006, 11:50:04 AM
Random number I made up for my example.

Ah, I see  :D  How would you complete the solution then?
Title: Re:iron thiocyanate calculations
Post by: kkrizka on January 09, 2006, 12:56:45 AM
Ah, I see  :D  How would you complete the solution then?

You mean calculate the unknown concentration? Well, since the absorbance reading is directly proportional to the concentration, halfing it would also half the concentration. So my number is for a 0.001M concentration of Fe(SCN)2+ For less nice numbers, just use cross multiplication.
Title: Re:iron thiocyanate calculations
Post by: plu on January 09, 2006, 06:55:21 AM
You mean calculate the unknown concentration? Well, since the absorbance reading is directly proportional to the concentration, halfing it would also half the concentration. So my number is for a 0.001M concentration of Fe(SCN)2+ For less nice numbers, just use cross multiplication.

Oy, but you would have to again assume a pathlength of 1.0 cm since you don't know the concentration of FeSCN2+ for either of the two trials  :-\
Title: Re:iron thiocyanate calculations
Post by: Borek on January 10, 2006, 04:38:33 AM
No. Try to solve the question using symbols and assuming pathlenthg l - it will cancel out at the end. That's why cross multiplication gives correct result immediately.