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Topic: Kc  (Read 5520 times)

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Offline dm164

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Kc
« on: March 23, 2009, 06:35:35 PM »
 I have two question that I don't get please help.

1. In one experiment, 20mL of 5.43x10-3M Fe(NO3)3 and 10mL of 6M HNO3 was placed in a 100mL volumetric flask and diluted to the mark with distilled water. What was the concentration of iron(III) nitrate in the final solution?

            If I read it right do I just do: [5.43x10^-3M Fe(NO3)3]*20/100 = 1.09x10^-3M Fe(NO3)3

2. If the solution in the previous problem orginally was 0.001M Fe(NO3)3, what is the equilibrium concentration of Fe3+ in the solution?
               
Thanks in advance
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Offline Vette Freak

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Re: Kc
« Reply #1 on: March 24, 2009, 10:31:50 AM »
Write out the balanced equations for both equilibrium reactions.  Then set up an equilibrium table using the moles of the reactants, x being the change in moles involved in reaching equilibrium.  Write the expressions for Kc for both equilibrium reactions (Kc1 and Kc2) and write the net equilibrium expression (Kcnet).  If you have access to a table to look up the Kc values for these two reactions (I think this is the key bit of information you are missing), you can then solve for Knet=Kc1*Kc2.  From there you can solve for x in your net equilibrium expression and calculate the concentration of Fe(NO3)3.  Repeat for part 2 using the new molarity of Fe(NO3)3.
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Offline Borek

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Re: Kc
« Reply #2 on: March 24, 2009, 11:19:41 AM »
I wonder if it really asks about equilibrium, or just about dilution.
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Offline Vette Freak

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Re: Kc
« Reply #3 on: March 24, 2009, 11:22:09 AM »
I was wondering the same, but given the subject line, I went with equilibrium.
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Offline Borek

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Re: Kc
« Reply #4 on: March 24, 2009, 11:27:23 AM »
Quote from: Vette Freak on March 24, 2009, 10:31:50 AM
Write out the balanced equations for both equilibrium reactions.

Can you list them?
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Offline dm164

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Re: Kc
« Reply #5 on: March 25, 2009, 03:48:01 PM »
Ok now I figured why I was confused, I couldn't solve it because I don't know the Kc, but I knew I wasn't supposed. I miss read the questions. There is another question that I left out, because I thought my 2nd question wasn't referring to it.

  The question is: Determine the quilibrium concentration of Fe(SCN)2+ in a solution with an abosrbance of .555 at 447nm with beer's law. I determined it to be 1.39x10^-4M Fe(SCN)2+

From this I can the equation Fe3+ + SCN-  ::equil:: Fe(SCN)2+

Using ICE 1x10-3 - -1.39x10-4 = 8.61x10-4 M Fe3+


Sorry for my mistake, but thanks for the help anyway.

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