Help with Equilibrium Lab Problem

Nov 2013
1
0
Minnesota
PLEASE READ THE INFO FIRST THEN HELP ANSWER MY QUESTION AT THE END THANKS.

This lab consisted of two parts.
The first part procedure :

1. Use a buret to transfer exactly 4.0 mL of 0.0025 M solution of Fe(NO3)3 (in 0.10 M HNO3) to a 100-mL volumetric flask.

2. Now obtain about 100 mL of 0.10 M HNO3 solution in a clean, dry 250 mL beaker. Add the HNO3 solution to the volumetric flask until the flask is filled halfway, stopper it and mix thoroughly. Add more 0.10 M HNO3 until the bottom of the meniscus almost coincides with the 100-mL mark on the flask. Add the last 0.5 mL of 0.10 M HNO3 from a medicine dropper to make sure that you do not go about the mark. Put the stopper in and miz thoroughly.

NOTE: Remember you have made a dilution here and will make further dilutions in test tubes 1-5.

So we added these volumes to obtain the equilibrium reaction :

Test Tube #1
1.0 mL Fe(NO3)3
5.0 mL 1.0 M KSCN
4.0 mL 0.10 M HNO3

Test Tube #2
2.0 mL Fe(NO3)3
5.0 mL 1.0 M KSCN
3.0 mL 0.10 M HNO3

Test Tube #3
3.0 mL Fe(NO3)3
5.0 mL 1.0 M KSCN
2.0 mL 0.10 M HNO3

Test Tube #4
4.0 mL Fe(NO3)3
5.0 mL 1.0 M KSCN
1.0 mL 0.10 M HNO3

Test Tube #5
5.0 mL Fe(NO3)3
5.0 mL 1.0 M KSCN
0.0 mL 0.10 M HNO3


The question is asking us to find the concentration of Fe(SCN)^2+ which will be the same as the starting concentration of Fe(NO3)3.

My question is : what is the starting concentration of Fe(NO3)3 ??
To find the concentration of Fe(SCN)^2+ I would just use the formula M1V1 = M2V2 right?
But then I don't know what to use for the initial concentration of Fe(NO3)3 if... it has been diluted in HNO3 ??

I don't quite understand.

I used 0.0025 M as the initial concentration for Fe(NO3)3 but now I am not so sure.