Could you please fill out the second table and please show me the calculations y

Question

Could you please fill out the second table and please show me the calculations you did. As well as answering the questions.
Thanks EXPERIMENT 3: CHEMICAL EQUILIBRIUM 6. Proceed as in the step for the remaining reaction solutions (rinse,fill, read). It is recommended to measure from the least concentrated to the most concer Note: Should you see a pattern for absorbance in your test tubes? 7. When you have recorded all of your data, the solutions should be poured into the acid waste tubs. MIXING DATA AND ABSORBANCE DATA. Prepared Solutions Fe' 2.00 x 10-3 M (in 0.500 M HNO, [HSCN) 2.00 x 10 s M (in 0.500 M HNO, [HNO,] 0.500 M (HINT: This should be constant!) Equilibrium [Fe(SCN) Tube mL Fe mu HSCN mL H Absorbance 500 4.00 1.00 2 500 200 3.00 3 5.00 3.00 2.00 5.00 400 1.00 000 5.00 5.00 CALCULATIONS Fe' (aq) HSCN(aq) Fe(SCN)2 (aq) H (aq) 1. Volume of 2.50 M HNO, required to prepare 25.00 ml of 0.500 M HNO 2. Equilibrium [Fe(sCN) 3, sample calculation of K

Explanation / Answer

For tube 1 calculation

1. VFe3+=5 ml =0.005 lit

Now in solution

V= Volume of total solution=10 ml

MV=MFe3+*VFe3+

So molarity of Fe3+ in solution= (MFe3+*VFe3+)/V=2*10-3*0.005/0.01=1*10-3 M

Molarity of HSCN in solution=(MHSCN*VHSCN)/V=2*10-3*0.001/0.01=2*10-4 M

HNO3 molarity is constant

So [H+]=0.5M

So ICT table

Fe3+   HSCN FeSCN2+ H+

Initial 1*10-3 2*10-4   0

Change -3.6*10-5 -3.6*10-5   3.6*10-5

Equilibrium(I+C) 9.64*10-4 1.64*10-4 3.6*10-5   0.5(constant)

Keq=[FeSCN2+][H+]/([Fe3+][HSCN])=(0.5*3.6*10-5)/(9.64*10-4*1.64*10-4)=113.85

Now doing same thing for all test tubes.

First find Molarity of each test tubes.

Mean Keq=221.0284

std deviation=199.6799

Tube vol Fe3+ vol HSCN [Fe3+ ] [HSCN] 1 5 1 0.001 0.0002 2 5 2 0.001 0.0004 3 5 3 0.001 0.0006 4 5 4 0.001 0.0008 5 5 5 0.001 0.001

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