FeSCN2 You mixed FeNO33 and KSCN together in part AThese two

[FeSCN]²⁺

You mixed Fe(NO₃)₃ and KSCN together in part A.These two reactants formed [FeSCN]²⁺ as a product:Fe³⁺ + SCN^- FeSCN²⁺

If you assume that Fe³⁺ is the limiting reagent, and that all of the iron reacted to form the product, then you can assume that the concentration of Fe³⁺ added to the test tube is the same as the concentration of the product:[Fe(NO₃)₃] = [FeSCN]²⁺

Using M₁V₁ = M₂V₂, find the concentration of the product, [FeSCN]²⁺, for each test tube and fill in table A1 above.Use Table 5.2 in your lab manual to find the final volume.

Show an example of how you calculated [FeSCN]²⁺ for test tube one. Show all units and circle your final answer.

Concentration of iron(lI) nitrate, [Fe(NO:)a]: 2.00x 10-4M Concentration of potassium thiocyanate, [KSCN]: 1.00 M Table A1: FescN* TestVolume of tube 2.00 104 M Absorbance [FesCNI+ step 1 Fe(NOs)3 (mL) 2 3 4 1.00 2.00 3.00 4.00 .012 002 012 021

Solution

We shall use the relation

M₁*V₁ = M₂*V₂

where M₁ = concentration of Fe²⁺ in the stock solution; V₁ = volume of stock solution taken; M₂ = concentration of Fe³⁺ in the stock solution = concentration of [FeSCN]²⁺ and V₂ = final volume of the test solution.

We have V₂ = 10.00 mL; plug in values and obtain

(2.00*10^-4 M)*(1.00 mL) = M₂*(10.00 mL)

====> M₂ = (2.00*10^-4 M)*(1.00)/(10.00) = 2.00*10^-5 M (ans).

Fill in the table.

Test Tube

Volume of 2.00*10^-4 M Fe(NO₃)₃ (mL)

Absorbance

[FeSCN]²⁺ (M)

1

1.00

2.00*10^-5

2

2.00

4.00*10^-5

3

3.00

6.00*10^-5

4

4.00

8.00*10^-5

Test Tube	Volume of 2.00*10-4 M Fe(NO3)3 (mL)	Absorbance	[FeSCN]2+ (M)
1	1.00		2.00*10-5
2	2.00		4.00*10-5
3	3.00		6.00*10-5
4	4.00		8.00*10-5