[HeroWH]

I.60 mL of Fe(NO3)3 (0.20 M in 0.50M HNO3) for standard solutions
II. 70 mL of Fe(NO3)3 (2.0 x 10^-3 M in 0.50 M HNO3) for unknown solutions
III. 60 mL of KSCN (2.0 x 10^-3 M in 0.50 M HNO3) for both standard and unknown solutions
IV. 100 mL of HNO3 (0.50 M) as solvent
Find the volume of solution III that gives the desired initial concentration of KSCN when the final volume of 100.00 mL is reached

SCN- concentration of Standard solution A: 1.5 x 10^-4 M
SCN- concentration of Standard solution B: 1.0 x 10^-4 M
SCN- concentration of Standard solution C: 5.0 x 10^-5 M
SCN- concentration ofStandard solution D: 2.0 x 10^-5 M
Each standard solution is prepared in a 100 mL volumetric flask by:
a. 50.0 mL of solution I. This will give you excess fe3+ in the final solution
b. the calculated volume of solution III measured with a volumetric pipette, that gives the desired initial concentration of KSCN when the final volume of 100.00 mL is reached.
c. Enough of solution IV to bring the total volume up to the 100.00 mL line on the volumetric flask.

Guys help, ive tried everything. Im using c1v2=c2v2 im using the solution I as a reference if im doing it right only 50.0 ml of solution I should be present but im not getting that. Someone pls *delete me*

[Borek]

Show your calculations (even if they are not getting the expected results) for any of the solutions.

[HeroWH]

Well ive tried every variation I could think of calculating moles separately and then using the molarity to determine volume. But this is the first thing I tried, so lets say I was trying to find out the volume of solution III needed to prepared standard solution a which SCN- ion concentration is 1.5x10^-4 M. So, 1.5x10^-4M x 0.1L/2.0x10^-3 M=7.5 mL.
But if i ran that same calculation on the Fe(NO3)3 I would expected to get 50.0 mL but I dont. Am I  missing something? Throw me a hint???

[Borek]

to find out the volume of solution III needed to prepared standard solution a which SCN- ion concentration is 1.5x10^-4 M. So, 1.5x10^-4M x 0.1L/2.0x10^-3 M=7.5 mL.

Looks OK to me.

But if i ran that same calculation on the Fe(NO3)3 I would expected to get 50.0 mL but I dont. Am I  missing something? Throw me a hint???

Why do you expect 50 mL of iron solution?

BTW, looks to me like there there is something wrong with the wording:

Each standard solution is prepared in a 100 mL volumetric flask by:
a. 50.0 mL of solution I. This will give you excess fe3+ in the final solution

You have only 60 mL of the solution I, so perhaps you shoudl use just 5 mL of the solution I per each standard solution?

Whichever it is, you are expected to prepare solution that has excess of Fe³⁺.

[HeroWH]

Turns out the Fe(NO3)3 is irrelevant. Thanks for the help though!

[Borek]

Turns out the Fe(NO3)3 is irrelevant.

Which is why I asked about 50 mL...