Question

Can somone help me with my lab? It is a M1V1=M2V2 equation. Just need help setting it up.

We labeled six 50.00mL volumetric flasks 1-6. Then pipet 10.00 mL of .200 M Fe(NO3)3 solution into each volumetric flask. Then pipet 1.00, 2.00, 3.00, 4.00, 5.00 mL of 2.00 x 10^-3 M NaSCN solution into flasks 2 through 6 respectively. We then added sufficient 0.10 M nitric acid to each flask to make the total volume 50.00 mL.

The equation we are to use is M1V2=M2V2 to find the initial [SCN-], M. If you can help me set it up for each flask number that would be great! :) Thanks!

Answer 1

\( Fe^{3+}+SCN^- \rightleftharpoons [Fe(SCN)]^{2+}\) ; formation constant: \(K_f\)=?

so find how many moles of SCN \((n_{SCN^−})\) are left after the complexation?

HINT: write an equilibrium expression

Your final volume is 50 mL so 0.05 L with unknown molarity "x M"

\(xM=\dfrac{n_{SCN^−}}{L_{solution}}→xM=\dfrac{n_{SCN^−}}{0.05L}\)

Answer 2

Oh, okay gotcha. So for the first equation would it be
(2x10^-3)(1mL)/.05L?

Answer 3

not quite. you have to find how much SCN is left after the complexation of \([Fe(SCN)]^{2+}\) occurs.

Answer 4

How do I find that?

Answer 5

write an equilibrium expression (make an ICE table) and find how much of it is left after everything has equilibrated.

Answer 6

Would you mind doing the first equation for me? I should be able to understand it after that.

Answer 7

the equilibrium expression is:

\(K_f=\dfrac{[Fe(SCN)^{2+}]}{[Fe^{3+}][SCN^-]}\)

okay, at 25 degrees celsius \(K_f^{[Fe(SCN)]^{2+}}=8.9×10^2\)

\(8.9×10^2=\dfrac{[Fe(SCN)^{2+}]}{[Fe^{3+}][SCN^-]}\)

"Then pipet 10.00 mL of .200 M Fe(NO3)3 solution into each volumetric flask"

\(n_{Fe^{3+}}=L_{Fe(NO_3)_3}*M_{Fe(NO_3)_3}=(0.001\;L)(0.200 M)=0.0002\;moles\)

new molarity: \(M_{Fe^{3+}}=\dfrac{0.0002\;moles}{0.05L}=0.004 M\)

flask 2: "Then pipet 1.00 mL o f 2.00 x 10^-3 M NaSCN"

(through a similar calc. to the one above) \(M_{SCN^-}=0.00004\;M\)

\(Fe^{3+} + SCN^- \rightleftharpoons [Fe(SCN)]^{2+}\)
I 0.004 0.00004 0

C -x -x +x


E 0.004-x 0.00004-x x

\(8.9×10^2=\dfrac{x}{(0.004-x)\color{red}{(0.00004-x)}}\rightarrow x=0.0000311743708865\)

\([SCN^-]=\color{red}{(0.00004-x)}=(0.00004-0.0000311743708865)\)

\([SCN^-]=0.000008825629=8.8*10^{-6}\) for flask 2.

Answer 8

by the way i didn't mention that i substituted the values from E (from the ICE table) into the initial equilibrium expression \(K_f=\dfrac{[Fe{SCN)]^{2+}}}{[Fe^{3+}][SCN^-]}\)