Question

@aaronq

Answer 1

Start by writing out your equilbrium expression first.

Answer 2

I mostly need help getting started. My instruction is to "determine the equilibrium concentration of FeNCS^2+ in each of the test tubes. After that I know how to do the rest, I'm just stuck on that one part :|

Answer 3

Sorry my handwriting is so light ! let me know if you can't read it

Answer 4

yeah, it's very light. I can see it if i magnify it.

i mean, as abb0t said, write the equilibrium expression.

Whats the equation for the reaction to start with?

Answer 5

ah sorry about that! the equation is Fe^3+ + SCN- <--> FENCS^2+

Answer 6

so the equal. Expression would be [FeCNS^2+]/[Fe^3+] [SCN]

Answer 7

equil*

Answer 8

no worries! yep that's right.

Now, to keep things organized, we make an I.C.E. table; were working in concentrations. I'll do it in general terms

\( [Fe^{3+}] ~~~~~~~~~~~~~~~~~~~~~~~~ [SCN^-] ~~~~~~~~~~~~~~~[FeNCS^{2+}]\)

I \( ~~~~~~~~[Fe^{3+}] ~~~~~~~~~~~~~~~~~~~~~~~~~~ [SCN^-] ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[0]\)
C \( -[FeNCS^{2+}]~~~~~~~~~~~~~~~~~~~-[FeNCS^{2+}]~~~~~~~~~~~~~~~+[FeNCS^{2+}]\)
E \( ~~[Fe^{3+}]-[FeNCS^{2+}] ~~~~~ [SCN^-]-[FeNCS^{2+}] ~~~~~~~~~[FeNCS^{2+}]\)



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

Answer 9

so all you're doing is subtracting the concentration of \(FeNCS^{2+}\) from the initial concentrations of Fe(III) and \(SCN^-\)

Answer 10

oh ok. How would i go about determining the Equil. [FeNCS^2+] in column 2 for starters

Answer 11

I understand what you mean but yet still have some confusion how to get started

Answer 12

so the concentration of \(FeSCN^-\) would be found from the absorbance values

so in the first column, you should have Eq \([FeSCN^-]=0\)

Answer 13

sorry that's not the right compound

Answer 14

i meant \(FeNCS^{2+}\)

Answer 15

gotcha. ok, I was told for the first column i would not calculate K eq.

Answer 16

how would i calculate the equal. of FeNCS^2+ for the second. with the information i have already

Answer 17

Okay, so you need the extrapolate from calibration curve in part A

Answer 18

right ahead of you! let me send you the graph for part A

Answer 19

This is too much for me. Lol.

Answer 20

LOL @abb0t


so, you need to find the concentration, use the equation for the line, the y value is the abs and the x, the concentration of FeNCS^2+

y=0.0328x+0.0306

Plug in y=0.053

0.053=0.0328x+0.0306

x=0.682926

so that's the concentration of FeNCS^2+ at eq. for trial two

Answer 21

I got a lightbulb moment! That looks familiar I understand why you did that! :)

Answer 22

you have Ms. Preetha on your question, i dont think you need me @toxicsugar22

Answer 23

@aaronq I'm going to do the rest on this row and Will let you know how i do. I may have a question about the remaining rows but i will let you know :D you're a great help

Answer 24

You're so annoying, toxic! Stop!

Answer 25

sounds good @elaornelas

Answer 26

@aaronq ok ran into a little problem!

Answer 27

To find the equil. [Fe^3+] = [Fe^3+] initial - [FeNCS^2+] equil
I do so and i got a negative answer. I know that isn't right :|

Answer 28

Did you use quadratic formula? Sometimes you get two answers, a + and -, value.

Answer 29

it shouldn't give you a negative answer. abb0t i dont think she needs to solve any quadratics, the concentrations can be found by just subtracting the eq. concentrations from the initial.

what values did you use?

Answer 30

(Solving for column 2.) From part B, I used the initial [Fe^3+] = 1.00x10^-3
& equil. [FeCNs^2+] = 0.682..
equil. [Fe^3+] = [Fe^3+] initial - [FeNCS^2+] equil
= 1.00x10^-3 - 0.682 = -0.679

Answer 31

oh i just noticed the label on the graph says \([FeNCS^{2+}1.0*1.0^{-5}\)

so i'm guessing you need to multiple the x value you get by \(1.0*1.0^{-5}\)

so for the second trial, you'd get \([FeNCS^{2+}= 0.0000068292682927\)

Answer 32

oh goodness!! Totally didn't do that

Answer 33

ok let me try once more

Answer 34

that should solve that problem

Answer 35

okays

Answer 36

I believe I'm done @aaronq :)

Answer 37

awesome ! :)