Question

Does the result of the calculation in Question 3 justify our original assumption that all of the SCN- is in the form of FeNCS^2+?

(Question 3:
Use the mean value off Keq to calculate the SCN- concentration in a solution whose initial Fe3+ concentration was 4.00x10^-2 M (.04M) and initial SCN- concentration was 1.00x10^-3M (.001M). Is all the SCN- in the form of FeNCS2+?
Mean Value = 48
Solved for SCN- concentration = 0.0006538 )

Answer 1

Sooo, \(SCN^-~unreacted=\dfrac{[SCN^-]_f}{[SCN^-]_i}*100\%\)

\(SCN^-~unreacted=\dfrac{0.0006538~M}{0.001~M}*100\%=65.38\%\)

Answer 2

which seems like a lot considering that the K value is 48

Answer 3

wait! oops i just realized i wrote the wrong SCN concentration value

Answer 4

I think you missed a step. After you solved for x in the eq. expression, you had to subtract

.001-0.0006231855734295

Answer 5

the SCN- concentration = 0.00346M

Answer 6

oh okay

Answer 7

is that the final? i think you messed up somewhere because its more than the initial conc.

I got \([SCN^-]_f=0.0003768~M\)

Answer 8

ah let me double check :/

Answer 9

okays

Answer 10

hm ok something must be wrong. I am still getting 0.0003462

Answer 11

0.0006538 is the x value I found after solving for the quadratic equation. How is it you got a different one? @aaronq

Answer 12

It's pretty close, 0.0006232, i do it with the computer so most likely due to rounding. It's the almost the same thing anyway

Answer 13

oh okay I see! ill try double checking my work anyway. Could you repeat again how I would answer this problem not entirely sure on this one

Answer 14

So we're just finding the fraction that didn't react.

We take what were left with at equilibrium and divide it by what we started with. This way we can get a relative measure of how much we have left. Does that make sense?

Answer 15

oh ok I understand! so if we follow the same step as you did previously we should get around 34 percent or so?

Answer 16

yes, exactly. i got 37.681% but again it's the rounding. I dont think it really matters.

You should comment on how not all of the SCN is complexed with the iron.

Answer 17

@aaronq ah ok I get what you did! This question has two more parts, would you be able to help explain those also? :/

Answer 18

sure! i gotta get going in a few mins though

Answer 19

oh ok! These don't require problem solving! They ask:

Answer 20

"Question 4 Asks: Does the result of the calculation in Question3 justify our original assumption that all of the SCN- is in the form of FeNCS^2+?

& Question 5 Asks: Based upon your answer to question 4 is the measured value of Keq too high or too low?"

Answer 21

for #4 that would be no because a small amount still remains ?

Answer 22

yeah, quite a substantial amount (more than one third) remains, so the assumption is not supported by the results.

for #5, you'd wanna think about what the value of K really is (it's a value for the ratio of products to reactants) and how it would change if it was a larger number or a smaller number.

If the reaction should be such that no SCN is left not complexed to Fe, then the value of K you calculated must've been too low.

Answer 23

that makes a lot more sense!

Answer 24

awesome!

Answer 25

Ill review this problem again to make sure i fully understand. Thank you for explaining that to me! Appreciate it :)

Answer 26

no problem at all! take care :)