Question

Medal and fan
An aqueous solution of .25 M silver nitrate, AgNO3, and .25M iron(11) nitrate, Fe(NO3)2, are allowed to come to equilibrium in the following chemical reaction:
Ag^+(aq) + Fe^2+(aq) <=> Fe^3+(aq) + Ag(s)
If the equilibrium constant, Kc, for the reaction is 3.0 x 10^-3, what is the equilibrium concentration of Fe^3+?

Answer 1

I got 7.5 * 10^-4 but I'm not sure if that's correct or not.

Answer 2

lets see

Answer 3

\[Kc= \frac{ product }{ reactant }\]
\[Kc= \frac{ [Fe^3+] * [Ag ] }{[Ag^+] * [Fe^2+] }\]

Answer 4

Fe3 + and Ag are unknown so i'll take each as X
\[Kc= \frac{ [x]*[x] }{ [0.25]*[0.25] }\]
\[3.0 x 10^-3= \frac{ x^2 }{ 0.25^2 }\]

Answer 5

now we take square root of everything, meaning the whole equation

Answer 6

I guess what I don't understand is how you use those concentrations as reactants when silver nitrate and iron nitrate aren't in the balanced equation..

Answer 7

well i am showing the working to you :) so that you get what to do, of course you should balance them

Answer 8

since you already did it, u can see if u r on the right track right?

Answer 9

or u want with balanced?

Answer 10

No, what I'm saying is that I thought that the Ag and Fe on the left side were the reactants? I don't see how silver nitrate and iron nitrate concentrations can be used as the reactants in that equation.

Answer 11

Ag^+(aq) + Fe^2+(aq) <=> Fe^3+(aq) + Ag(s)

Fe^3+(aq) + Ag(s) is product
Ag^+(aq) + Fe^2+(aq) is reactant


Ag^+(aq) is from AgNO3
Fe^2+(aq) is from Fe(NO3)2

Answer 12

look Kc is about concentrations
if Ag+ has a concentration of 0.25 M
then that means the whole solution is same concentration

Answer 13

in this reaction they are just showing you that Ag+ is oxidized to Ag and Fe2+ is reduced to Fe3+

Answer 14

I got you now, sorry, I never knew that.

Answer 15

that's what they are showing you

Answer 16

no problem :) it's okay

Answer 17

so u get the whole thing now?

Answer 18

I have 1.36 * 10^-2?

Answer 19

as ur answer u mean?
btw the reaction is balanced actually

Answer 20

I know it is, I didn't mean that it was unbalanced.

Yes x= 1.36*10^-2

Answer 21

yup :)

Answer 22

So that would be the answer?

Answer 23

this is your answer actually

Answer 24

wait let me check of a sec to confirm

Answer 25

Okay.

Answer 26

yup that's right :)

Answer 27

u can check it you know
but round of 1.37 *10^-2

Answer 28

i mean to 1.37 *10^-2

Answer 29

This was actually a multiple choice answer problem, and none of the answers match. The closest to my answer is 1.2 * 10^-2

Answer 30

can u show me the options?

Answer 31

Yep, hold on. In addition to that we have 1.9*10^-14, 7.5*10^-4, and 4.8*10^-2

Answer 32

they are too big and too small
i wonder if im meaning mistake anywhere

Answer 33

oh yeah
wait

Answer 34

no, idk but still can't get why it's like that

Answer 35

i tried putting 0.25-x instead of 0.25

Answer 36

That wouldn't make a difference though since k is so small.

Answer 37

yeah i'll tag someone who's good at this
since im not really friends with this equilibrium concentration

Answer 38

@aaronq @Kainui

Answer 39

Awesome, thankyou(:

Answer 40

they both are amazing at this, but now they seem to be offline
i'll try checking out whats wrong on my own :)

Answer 41

@Miracrown Would you have any idea how to solve this?

Answer 42

@alphadxg

Answer 43

@abb0t

Answer 44

I don't understand what you're lost in? Someone explained it already.

Answer 45

The answers don't correspond with the multiple choice options.

Answer 46

@IamStark thanks for finding someone for me(:

Answer 47

"if Ag+ has a concentration of 0.25 M
then that means the whole solution is same concentration " is this statement correct ?

Answer 48

See, I don't know, it didn't seem right to me, but I take online AP Chem and sometimes things aren't thoroughly explained in the lesson calculations.

Answer 49

I'd ask my teacher, but they take ages to get back to you and I'm a bit pressed for time.

Answer 50

@Somy can u xplain how...i am not able to get it :(

Answer 51

@iPwnBunnies

Answer 52

@mathslover

Answer 53

is there any dissociation constant given in the question @seehearcreate ?

Answer 54

Nope just the concentration equilibrium constant.

Answer 55

Is answer 7.5 X 10^-4 ?

Answer 56

omg im so sorry i was away

Answer 57

That's the correct answer? The answer I had originally when asking the question?

Answer 58

how did u get it @Abhisar ? it is in the options

Answer 59

U mean its 7.5 X 10^-4 ??

Answer 60

I think its the correct answer

Answer 61

Is it ???

Answer 62

show the working already =_=

Answer 63

I don't know, it doesn't give me the answers till the end.

Answer 64

not in the end I mean, until the teacher checks the homework.

Answer 65

Looks like u guys were struggling for a long time !

Answer 66

I'm confused, is the answer that I originally had the correct one? The 7.5*10^-4?

Answer 67

Could you show the work so I can compare it with mine, since we got the same answers?

Answer 68

one sec....

Answer 69

Alright, thanks(:

Answer 70

r u sure the first option is 1.9 * 10^-14 or 1.9 * 10-4 ?

Answer 71

i wonder what exactly went wrong hmmmmm

Answer 72

It's 1.9 * 10^-4

Answer 73

its the correct answer

Answer 74

Take conc of solid as 1

Answer 75

wow

Answer 76

wait

Answer 77

conc of Ag = 1

Answer 78

Okay, but I still don't see how you came to that answer, can you show me the calculations you did?

Answer 79

rest as @Somy said

Answer 80

Oh yeah, I got it. So for future, will all pure solids be in a concentration of 1 unless stated otherwise?

Answer 81

\[K_{c} = \frac{ [Fe^{+3}]X[Ag]}{[Ag^{+}] X [Fe^{+2}]}\]

Answer 82

and [Ag] = 1

Answer 83

this is the question

Answer 84

Active mass of the solid substances is always unity

Answer 85

u will never see anything else than 1 for solid \(\color{green}{\huge\ddot\smile}\)

Answer 86

Awesome, thankyou, I have to medal @Somy though because she provided the most help and stayed with me through it haha

Answer 87

yeah she deserve it..even i have given her 1

Answer 88

BTW, out of curosity, if some one wants to know then i'll like to tell that in equilibrium reaction, the concentration of solids remain constant no matter how much solid is taken.
Therefore, by convention, the concentration of solids are taken as unity.
In case, a liquid is present in equilibrium with some gas, by covention, it is taken as unity.

Answer 89

owhhhhh

Answer 90

i didn't know such thing hhh

Answer 91

sorry for kinda confusing u more T_T @seehearcreate

Answer 92

Haha noo, you were on the right track, you just didn't know about the unity...neither did I lol

Answer 93

I was completely off entirely

Answer 94

owh i guess this is right also, which grade are u in btw?

Answer 95

12th. I did online, so I couldn't graduate with the rest of my class until my online classes were finished.

Answer 96

i just finished my AS level gr12 and this is not in my syllabus so i guess that's why i didn't know it hhhh