Finding Equilibrium concentrations with added solutes?

  • #1
HF(aq) ↔ H+ + F-

Determine equilibirum concentrations with Initital concentrations and Kc.

Kc = 7.2 x 10-4
[HF] = 1.00M
[NaF] = 0.50M

I understand how to do an ICE table and put together questions, but what do you do with [NaF] when it is not part of the chemical equation? How to I factor it into it? Thanks!
 

Answers and Replies

  • #2
What you have to remember is that ionic salts dissociate into their ions when in solution.

Consider the dissociation reaction of the NaF:

NaF → F- + Na+

Hopefully that should help you understand what to do.
 
  • #3
What you have to remember is that ionic salts dissociate into their ions when in solution.

Consider the dissociation reaction of the N
NaF → F- + Na+

Hopefully that should help you understand what to do.
So the if it all dissociates then the initial concentration for F- should be 0.5M. I use that molarity but it still doesn't seem to work out. I can't seem to get anything to equal the equilibrium constant.
 
  • #4
So the if it all dissociates then the initial concentration for F- should be 0.5M. I use that molarity but it still doesn't seem to work out. I can't seem to get anything to equal the equilibrium constant.
It is possible that I am misunderstanding. You were asked to find the equilibrium concentrations correct?

so you should have an ice table similar to this:
HF(aq) ↔ H+ + F-
i 1M 0M 0.5M
c -x +x +x
e 1-x x 0.5+x

from here you should be able to create your equilibrium expression of Kc = [products]e/[reactants]e

then if substitute in your expressions from ice table and the equilibrium constant you should be able to calculate x and then determine all of your equilibrium concentrations.

You should also note that it is probably acceptable to assume that 1-x ≈ 1 and that 0.5+x ≈ 0.5. this makes your calculations much simpler.

I can't seem to get anything to equal the equilibrium constant.
You were already given the equilibrium constant. you are solving for the equilibrium concentrations are you not?
 
  • #5
I can setup my ice table like that fine, but where does [NaF] come into play, do I just add 0.50M to F- when I have equilibrium concentrations??
 
  • #6
thank you for the help btw :) !
 
  • #7
Borek
Mentor
28,600
3,078
but where does [NaF] come into play
Have you checked the ICE table Ryan Schmidt gave you? 0.5 is already in the I (initial) line, taken care of.

@Ryan Schmidt - please don't go too far with help, by forum rules we guide people to the answer, not give it to them.
 

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