Solving for Concentration of Ethanoate Ions in Ethanoic Acid Solution

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Discussion Overview

The discussion revolves around calculating the concentration of ethanoate ions in a solution of ethanoic acid using the acid dissociation constant (Ka). Participants explore the application of the ICE (Initial, Change, Equilibrium) method in this context, addressing challenges in setting up the calculations correctly.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the problem of finding the concentration of ethanoate ions in a 0.18 mol/L solution of ethanoic acid, using Ka = 1.76 x 10-5.
  • Another participant suggests that the ICE method is suitable and should work as expected.
  • A participant attempts to set up the ICE table but arrives at a concentration of 0.1799999 mol/L, questioning the discrepancy with the expected answer of 1.8 x 10-3 mol/L.
  • Another participant points out that the ICE table is incomplete and that the Ka expression is incorrectly set up, suggesting that the dissociation reaction should start with the undissociated substance as the reactant.
  • A participant proposes a revised ICE table but struggles with the Ka expression, questioning the placement of products and reactants in the equation.
  • There is confusion regarding the correct identification of products in the dissociation reaction, with participants discussing the roles of hydronium and acetate ions.
  • Participants express uncertainty about the correct setup of the Ka expression and the ICE table conventions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct setup of the ICE table and the Ka expression. There are multiple competing views on how to approach the problem, and the discussion remains unresolved.

Contextual Notes

Participants highlight limitations in their understanding of the dissociation reaction and the proper conventions for setting up the ICE table and Ka expression. There are unresolved questions about the correct identification of products and the implications for the calculations.

dav1d
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Homework Statement


What concentration of ethanoate (acetate) ions is
expected in a 0.18 mol/L solution of ethanoic
(acetic) acid? Ka is 1.76 x 10-5 at 25°C.


Homework Equations



Ka=products/reactants

The Attempt at a Solution



I know that Ka=1.76 x 10-5=products/reactants but I really don't know what to do from there. I have tried the ICE box method; initial, change, equilibrium but that doesn't seem to work. My textbook only covers the ICE box method and the basic formula, help is appreciated!
 
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Show your take with the ICE method, not only it is perfectly suitable here but it also works exactly as expected.
 
hydrogen acetate acetic acid
initial: 0 0 0.18mol/L
change: - +x -x
equilibrium: 0 0+x 0.18-x

1.76x10^-5=(0.18-x)/x
x=.1799999 mol/L

but apparently the answer is 1.8x10^-3 mol/L.

Am I missing something?
 
Last edited:
dav1d said:
hydrogen acetate acetic acid
initial: 0 0 0.18mol/L
change: - +x -x
equilibrium: 0 0+x 0.18-x

1.76x10^-5=(0.18-x)/x

Am I missing something?

Yes, your ICE table is incomplete (what happens to H+ concentration during acid dissociation?) and your Ka expression is umop apisdn.

To some extent it doesn't matter, but the convention is to write dissociation reaction starting with undissociated substance as reactant, and to make ICE table for such a reaction. You reversed everything.
 
so would it be

hydrogen acetate acetic acid
initial: 0 0 0.18mol/L
change: +x +x -2x
equilibrium: x x 0.18-x

1.76x10^-5=(o.18-x)/x^2
that still gives an incorrect 0.18 or -1.76...

What do you mean the Ka is upside down? Isn't it products/reactants?
So how should I write the table using the 'normal' convention?
 
What are products of the dissociation reaction?
 
Borek said:
What are products of the dissociation reaction?

hydrogen and acetate are dissociated from the acetic acid.
 
dav1d said:
Isn't it products/reactants?

dav1d said:
hydrogen and acetate are dissociated from the acetic acid.

So if hydronium (not hydrogen!) and acetate are products, why do you put them in denominator?
 
Borek said:
So if hydronium (not hydrogen!) and acetate are products, why do you put them in denominator?

lol I guess I should have written the equation first. will update soon.
 

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