Solving for Concentration of Ethanoate Ions in Ethanoic Acid Solution

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SUMMARY

The concentration of ethanoate ions in a 0.18 mol/L solution of ethanoic acid can be calculated using the acid dissociation constant (Ka) of 1.76 x 10-5 at 25°C. The ICE (Initial, Change, Equilibrium) method is essential for this calculation. The correct setup involves recognizing that hydronium ions (H3O+) and acetate ions (CH3COO-) are the products of the dissociation of acetic acid. The final concentration of ethanoate ions is determined to be 1.8 x 10-3 mol/L, correcting the initial miscalculation.

PREREQUISITES
  • Understanding of acid dissociation constants (Ka)
  • Proficiency in using the ICE table method for equilibrium calculations
  • Knowledge of the dissociation reaction of weak acids
  • Familiarity with hydronium ion concentration in acid solutions
NEXT STEPS
  • Study the derivation and application of the ICE table method in equilibrium problems
  • Learn about the dissociation reactions of other weak acids and their corresponding Ka values
  • Explore the concept of equilibrium constants and their significance in chemical reactions
  • Investigate the role of hydronium ions in acid-base chemistry
USEFUL FOR

Chemistry students, educators, and anyone involved in acid-base equilibrium calculations will benefit from this discussion, particularly those studying weak acids like acetic acid.

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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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