Percent Dissociation of Acetic Acid

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Discussion Overview

The discussion revolves around calculating the pH and mass of acetic acid in a solution that is 11% ionized at 25 degrees Celsius. Participants explore the relationships between concentration, dissociation, and the acid dissociation constant (Ka) in the context of a homework problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses uncertainty about how to begin the calculations without knowing the concentration of acetic acid.
  • Another participant suggests that there are equations related to the 11% dissociation and the stoichiometry of the dissociation that need to be considered.
  • A participant proposes letting x represent the number of moles of acetic acid in 1 liter of solution and asks how to express the moles of acetic acid, acetate, and hydronium ions after 11% dissociation.
  • There is mention of a formula relating concentration and degree of dissociation, with a participant questioning its name and the reason for squaring the degree of dissociation.
  • One participant confirms a calculation approach involving the concentrations of the species after dissociation.
  • A participant references an answer key that aligns with the formula discussed, suggesting it is correct.
  • Another participant acknowledges the answer key but expresses confusion about its application.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the application of the formula for calculating concentration based on degree of dissociation, as there are differing interpretations of the answer key and its implications.

Contextual Notes

There are unresolved questions regarding the derivation of the formula mentioned and the specific definitions of terms used in the calculations. The discussion also reflects varying levels of understanding of the concepts involved.

Who May Find This Useful

Students working on acid-base chemistry problems, particularly those involving dissociation and pH calculations, may find this discussion relevant.

Teemo
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Homework Statement


In a particular solution, acetic acid is 11% ionized at 25 degrees Celsius. Calculate the pH of the solution and the mass of acetic acid dissolved to yield 1.00 L of solution.
Ka of CH3CO2H: 1.8x10^(-5)

Homework Equations


Ka=([CH3CO2-][H3O+])/[CH3CO2H]
pH=-log[H3O+]

The Attempt at a Solution


I'm not sure how exactly I should begin, as I do not know the concentration of the acetic acid, so I don't know how to solve for [H+]
 
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There are two equations that you have not listed yet. One will describe 11% dissociation. The other will be a direct effect of the dissociation stoichiometry - what can you tell about amounts of H+ and CH3COO- present in the solution?
 
Teemo said:

Homework Statement


In a particular solution, acetic acid is 11% ionized at 25 degrees Celsius. Calculate the pH of the solution and the mass of acetic acid dissolved to yield 1.00 L of solution.
Ka of CH3CO2H: 1.8x10^(-5)

Homework Equations


Ka=([CH3CO2-][H3O+])/[CH3CO2H]
pH=-log[H3O+]

The Attempt at a Solution


I'm not sure how exactly I should begin, as I do not know the concentration of the acetic acid, so I don't know how to solve for [H+]
Suppose you let x represent the number of moles of acetic acid dissolved in 1 liter of solution. After the acetic acid ionizes 11%, in terms of x, how many moles of acetic acid are left. How many moles of CH3CO2- are formed, and how many moles of H3O is formed? What are the new molar concentrations of these three species? Using Ka, what is the value of x?
 
Right, I have the answer key which writes that the formula is C=Ka/(degree dissociation)^2, however I'm not sure what this formula is called and why the degree of dissociation is squared.

Chestermiller said:
Suppose you let x represent the number of moles of acetic acid dissolved in 1 liter of solution. After the acetic acid ionizes 11%, in terms of x, how many moles of acetic acid are left. How many moles of CH3CO2- are formed, and how many moles of H3O is formed? What are the new molar concentrations of these three species? Using Ka, what is the value of x?

How exactly would this work? Would it be ([.11x][.11x])/[.89x], where x is original concentration of acetic acid?
 
Teemo said:
How exactly would this work? Would it be ([.11x][.11x])/[.89x], where x is original concentration of acetic acid?
Yes.
 
Teemo said:
Right, I have the answer key which writes that the formula is C=Ka/(degree dissociation)^2, however I'm not sure what this formula is called and why the degree of dissociation is squared.

It is an approximated version of the http://en.wikipedia.org/wiki/Ostwald_dilution_law

You should be able to do derive it by yourself using hints that were already posted.
 
Thanks so much :D, I guess my answer key is wrong haha.
 
Your answer key is correct, have you checked the wikipedia page I linked to? It derives exactly the same formula, just uses it to calculate degree of dissociation, not C.
 

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