Percent Dissociation of Acetic Acid

[Teemo]

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+]

 

[Borek]

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?

 

[Chestermiller]

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?

 

[Teemo]

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.

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?

 

[Chestermiller]

How exactly would this work? Would it be ([.11x][.11x])/[.89x], where x is original concentration of acetic acid?

Yes.

 

[Borek]

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.

 

[Teemo]

Thanks so much :D, I guess my answer key is wrong haha.

 

[Borek]

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.