[DesertRose]

Hi,
When given that the acid dissociation constant of ethanoic acid at 25 degrees celcius is 1.7 * 10^-5, how do you calculate the equilibrium concentration of ethanoic acid in a solution that has a pH of 5?

I wrote the equation and found the concentration of Hydrogen ion using logs. i got [H+]= 10^-5
 but i do not know what to do from there. i'm really confused. help please?

Any assistance will be greatly appreciated. Thanks a lot!

[Corribus]

First step for just about everything is to write out a balanced equation. 

[DesertRose]

Hi, my equation is:
CH3COOH (reversible arrows)------> (CH3COO-) + (H+)

using [H+] = antilog(-pH)
i found [H+]= 10^5

And i'm stuck from here....

[Corribus]

So now what?  You are given another critical piece of information.  How can you use it to find the equilibrium concentration of ethanoic acid?

EDIT: And be careful with your signs.  Is [H⁺] really 10⁵?

[DesertRose]

My mistake, [H+] = 10^-5

i don't know if this is correct, but i still tried something. using the mole ratio from the equation , which is 1 : 1 : 1, i used initial concentration and final  concentration. the used initial conc. of both products as 0, but i don't know the initial conc. of the reactant. for final conc. since the products are 1:1 mole ratio, i put the conc. for each as 10^-5.

am i going wrong? i'm really confused

[Corribus]

Alright - the acid dissociation constant is just an equilibrium constant.  So: How do you relate an equilibrium constant to concentrations of products and reactants at equilibrium?

If you don't know, then you should read in your chemistry book about equilibrium constants.

[DesertRose]

the equilibrium constant, Ka instead of Kc in this case (correct me if i'm wrong)
Ka= conc. products/conc. reactants
 Ka = [H+][CH3COO-] /[CH3COOH]
     
substituting the values of Ka and H+ .
(1.7 x 10^-5)  = (10^-5) [CH3COO-] /[CH3COOH]

i don't know how to find the value of [CH3COO-] to make [CH3COOH] the only unknown.
if it is in the equation the mole ratio of H+ to CH3COO- is 1:1, then does that mean that the value of [CH3COO-] = [H+] = (10^-5) ?

[Corribus]

if it is in the equation the mole ratio of H+ to CH3COO- is 1:1, then does that mean that the value of [CH3COO-] = [H+] = (10^-5) ?

Yes.  This is why writing out a balanced equation is always an important first step.

[DesertRose]

yayy.....so i got my answer to be:
[ CH3COOH] = 5.882 x 10^-6

i have a question about this though, in my chemistry book, they tend to work out questions in equilibria with the use of a table, with mole ratio, initial conc and equilibrium conc. as headings on the side of each row, with the reaction equation at the top....this was getting me a bit confused...when will you use the table format then?

Thank you sooo much Corribus

[Corribus]

Honestly it would help me to see what the actual wording of your question is.  But you would use the approach you mention if the question was something like:

If you dissolve 1 gram of acetic acid in 1 L of water at 25 C, what will be the pH?  (Ka for acetic acid is whatever.)

You need to use a table here because you're not given equilibrium concentrations - you're given a starting amount BEFORE equilibrium is reached, and you need to see what the equilibrium concentration of [H] is to see what the pH is.  So you'd need to see how much of your starting material is lost (and products gained) once equilibrium is reached.

However if you're told what the pH is up front then equilibrium has already been reached and you can find equilibrium concentrations of the other species in solution by simple application of the equilibrium constant.  You're not given a starting amount so you don't have to worry about find how much is lost during equilibrium.  (Now, if they asked you what the amount of ethanoic acid was that they started with to get a pH of 5, then you'd have to back calculate it using the table approach you mentioned.)

Does that make sense?

[DesertRose]

 You're not given a starting amount so you don't have to worry about find how much is lost during equilibrium. 

So just making sure, when doing calculations of this sought, you're not really concerned with the initial conc. of the reactants, but the equilibrium/ final concentration values since those are used to calculate Kc values?

[Corribus]

Well essentially everything you're given in this problem (as far as I can tell) are equilibrium values, so since everything is already at equilibrium you don't have to worry about what you started with.  However if you're given a pre-equilibrium condition, or if you are asked to determine a pre-equilibrium condition, then you need to use the tabular approach that you've learned.

[DesertRose]

I understand now

Thank you sooooo much Corribus!!!