Percentage Dissasociation?

[oneoneoneone]

What did you guys put?

[daniel034]

9.93 x 10^-13 so i'm guessing i'm completely wrong

[baddin]

Yeah I got ~100% and i was like WTFFFFF is this shit  . So i rubbed it out and left that question blank.

[Jordzs]

9.93 x 10^-13 so i'm guessing i'm completely wrong

I think I got close to your answer than the 0% or 100%... hmm

[VCEstudentguy]

Well all the proper steps in doing that type of question suggest the ~100%, it's only because it's a really low pH to have for a solution of that type of acid, like a normal pH for that type of acid would be around 6..

[thushan]

What was the question?

[kimk2kr]

i got 3.34x10^-4%.... :/

[asdfgh]

It's approximately 99.6%. It makes sense though, cause even though it is a weak acid, the pH was relatively high for a weak acid (4.76 instead of the usual 3~4). This means there must have been a larger percentage ionisation to form extra sorbate ionsations to ensure the equillibrium is balanced. The extra pH less means at least a 10 fold less of hydrogen concentration, so a high ionisation is likely.

Also keep in mind that the question did say something about adding sorbate ions - so this isnt the natural ionisation percentage in just pure water either.

[meemz]

i got 99.6%, so did a few others at my school

[malekv]

9.93 x 10^-13 so i'm guessing i'm completely wrong

I got that too...hmmm

[thushan]

i got 50%, pH = 4.76, hence [acid] = [conjugate base]

Ka = [H+][sorbate]/[acid] = 1.73 x 10^-5

We know that pH = 4.76, hence [H+] = 1.73 x 10^-5

Hence [sorbate]/[acid] = 0.99(55)

This means that there is an equal amount of sorbate and sorbic acid in the solution, hence the percentage dissociation is 50% (to 2 sf).

[illuminati]

have you got the exam/question thush?

[johnfk]

Thushan: how did you get 50%? I had exact same working out and got 99.6%...

[Hazdog94]

Initially I had 99.6 or whatever but I knew this wasn't right because it was a weak acid, I realised that the concentration I had found for the acid was the one at equilibrium and I needed the initial concentration so I added the concentration of the h+ to the acid because it was 1:1 mole ratio which ended up giving me 49.something, not sure if this is right though

[asdfgh]

i got 50%, pH = 4.76, hence [acid] = [conjugate base]

Ka = [H+][sorbate]/[acid] = 1.73 x 10^-5

We know that pH = 4.76, hence [H+] = 1.73 x 10^-5

Hence [sorbate]/[acid] = 0.99(55)

This means that there is an equal amount of sorbate and sorbic acid in the solution, hence the percentage dissociation is 50% (to 2 sf).

It's correct working out, but to find % ionisation the formula was (sorbate)/(acid) x 100....so approximiately 99.6%