Percentage Dissasociation?

[Tonychet2]

this question is the killer man i spent last 25 minutes trying to get it and got it wrong GG!

[t35t]

so is the percentage of disocciation 99.6%?  I don't like to go against what thushan says...

on that note- i think i may have overcomplicated the question for myself. In thushan's ATAR notes section there was an entire question about how ka x kb =kw so i spent like 10 minutes going from ka to kb.... eventually I got 99.6% for both methods however, when i realized that i could just use the ka and its related equation...
=/ i hope its 99% or at the very least, marks are awarded for working out...

[pi]

I tink this is what thushan is getting at: If [sorbate]/[acid] ~ 1, therefore there is approximately an equal concentration of sorbate and acid. Hence 50%.

[thushan]

@asdfgh

%(ionisation) = [sorbate](equilibrium)/[acid](initial), not [sorbate](equilibrium)/[acid](equilibrium)

[illuminati]

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%

If it specified that you use that formula its fine
but if it didn't specify, the formula only holds true because you assume its a weak acid
and you assume that the acid doesn't dissociate to a large extent
But in this case since it does....
then the percentage ionisation needs to be calculated thushan's way

[Suzanne]

Thats what you get for shaking the bloody table every 5 seconds tony 

[thushan]

LOL you go to school with tony?

[thea.]

How much will I get penalised by if I put wrong significant figures? :/ Put 99.55% (silly)
Will I get penalised one mark for the whole exam or one mark on the question?

[oneoneoneone]

I dont think they will penalise sig figs on this question because it involved pH and the rules for pH arent taught in schools often.

[thushan]

Guys - to replicate the scenario:

Suppose you dissolved some sorbic acid in water, you'll get a pH of say 2-3 (suppose its a really really small amount of sorbic acid, enough so it can itself dissolve in water). The percentage ionisation will be say 0-2% then.

Now, to increase the pH to 4.76, just stick on OH- ions. This will cause progressively more sorbic acid molecules to be converted to sorbate ions as per the equilibrium:

HSorb <--> Sorb- + H+ (yes I'm lazy)

where H+ is being used up by the OH- ions being stuck into the solution, thereby increasing the percentage ionisation. Turns out by the time the pH is 4.76, the percentage ionisation has increased to 50%.

[thushan]

I dont think they will penalise sig figs on this question because it involved pH and the rules for pH arent taught in schools often.

Concur, I doubt that this is the sig figs question (they only look at one question for sig figs) - unless one of your other questions has a ridiculous number of sig figs.

[t35t]

even though its 50%, what if enough people in the state put 99%?

[Suzanne]

haha yeah i do... he showed us a video of you playing piano before the exam

[pi]

even though its 50%, what if enough people in the state put 99%?

Then most of the state gets it wrong

[asdfgh]

@asdfgh

%(ionisation) = [sorbate](equilibrium)/[acid](initial), not [sorbate](equilibrium)/[acid](equilibrium)

Oh crap I think you're right..that's really tricky. Haha assumed the change in concentration of sorbic acid wouldnt be very high either.. dam ._.