Percentage Dissasociation?

[jadams]

I got a similar answer to Thushan...I think 49.9%, nts about where I stopped for sigfigs...

[nancybaby]

99.6%

The question with COCL2 was hard...:S what did u guys get?

[diam0nds]

I got 99.6 but I didn't even understand the question lol I thought it seemed wrong when I did it, everyone else from my school who I talked to got realllllly small numbers

[Fantasia94]

For the 'choose your chemical questions'(btw I did ammonia) could you say that the potential environmental damage that is caused from the production of ammonia is the production of sulfur oxides. could you also say(the next question that followed this one) that one could prevent such environmental damage by reacting the sulfur oxides to form other harmless sulfur compounds. 

[ZanyCat]

99.6%

The question with COCL2 was hard...:S what did u guys get?

Yeah I couldn't get any of the bits below the question where you had to explain why it stayed at 1.0 mol.

If I wrote out the K formula, put 0.333 as [COCl2] and then multiplied by the K value to get [CO][Cl2] = *a number*, would that get me 1 or 2 of the 4 marks? I didn't realise the concentration of both were the same.

[charmanderp]

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).

A bit of logic goes a long way! How many marks would we lose for putting down 99.6%?

[Tonychet2]

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).

A bit of logic goes a long way! How many marks would we lose for putting down 99.6%?

2, u get 1 mark for working out H+ and the other mark for using his method and the last mark for realising that its 50%

[woliaj]

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%

Right but the (acid) is not the concentration of acid at equilibrium which is what you used, its the concentration of acid before the reaction commences

[charmanderp]

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%

Right but the (acid) is not the concentration of acid at equilibrium which is what you used, its the concentration of acid before the reaction commences

This exactly. When you solve for [sorbic acid] using Ka you find the equilibrium concentration, rather than the concentration of acid dissolved in solution which is what is needed.

[Somye]

ahahahah 50% occurred to me during the exam, and wouldve put it down if it asked ionisation, but I thought dissociation simply referred to the equilibrium conc. / wasn't sure if my logic was flawed....

well, serves me right for over thinking it...

[jadams]

This exactly. When you solve for [sorbic acid] using Ka you find the equilibrium concentration, rather than the concentration of acid dissolved in solution which is what is needed.

I ended up solving for that [sorbic acid] equilibrium concentration, then used percentage dissociation as = [H+] given from ph /  (  [sorbic acid] at equilibrium + [H+] ) because that total denominator would be the initial concentration of sorbic acid. My final answer was 49.9%....would that end up with full marks for the question?

[charmanderp]

This exactly. When you solve for [sorbic acid] using Ka you find the equilibrium concentration, rather than the concentration of acid dissolved in solution which is what is needed.

I ended up solving for that [sorbic acid] equilibrium concentration, then used percentage dissociation as = [H+] given from ph /  (  [sorbic acid] at equilibrium + [H+] ) because that total denominator would be the initial concentration of sorbic acid. My final answer was 49.9%....would that end up with full marks for the question?

Most likely, especially given what a low percentage of the state would have gotten that question right.

[kenhung123]

Hahaha. I remember my chem teacher used to say "its the concentration of the solution INITIALLY" and I was like "what's the difference?"

[Fantasia94]

for the question that asked you to write an equation for methanoic acid or i forgot what it was,  would you lose the whole mark if you forgot to include electrons in equations when everything else was balanced and states were included too?

[thushan]

for the question that asked you to write an equation for methanoic acid or i forgot what it was,  would you lose the whole mark if you forgot to include electrons in equations when everything else was balanced and states were included too?

The methanol redox qn? :/ yeah you'd lose the whole mark