Chem MCQ

r7kytt

I put D, correct?

BD98E81A-DFFB-4355-A81E-3919E2034E54.jpg.jpeg

Reply 1

Pigster

Original post by r7kytt

I put D, correct?

BD98E81A-DFFB-4355-A81E-3919E2034E54.jpg.jpeg

What values do you have for [HA], [H+] and [A-] at equilibrium?

Reply 2

lordaxil

No - that's too high for a weak acid.
Check your definition of K_a and try again.

Reply 3

r7kytt

Original post by Pigster

What values do you have for [HA], [H+] and [A-] at equilibrium?

(0.0396)^2 / 0.040

Reply 4

Pigster

Original post by r7kytt

(0.0396)^2 / 0.040

If you had 1 dm3 of your acid, then n(HA) =[HA], hence...

n(HA)initial = 0.04

HA -> A- + H+

n(A-)equil = 0.0396
n(HA)equil = n(HA)initial = 0.04

Does that all seem sensible?
Also, is that 1% dissociation?

Reply 5

r7kytt

Original post by Pigster

Honestly I have no clue

Reply 6

Pigster

Original post by r7kytt

Honestly I have no clue

Try again.

If you start with 0.04 mol of HA and produce 0.0396 mol of A-, does that sound like 1% dissociation to you?

Just to check, do you know what dissociation is?

Reply 7

r7kytt

Original post by Pigster

Try again.

If you start with 0.04 mol of HA and produce 0.0396 mol of A-, does that sound like 1% dissociation to you?

Just to check, do you know what dissociation is?

Yeah, so acids can dissociate in aq conditions to release H+ ions, but isn’t 1% times by 0.99?