Jump to content

# Chemistry Paper 1 Question Help (buffers)

## Recommended Posts

Hey guys,

I was doing this question from the November 2014 Paper 1 but for some reason, my answers don't match any of the choices and I had to use a calculator for square roots even though we can't use them on the exam, so I must have messed up somewhere.

What I did is that the initial concentration of propanoic acid is 1.0M but it is diluted to double the volume so the final concentration is 0.5M. Now,

Ka = [H+][A-]/[HA]

[H+]^2 = Ka x [HA]

[H+] = square root(Ka x [HA])

= square root(1.30 x 10^-5 x 0.5)

= square root(0.65 x 10^-5)

I couldn't proceed from here. 0.65 x 10^-5 is an option but that's [H+]^2.

I guess you could use the henderson hasselbalch equation but that would give me pH and I'd have to calculate logs which we can't do.

So how do we do this by hand, and what did I go wrong in?

Thanks

#### Share this post

##### Share on other sites

Hey guys,

I was doing this question from the November 2014 Paper 1 but for some reason, my answers don't match any of the choices and I had to use a calculator for square roots even though we can't use them on the exam, so I must have messed up somewhere.

What I did is that the initial concentration of propanoic acid is 1.0M but it is diluted to double the volume so the final concentration is 0.5M. Now,

Ka = [H+][A-]/[HA]

[H+]^2 = Ka x [HA]

[H+] = square root(Ka x [HA])

= square root(1.30 x 10^-5 x 0.5)

= square root(0.65 x 10^-5)

I couldn't proceed from here. 0.65 x 10^-5 is an option but that's [H+]^2.

I guess you could use the henderson hasselbalch equation but that would give me pH and I'd have to calculate logs which we can't do.

So how do we do this by hand, and what did I go wrong in?

Thanks

Since the acid is diluted by a factor of 2, the concentration of the acid must decrease by half. So like you said, [HA] = 1M / 2 = 0.5M. Similarly, the concentration of propanoate ions in the salt must also decrease by a half, i.e. [A-] = 0.5M / 2 = 0.25M.

It's obvious that Ka = [H+][A-] / [HA] (note that this is just another way of writing the Henderson-Hasselbalch equation). Rearranging this equation gives: [H+] = Ka * [HA] / [A-]. By plugging in the values, we'll get [H+] = Ka * 0.5 / 0.25 = 2 * Ka = 2.6*10^(-5)M. So the answer is A; and the markscheme says that as well.

I noticed that this is a bit inaccurate because we would expect the Le Chartelier principle to shift the equilibrium towards the production of C2H5COOH due to the increase in C2H5COOâ€“ when the 2 solutions are mixed together. However, by searching a bit on google, I found out this is completely OK because it is exactly the limitation of using the Henderson-Hasselbalch equation. The equation can only be used as an estimate when the amount of acid in ionized form is negligible compared to the amount of acid in non-ionized form.

I'm not too sure of my workings though, as I've never something like this before. Well, but at least I tried. Hope somebody can point out whether it's correct or not. Btw, good luck with the exam

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account

## Sign in

Already have an account? Sign in here.

Sign In Now

×

## Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.