Just one sec...
 You are Here: Home

Finding the concertration of a acid when given the Ka and Ph

Announcements Posted on
Take our short survey, ยฃ100 of Amazon vouchers to be won! 23-09-2016
1. I don't understand how u find the concertration using Ka. Also how to calculate buffer solutions.

Any examples would be helpful. Thank you
2. (Original post by Hazel99)
I don't understand how u find the concertration using Ka. Also how to calculate buffer solutions.

Any examples would be helpful. Thank you
Do you know the definitions of Ka and pH?
3. (Original post by MexicanKeith)
Do you know the definitions of Ka and pH?
Acid dissociation constant
Nurmerical scale of acidity and alkalinity
4. (Original post by Hazel99)
Acid dissociation constant
Nurmerical scale of acidity and alkalinity
That's true, but they both also have specific formula which define them

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

pH = -log[H+]

are those familiar?
5. (Original post by MexicanKeith)
That's true, but they both also have specific formula which define them

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

pH = -log[H+]

are those familiar?

Yeah, I have used them before but sometimes in the question it will give a concertration (e.g. 0.100 mol dm-3). So where does that concertration go in the formula ?. Also when's it saids a time 298k that doesn't need to be in the equation?
6. (Original post by Hazel99)
Yeah, I have used them before but sometimes in the question it will give a concertration (e.g. 0.100 mol dm-3). So where does that concertration go in the formula ?. Also when's it saids a time 298k that doesn't need to be in the equation?
Do you know what the different terms in the two equations mean, particularly the parts in square brackets?
7. (Original post by MexicanKeith)
Do you know what the different terms in the two equations mean, particularly the parts in square brackets?
Concertration of hydrogen ions?
8. (Original post by Hazel99)
Concertration of hydrogen ions?
That's along the right lines, [H+] is the concentration of protons
[A-] is concentration of conjugate base
[HA] is the concentration of the undissociated acid

I'll give you an example

Ka of ethanoic acid is 1.8x10-5
pH of a solution of ethanoic acid is 2.523
find the concentration of the acid solution?

So the concentration its asking you for is [HA]

First you can use pH to find [H+]
pH = -log[H+]
rearranges to [H+] = 10-pH

so here [H+] = 10-2.523 = 0.003 mol dm-3

Than you can use the definition of Ka to find what you want.
First you have to realise that all the protons (H+) and all the conjugate base molecules (A-) in this question come from dissociation of (HA) molecules so there must always be the same amount of H+ and A- ( ie [H+] = [A-] )

so
Ka = [H+][A-] / [HA]
becomes
Ka = [H+]2 / [HA]

putting in the numbers you know for Ka and for [H+]
1.8x10-5= (0.003)2 / [HA]

rearranging gives

[HA] = (0.003)2 / 1.8x10-5 = 0.5 mol dm-3

So the acid solution is 0.5 mol dm-3

Hope from that example you can see that if you know two out of three of the following bits of information:
[HA]
pH
Ka

than you can use the two equations too work out the third bit of information!

Sorry for the long post, hope it helped!
9. (Original post by MexicanKeith)
That's along the right lines, [H+] is the concentration of protons
[A-] is concentration of conjugate base
[HA] is the concentration of the undissociated acid

I'll give you an example

Ka of ethanoic acid is 1.8x10-5
pH of a solution of ethanoic acid is 2.523
find the concentration of the acid solution?

So the concentration its asking you for is [HA]

First you can use pH to find [H+]
pH = -log[H+]
rearranges to [H+] = 10-pH

so here [H+] = 10-2.523 = 0.003 mol dm-3

Than you can use the definition of Ka to find what you want.
First you have to realise that all the protons (H+) and all the conjugate base molecules (A-) in this question come from dissociation of (HA) molecules so there must always be the same amount of H+ and A- ( ie [H+] = [A-] )

so
Ka = [H+][A-] / [HA]
becomes
Ka = [H+]2 / [HA]

putting in the numbers you know for Ka and for [H+]
1.8x10-5= (0.003)2 / [HA]

rearranging gives

[HA] = (0.003)2 / 1.8x10-5 = 0.5 mol dm-3

So the acid solution is 0.5 mol dm-3

Hope from that example you can see that if you know two out of three of the following bits of information:
[HA]
pH
Ka

than you can use the two equations too work out the third bit of information!

Sorry for the long post, hope it helped!
Thank you it helped a lot
I just get a bit confused on the steps

Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: September 20, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

What uni's really like

How it is: now and next