Chemical Forums

Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: claire18 on March 22, 2010, 04:43:23 PM

Title: PH of a buffer made by mixing two solutions
Post by: claire18 on March 22, 2010, 04:43:23 PM

Hi, could anyone help with this. I know its properly simple but I cant get my head round it. Thanks

Find the PH of a buffer made by mixing 100cm3 of 0.100 molar benzoic acid (PKa = 4.202) and 200cm3 of 0.050 molar sodium benzoate. 

Title: Re: PH of a buffer made by mixing two solutions
Post by: Borek on March 22, 2010, 04:47:15 PM

What equation is used for buffer pH calculation?

Title: Re: PH of a buffer made by mixing two solutions
Post by: claire18 on March 23, 2010, 08:45:28 PM

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

or

PH = PKa + log ([A-]/[HA])

Title: Re: PH of a buffer made by mixing two solutions
Post by: sjb on March 24, 2010, 03:00:09 AM

So what, for instance are [A] and [HA] here?

Title: Re: PH of a buffer made by mixing two solutions
Post by: claire18 on March 24, 2010, 09:02:00 AM

I think 0.100 and 0.050 but i really dont know to be honest  :-[

Title: Re: PH of a buffer made by mixing two solutions
Post by: Borek on March 24, 2010, 09:37:50 AM

What is A^-, what is HA, how are they related to the substances given in question?

Title: Re: PH of a buffer made by mixing two solutions
Post by: claire18 on March 24, 2010, 12:32:35 PM

[A−] is the concentration of the conjugate base and [HA] is the concentration of the acid.

What i don't get is which is which, sorry

Title: Re: PH of a buffer made by mixing two solutions
Post by: Borek on March 24, 2010, 03:05:18 PM

Try to apply definitions.

Note, that even looking at symbols (HA & A^-) should give you a strong hint...

Title: Re: PH of a buffer made by mixing two solutions
Post by: claire18 on March 24, 2010, 08:23:15 PM

A- is 0.050 and HA is 0.100

i think!

Title: Re: PH of a buffer made by mixing two solutions
Post by: Matias Ekstrand on March 25, 2010, 01:17:04 AM

You're right about which chemical is the acid and which is the conjugate base, but make sure to take into account that you mix two solutions and change the volume (which changes the concentrations).

Title: Re: PH of a buffer made by mixing two solutions
Post by: claire18 on March 25, 2010, 02:44:39 PM

So [HA] is 10 and [-A] is 10, is this right and what i do next?

Sorry about this

Title: Re: PH of a buffer made by mixing two solutions
Post by: sjb on March 25, 2010, 04:06:17 PM

No, [HA] was 0.1 molar (the benzoic acid), making 0.01 moles in total. What is the total volume of solution now?

Title: Re: PH of a buffer made by mixing two solutions
Post by: Borek on March 25, 2010, 05:58:06 PM

Quote from: claire18 on March 25, 2010, 02:44:39 PM

So [HA] is 10 and [-A] is 10, is this right and what i do next?

You are still just guessing till someone tells you you hit the target. You won't learn anything this way.

What are definitions of acid and conjugate base?

Title: Re: PH of a buffer made by mixing two solutions
Post by: claire18 on April 21, 2010, 01:09:50 PM

PH = PKa + log [acid/salt]
PH = 4.202 + log [0.100/0.050]
PH = 4.50

is this right?

Title: Re: PH of a buffer made by mixing two solutions
Post by: Borek on April 21, 2010, 01:50:15 PM

No.

Title: Re: PH of a buffer made by mixing two solutions
Post by: claire18 on April 21, 2010, 01:52:57 PM

What can i do to make it right?

Title: Re: PH of a buffer made by mixing two solutions
Post by: Matias Ekstrand on April 21, 2010, 02:58:46 PM

Your equation is wrong:
http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation

and this:

Quote from: Matias Ekstrand on March 25, 2010, 01:17:04 AM

You're right about which chemical is the acid and which is the conjugate base, but make sure to take into account that you mix two solutions and change the volume (which changes the concentrations).

Title: Re: PH of a buffer made by mixing two solutions
Post by: claire18 on April 21, 2010, 03:46:41 PM

thanks, is my answer way off then?

Title: Re: PH of a buffer made by mixing two solutions
Post by: claire18 on April 21, 2010, 06:53:57 PM

0.050 x 200 = 10
0.100 x 100 = 10

4.202 + log (10/10)
PH= 4.202

Is this one right?

Title: Re: PH of a buffer made by mixing two solutions
Post by: Matias Ekstrand on April 22, 2010, 12:23:57 AM

Not really.

Do you know how to make calculations on a dilution (http://www.friendlyscience.com/index.php?option=com_content&view=article&id=43&Itemid=55)?

Title: Re: PH of a buffer made by mixing two solutions
Post by: Borek on April 22, 2010, 03:00:36 AM

The real question is - if it is right, do you understand why? I doubt.