Chemical Forums

Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: Starc on March 01, 2009, 12:46:44 AM

Title: Converting to Buffer
Post by: Starc on March 01, 2009, 12:46:44 AM

Hello,

If I have 450 mL of 0.0596 M NaH2PO4 solution, how many grams of K2HPO4 must be added to convert this solution into a pH 7.35 buffer?

I believe that buffer capacity increases as the concentration increases, would the common ion effect or Henderson-Hasselbalch equation be used?

Title: Re: Converting to Buffer
Post by: Arkcon on March 01, 2009, 07:19:33 AM

Since you want a quantitative result, that is, your answer has a number in it, you'll want the one of your two options that is a formula, and not the one that is just a definition/theory/description.  Can you tell them apart, and let us know, which is which?

I'm not trying to confuse you by being obscure, but this is how we help on this forum, we give hints, that help you learn, so you're better off for your next question.

Title: Re: Converting to Buffer
Post by: Starc on March 01, 2009, 11:31:20 AM

Yes, I realize that the common ion effect is a theory and the HH equation is a formula, I was just wonder if it applied to the situation. So are you saying that the HH equation is the right formula to use to get the quantitative result? Can it be used with a solid?

Title: Re: Converting to Buffer
Post by: Borek on March 01, 2009, 11:48:05 AM

Quote from: Starc on March 01, 2009, 11:31:20 AM

Can it be used with a solid?

Not directly, but you have volome, once you will know concentration you will be able to calculate number of moles and mass...

Title: Re: Converting to Buffer
Post by: Starc on March 01, 2009, 10:25:24 PM

That makes sense but what value should i be using for pKa? I cant find a match to this situation.

Title: Re: Converting to Buffer
Post by: AWK on March 02, 2009, 01:03:39 AM

Write down an expression for equilibrium and all should be clear

Title: Re: Converting to Buffer
Post by: Starc on March 02, 2009, 01:40:55 AM

I can't get the equation finished....

K2HPO4 + NaH2PO4 <-->

I am guessing that there is some form of phosphorous acid and its conj base but i can't figure it out...

Title: Re: Converting to Buffer
Post by: AWK on March 02, 2009, 02:16:33 AM

This is the equilibrium you need
H₂PO₄^- + H₂O = HPO₄^2- + H₃O⁺

Title: Re: Converting to Buffer
Post by: Starc on March 02, 2009, 08:50:55 PM

How does this sound?

7.198=pKa

7.35=7.198 + Log([HPO4]/[.0596])

[HPO4]=.084575 M

X mol/.450L = .084575 M

=.038058mol= 6.63g K2HPO4

Title: Re: Converting to Buffer
Post by: Borek on March 03, 2009, 03:05:33 AM

Looks OK.

Title: Re: Converting to Buffer
Post by: Starc on March 03, 2009, 03:54:26 AM

Thanks!