[heidi_2]

Calculate the volume of 0.10M phosphate solution to mix to prepare 100mL of a buffer with pH 6.0 starting with 0.10M stock solutions of NaH₂PO₄ and Na₂HPO₄. The pKa for this reaction is 7.21.

Okay, so this is what I did: x=Na₂HPO₄ and .100L-x= NaH₂PO₄:

HH equation: pH=pKa+log([A-]/[HA]) so I let:

A-=.1x/.1L and HA=.1x/(.1-x), plugged that into the HH equation (.1L cancel) so
6=7.21+log([.1x]/[.1(.1-x)] and solved for x which was .0058. For the longest time it didn't make sense to me, but then maybe I thought it's in mL so it would equal 5.8mL so I would have 5.8mL of the HPO4 and 94.2mL of H2PO4.

[Borek]

Instead of guessing what you did afterwards define precisely what you mean by x when you first put that in the HH equation.

[heidi_2]

Instead of guessing what you did afterwards define precisely what you mean by x when you first put that in the HH equation.

I did define x.. x is the volume of Na2HPO4.

[Borek]

x=Na₂HPO₄

Strange way of writing "it is a volume". Besides, later you wrote

but then maybe I thought it's in mL

so apparently even if you assumed it is a volume, you were not sure about units you were using.

Each time you see such a situation (result looks wrong and counterintuitive) start with precisely and systematically checking all your assumptions and all units used in calculations. Actually fact that you had doubts about the number you got is quite good, it never hurts to be critical about your own work!

[heidi_2]

x=Na₂HPO₄

Strange way of writing "it is a volume". Besides, later you wrote

but then maybe I thought it's in mL

so apparently even if you assumed it is a volume, you were not sure about units you were using.

Each time you see such a situation (result looks wrong and counterintuitive) start with precisely and systematically checking all your assumptions and all units used in calculations. Actually fact that you had doubts about the number you got is quite good, it never hurts to be critical about your own work!

In my head I was thinking the units were mL which is why when I got the .0058 I initially thought I was wrong, but then I realized I was still working with L. I even had .0058L written down on my paper while I was doing it.

x=mL of Na2HPO4 would have been a better way to put it. So ultimately, my .0058L is converted into the amount I need in mL which is 5.8mL. In turn since I need 100mL, 100-5.8=94.2mL of NaH2PO4.

[Borek]

So ultimately, my .0058L is converted into the amount I need in mL which is 5.8mL. In turn since I need 100mL, 100-5.8=94.2mL of NaH2PO4.

And these numbers look about OK.

[heidi_2]

In order to make a phosphate buffer at pH 7.5, I used 67mL of Na₂HPO₄ and 33mL NaH₂PO₄. I took 40mL of that mixture which is now a 0.10M buffer.

I have to find the moles of conjugate base (Na₂HPO₄) that are present in the 40mL...

I thought all I had to do was 0.004L*0.10M but then I have another question that asks the same thing but at pH 6 so I feel like the pH/amount of solutions I used play a factor in the calculation but I'm not sure how to go about that.... Thank you!

*Okay, actually I used the HH equation, 7.5=7.2(this was given)+log[A/HA]. I let Na2HPO4=y and NaH2PO4=x. so y/x=10^(.3), so y=1.99x. Plugging this into x+y=.10M, x+1.99x=.10M, x=.0334. I solved for y which equals, .0666. Then I multiplied .0666 by the .004L to get my answer (2.66*10^-4)

[Arkcon]

Hope you don't mind, I merged these two posts because they get to the answer in the same way.  The work you've done before could help you solve this one.

[heidi_2]

Hope you don't mind, I merged these two posts because they get to the answer in the same way.  The work you've done before could help you solve this one.

Alright.. I totally forgot about this problem. The work I did for this problem and the work I did for the question prior are similar in a way, I can see that.