# Volumes of a Buffer using Henderson-Hasselbalch?

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1. Sep 11, 2017

### Heidi_2

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
pH=pHa+log( [A-]/[HA])

3. The attempt at a solution
When I did it by myself, I let x+y=1 and x=Na2HPO4 and y=NaH2PO4, solved for x using the HH equation but got weird numbers that didn't make sense (like .984). Then I tried x=Na2hPO4 and .100L-x=NaH2PO4:

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 and once again got a very small number that doesn't make sense (.0058).

Now I'm lost and flustered. Thank you before hand for any help offered!

*after staring at this for a while, my final answer of .0058 is L, which is 5.8mL, therefore 5.8mL of Na2HPO4 and then 94.2mL of NaH2PO4 would be needed to achieve a pH of 6?

2. Sep 16, 2017

### epenguin

I'm not flustered. What's my secret? Getting the right answer doesn't really matter to me, so I can distance myself from it and relax.

If I had been more relaxed and distanced more of the time I might have studied a bit better in my time.

I suggest you complete your equation to include explicitly that the ratio of the phosphate Ions is the ratio of the two salts. That is as much chemistry as is involved, after that it's just arithmetic. Stuff that you could do 10 or 15 years ago

You haven't stated the ratio of salt moles that you calculated (direct from the formula I suggested you get). Find that. Take, say, 100 ml of one salt, then from the ratio how many ml of the other salt do you need to make the solution of pH 6.0? Mixed they will make a volume of more than 100 ml. By what factor do you need to reduce both volumes to make 100 ml?

Last edited: Sep 16, 2017