[PieceOfSpace]

How many moles of NaOBr have to be added to 125 mL of 0.160M HOBr to produce a buffer solution with [H+] = 5.00 x 10^-9M? There is almost no volume change. Ka= 2.3 x 10^-9 for the dissosciation of HOBr.

I tried using the Henderson Hasselbalch equation and I got over 9,000 moles of NaOBr so I am sure I did this incorrectly, but you do use that equation since it is a buffer problem, right?

[kingchemist]

Think so. But maths is a little rusty - see step with * below - this may not be correct.

I got 2.623 moles per litre which would be (2.623/1000) x 125 moles in 125 mL

If H+ = 5.0 x 10-9 M and Ka = 2.3 x 10-9, then pH = 8.301 and pKa = -8.638

pH = pKa + log([OBr-]/[HOBr])
8.301 = -8.638 + log([OBr-]/0.160)
-0.337 = log [OBr-] - log 0.160      * If you are dividing in logs, this is the same as subtracting???
log [OBr-] = -.337 + 0.796
log [OBr-] = 0.419   
[OBr-] = 2.623 mole/L
          0.328 mole in 125 mL   
hey

[PieceOfSpace]

Thanks, I forgot one could even just subtract the logs if it is easier. However the stoichiometry part (at least just the L and mL and gram part) I still need work on.
Thanks again.
hey