[tiny101]

For 100.0 mL of a buffer that is 0.50 M in HOCl and 0.40 M in NaOCl, what is the pH after 10.0 mL of 1.0 M NaOH is added.  Ka  for HOCl = 3.5E-8

First I calculated the initial pH of the buffer before the addition of NaOH. To do so, I used the Henderson-Hasselbach Equation:

pH = pKa + log [OCl-]/[HOCl] = 7.456 + log 0.4/0.5 = 7.359

The next step is to determine which species reacts with which other species after the addition of NaOH. However, I really can't figure it out. I know after adding NaOH (but before anything reacts), we have: HOCl, Na+, OCl- and OH-.

Any help would be much appreciated,
thanks in advance.

[Borek]

The next step is to determine which species reacts with which other species after the addition of NaOH. However, I really can't figure it out. I know after adding NaOH (but before anything reacts), we have: HOCl, Na+, OCl- and OH-.

I see an acid and a strong base.

[tiny101]

If I understand correctly, the next step is to calculate the number of moles of HOCl left after it first dissociates in the water. I did that and came to find there are 0.050 moles left because it didn't dissociate very much.

Next, I found the number of moles of NaOH added = 0.01 mol. The limiting reagent is NaOH, and thus there are 0.01 moles of HOCl left and 0 moles of NaOH.

I suppose now I should do another ICE table to determine the concentrations of HOCl , H+ and OCl- ? However, even after taking into consideration that the total volume has been changed, the finished ICE table leads to an impossible Henderson-Hasselbach equation which basically includes log(1/0).

Any ideas?
Thank you again.

[Borek]

No need for ICE. Assume there is no further dissociation taking place. From limiting reagent you know amount of acid left, now find the amount of conjugated base and put these into Henderson-Hasselbalch equation.

Note: if there is an excess of base, you may end with situation when there is no acid left. Then pH is given by this excess strong base.