Chemistry
posted by Trixie on .
The pH of 0.50 M HBrO(aq) is 4.50. Calculate the change in pH when 6.50 g of sodium hypobromite is added to 110. mL of the solution. Ignore any change in volume. Ka of HBrO is 2E9
I found the amount of moles used for each (.0547 mol NaBrO, and .055 mol HBrO soln)
I'm not totally sure how to find the new pH. Do I set up an ICE chart? If I do...I'm not sure how to set up the expression for the addition of NaBrO to HBrO...

I am trying to use pH=pKa + log(acid/base) to see where that gets me...I'm attempting to find the [base] using the given pH and pKa of 8.69 (for HBrO)...then use the concentration to find the new pH...

You can use the HendersonHasselbalch equation to calculate pH or you can set up the Ka expression and substitute (HBrO) where needed and use (NaBrO) as (BrO^) in the Ka expression. Calculate H^+ and pH from that.

I am pretty stuck...not sure if I used the HendersonHasselbach the right way and I'm a little confused on how to set up the Ka expression...

Nevermind I got it...I just mixed up my variables a little. thanks!

First, a correction. The HH equation is
pH = pKa + log (base/acid); your fraction is invered incorrectly.
Here are both ways:
HBrO ==> H^+ + BrO^
Ka = 2 x 10^9 = (H^+)(BrO^)/(HBrO)
Plug in (BrO^) 6.50grams/molar mass NaBrO and that divided by 0.110 L. (HBrO) = 0.5 M and solve for H^+ then convert to pH.
The HH equation is easier.
pKa = 8.70 rounded a little but you need to confirm that.
pH = pKa + log (base/acid)
Plug in Ka, (base) = 6.5/molar mass NaBrO and divide by 0.110 L. For acid plug in 0.5 M. I get pH about 7.7 or so.