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 2E-9

Question

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...

Answer 1

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...

Answer 2

You can use the Henderson-Hasselbalch 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.

Answer 3

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

Answer 4

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

Answer 5

First, a correction. The H-H 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.