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# Need help with acid base problem?

What is the pOH of a 0.08 M solution of NaOBr at 25°C? (Ka for HOBr is 2.5 x 10−9.)

I try setup HOBr goes to H+ and OBr with OBr concentration at 0.08 M but I don't know the Molarity for H+ and HOBr. What do I do? I know I have to get Kb later from the Ka given. I feel like I am missing something that I learned.

So, HOBr <---> OH^- + Br^+

ICE table

............HOBr.....H^+.....OBr

Initial.... 0.08...... 0.......... 0

Change.. -x....... +x........ +x

Equil.. 0.08-x.... x........... x

pKa = [H^+[OBr^-]

.........----------------------

...........[HOBr]

pKa = [x][x]

.......-------------- = 2.5*10^-9

.......[0.08-x]

'-x in 0.08-x is negligible. Leave it out! Cross multiply

x^2 = 0.08*2.5*10^-9

x = 1.414 * 10^-5 which is [H^+]

pH = -log[1.414*10^-5] = 4.84955 M, approx 4.85 M

pOH = 14 - 4.85 = 9.15 M