Please elaborate - do you mean number of moles of OBr- produced in neutralization equals number of moles of base added, or something else?

Yes that is what i meant.

Step by step of what I did

equiv

n(HOBr) = n(OH-) = n(OBr-)

=0.2 x 10/1000 = 0.1V

_{of NaOH}Hence V

_{of NaOH} = 20mL

Hence total Volume = 20 + 10 = 30mL

[OBr-] = 0.2 x 10/1000 x 1000/30 = 0.066666molL-1

K

_{b} = Kw/Ka = 4.90 x 10

^{-6}[OH-] = root of Kb x [OBr-] = 5.71 x 10-4

Hence pOH = 3.24

Therefore pH = 10.8

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The school told me to use the forumula of the markscheme but what it really doesnt make sense is that they used the fact that [OBr-] = [HOBr].

It doesn't mean concentration of HOBr is zero, as OBr- hydrolyzes. Concentration of HOBr is close to zero, but not equal to.

I do not understand this, as we are assuming that HOBr is fromed by ALL THE OBr- getting protonated by water, as [OBr-] and [HOBr] is the same