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 NaOHHence 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