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8>Poh=3.55
So,[OH-]=2.82*10^-4.
So,c*a=2.82*10^-4. where c is conc.=0.02 and a is degree of diss.
So,a=0.0141.
But,ca²=Kb ,where Kb is dissociation constant of BOH.
Thus,Kb=3.97*10^-6.
Now,during expt.,Meq of acid=1
Meq of base=2.
Thus,after neutralization,Meq. of salt formed=1 and meq of base left=1
Now,using Poh=PKb+log(meq of salt/meq of base),
you get Poh=5.401.
Thus Ph=14-5.401=8.598.

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