millimoles of HBr=50*0.04=2 millimoles of KOH=25*.107=2.675 millimoles of acid remaining=0 millimoles of base remaining=0.675 Kb=[OH-][K+]/[KOH] 1=[OH-]*2/0.675 [oh-]=0.675/2 =0.3375 pH=14+log(0.3375) =13.53 Solution millimoles of HBr=50*0.04=2 millimoles of KOH=25*.107=2.675 millimoles of acid remaining=0 millimoles of base remaining=0.675 Kb=[OH-][K+]/[KOH] 1=[OH-]*2/0.675 [oh-]=0.675/2 =0.3375 pH=14+log(0.3375) =13.53.