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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

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millimoles of HBr=500.04=2 millimoles of KOH=25.pdf

  • 1. 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