HA = H + A
c-cα cα cα
Ka=cα*cα/(c-cα)
c=0.01,P[H]=-lgcα=4,代入上式,離解度α=1%,電離常數(shù)Ka=1*10^-6
稀釋一倍后,c’=0.005,Ka僅隨溫度變化,故不變,仍代入上式,得α’=1.41%
[H]’=c'α’=0.005*1.41%=7*10^-5,稀釋后PH=5-lg7=5-0.845=4.15
混合后c'=0.005,溶質(zhì)為NaA.其水解常數(shù)Kb滿足Ka*Kb=Kw=10^-14,故Kb=10^-8
A + H2O= HA + OH
水解方程同樣滿足Kb=c'α'*c'α'/(c'-c'α')
代入Kb,c'值后,得α'=0.141%,[OH]=c'α'=0.005*0.141%=7*10^-6
故PH=14-POH=14-6+lg7=8.85