已知298K时某一元弱酸的0,010mol.L-1水溶液的ph=4,求HA的Ka和a

HA = H + A
c-cα cα cα
Ka=cα*cα/(c-cα)
c=0.01,P[H]=-lgcα=4,代入上式,离解度α=1%,电离常数Ka=1*10^-6
稀释一倍后,c’=0.005,Ka仅随温度变化,故不变,仍代入上式,得α’=1.41%
[H]’=c'α’=0.005*1.41%=7*10^-5,稀释后PH=5-lg7=5-0.845=4.15
混合后c'=0.005,溶质为NaA.其水解常数Kb满足Ka*Kb=Kw=10^-14,故Kb=10^-8
A + H2O= HA + OH
水解方程同样满足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