晨风123456
幼苗
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混合后是0.05mol/L的NaZ溶液
HZ = H+ + Z-,Ka=[Z-][H+]/[HZ]……①
H2O = H+ + OH-,Kw=[H+][OH-]……②
[H+] + [Na+] = [Z-] + [OH-](电荷守恒)……③
[Z-] + [HZ]=0.05mol/L=[Na+](物料守恒)……④
其中[H+]=10^-9mol/L,可得[OH-]=10^-5mol/L
由④:[Z-]=0.05-[HZ]
上面两式带入③:10^-9 + 0.05 = 0.05 - [HZ] + Kw/10^-9
化简得[HZ] = Kw/10^ - 9-10^-9 ≈ 10^-5mol/L
上式带入[Z-]:[Z-]=0.05 - 10^-5 ≈ 0.05mol/L
将[Z-]和[HZ]带入①:Ka= 10^-9×0.05/10^-5 = 5×10^-6
1年前
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