Ka(HF)=3.6X10^-4=[F-][H+]/[HF] 设[F-]为x
由于2溶液混合,则浓度变为原来一半,
则x^2/(5*10^-4-x)=3.6X10^-4
此时[F-]为6.4*10^-4
[F-]^2*[Ca]=4.096*10^-10大于Ksp(CaF2)=1.46X10^-10
综上,会沉淀Ka(HF)=3.6X10^-4=[F-][H+]/[HF] [F-]^2/(10^-3/2)=3.6X10^-4[F-]^2=1.8X10^-7[F-]^2*[Ca]=1.8X10^-10大于Ksp(CaF2)=1.46X10^-10会沉淀