Question

已知HF,CaCl2的浓度,判断是否有CaF2沉淀

Answer 1

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