设氟化钙的溶解度为zCaF2(s) ⇋ Ca2+ + 2F-z 2z则溶液中[F-]=2z+0.010Ksp=[Ca2+][F-]^2=z(2z+0.010)^2=3.4*10^-11解得 z=3.4*10^-7所以CaF2在0.010mol/L的NaF溶液中溶解度为3.4*10^-7mol/L.
