If we are mixing equal volumes of #"KF"# and of #"Ca"("NO"_3)_2#, the concentrations of each after mixing will be half the concentrations before mixing.
Thus, after mixing, #["Ca"^"2+"] = 2 × 10^"-4" color(white)(l)"mol/L"#.
To solve this problem, we must use the solubility product constant expression.
#"CaF"_2 ⇌ "Ca"^"2+" + "2F"^"-"#
#K_text(sp) = ["Ca"^"2+"]["F"^"-"]^2 = 3.9 × 10^"-11"#
Let #x = ["F"^"-"]#
Then,
#K_text(sp) = (2 × 10^"-4")x^2 = 3.9 × 10^"-11"#
#x^2 = (3.9 × 10^"-11")/(2 × 10^"-4") = 2.0 × 10^"-7"#
#x = 4.4 × 10^"-4"#
∴ #["KF"] = ["F"^"-"] = xcolor(white)(l) "mol/L" = 4.4 × 10^"-4"color(white)(l) "mol/L"#
The concentration of #"KF"# was halved on mixing, so the original concentration of #"KF"# before mixing was
#2 × 4.4 × 10^"-4"color(white)(l) "mol/L" = 9 × 10^"-4" color(white)(l)"mol/L"#
