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"
