Question #9d048

1 Answer
Apr 22, 2017

Answer:

The minimum concentration of #"KF"# to start precipitation is #9 × 10^"-4"color(white)(l) "mol/L"#.

Explanation:

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"#