# Question 9d048

Apr 22, 2017

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

#### Explanation:

If we are mixing equal volumes of $\text{KF}$ 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.

$\text{CaF"_2 ⇌ "Ca"^"2+" + "2F"^"-}$

K_text(sp) = ["Ca"^"2+"]["F"^"-"]^2 = 3.9 × 10^"-11"

Let $x = \left[\text{F"^"-}\right]$

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 $\text{KF}$ was halved on mixing, so the original concentration of $\text{KF}$ before mixing was

2 × 4.4 × 10^"-4"color(white)(l) "mol/L" = 9 × 10^"-4" color(white)(l)"mol/L"#