Question #9d048

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Question #9d048

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Answer 1 · 2017-04-22T18:27:08

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

Explanation:

If we are mixing equal volumes of $KF$ $"KF"$ and of $Ca {({NO}_{3})}_{2}$ $"Ca"("NO"_3)_2$ , the concentrations of each after mixing will be half the concentrations before mixing.

Thus, after mixing, $[{Ca}^{2+}] = 2 \times 10^{-4} l mol/L$ $["Ca"^"2+"] = 2 × 10^"-4" color(white)(l)"mol/L"$ .

To solve this problem, we must use the solubility product constant expression.

${CaF}_{2} ⇌ {Ca}^{2+} + {2F}^{-}$ $"CaF"_2 ⇌ "Ca"^"2+" + "2F"^"-"$

$K_{sp} = [{Ca}^{2+}] {[F^{-}]}^{2} = 3.9 \times 10^{-11}$ $K_text(sp) = ["Ca"^"2+"]["F"^"-"]^2 = 3.9 × 10^"-11"$

Let $x = [F^{-}]$ $x = ["F"^"-"]$

Then,

$K_{sp} = (2 \times 10^{-4}) x^{2} = 3.9 \times 10^{-11}$ $K_text(sp) = (2 × 10^"-4")x^2 = 3.9 × 10^"-11"$

$x^{2} = \frac{3.9 \times 10^{-11}}{2 \times 10^{-4}} = 2.0 \times 10^{-7}$ $x^2 = (3.9 × 10^"-11")/(2 × 10^"-4") = 2.0 × 10^"-7"$

$x = 4.4 \times 10^{-4}$ $x = 4.4 × 10^"-4"$

∴ $[KF] = [F^{-}] = x l mol/L = 4.4 \times 10^{-4} l mol/L$ $["KF"] = ["F"^"-"] = xcolor(white)(l) "mol/L" = 4.4 × 10^"-4"color(white)(l) "mol/L"$

The concentration of $KF$ $"KF"$ was halved on mixing, so the original concentration of $KF$ $"KF"$ before mixing was

$2 \times 4.4 \times 10^{-4} l mol/L = 9 \times 10^{-4} l mol/L$ $2 × 4.4 × 10^"-4"color(white)(l) "mol/L" = 9 × 10^"-4" color(white)(l)"mol/L"$

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Question #9d048

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