How would you calculate the solubility of $L a F_{3}$ $LaF_3$ in grams per liter in the following solutions? a) 0.049 M $K F$ $KF$ solution b) 0.070 M $L a C l_{3}$ $LaCl_3$ solution

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How would you calculate the solubility of $L a F_{3}$ $LaF_3$ in grams per liter in the following solutions? a) 0.049 M $K F$ $KF$ solution b) 0.070 M $L a C l_{3}$ $LaCl_3$ solution

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Answer 1 · 2016-06-27T12:02:19

Here's what I got.

Explanation:

I'll show you how to solve part (a) and leave part (b) to you as practice.

Lanthanum trifluoride, ${LaF}_{3}$ $"LaF"_3$ , is considered Insoluble in water, so right from the start you know that when the salt is dissolved in water, an equilibrium is established between the undissolved salt and the dissolved ions.

${LaF}_{3 (s)} ⇌ {La}_{(a q)}^{3 +} + 3 F_{(a q)}^{-}$ $"LaF"_ ( 3(s)) rightleftharpoons "La"_ ((aq))^(3+) + color(red)(3)"F"_ ((aq))^(-)$

The position of the equilibrium is given to you by the value of the solubility product constant, $K_{s p}$ $K_(sp)$ .

More specifically, the smaller the $K_{s p}$ $K_(sp)$ , the more the equilibrium will lie to the left, i.e. the less of the salt actually dissolves in solution.

Now, you want to know how much lanthanum trifluoride can be dissolved in a solution that is $0.049 M$ $"0.049 M"$ in potassium fluoride, $KF$ $"KF"$ , which, unlike lanthanum trifluoride, is soluble in water.

In other words, potassium fluoride dissociates completely in aqueous solution to form potassium cations, $K^{+}$ $"K"^(+)$ , and fluoride anions, $F^{-}$ $"F"^(-)$

${KF}_{(a q)} \to K_{(a q)}^{+} + F_{(a q)}^{-}$ $"KF" _ ((aq)) -> "K"_ ((aq))^(+) + "F"_ ((aq))^(-)$

Since every mole of potassium fluoride that dissociates form $1$ $1$ mole of fluoride anions, you will have

$[F^{-}] = 1 \times [KF] = 0.049 M$ $["F"^(-)] = 1 xx ["KF"] = "0.049 M"$

The presence of the fluoride anions will affect the position of the equilibrium in accordance with Le Chatelier's Principle.

More specifically, the equilibrium will shift to the left in an attempt to decrease the extra concentration of dissolved fluoride anions. As a result, the solubility of the salt will decrease $\to$ $->$ this is known as the common-ion effect.

So, if you take $s$ $s$ to be the molar solubility of lanthanum trifluoride in your solution, you can use an ICE table to write

${LaF}_{3 (s)} ⇌ {La}_{(a q)}^{3 +} + 3 F_{(a q)}^{-}$ $"LaF"_ ( 3(s)) " "rightleftharpoons" " "La"_ ((aq))^(3+) " "+" " color(red)(3)"F"_ ((aq))^(-)$

$I a a a - a a a a a a a a a a a a 0 a a a a a a a a a 0.049$ $color(purple)("I")color(white)(aaacolor(black)(-)aaaaaaaaaaaacolor(black)(0)aaaaaaaaacolor(black)(0.049)$
$C a a a - a a a a a a a a a a (+ s) a a a a a a (+ 3 s)$ $color(purple)("C")color(white)(aaacolor(black)(-)aaaaaaaaaacolor(black)((+s))aaaaaacolor(black)((+color(red)(3)s))$
$E a a a - a a a a a a a a a a a a s a a a a a a a 0.049 + 3 s$ $color(purple)("E")color(white)(aaacolor(black)(-)aaaaaaaaaaaacolor(black)(s)aaaaaaacolor(black)(0.049 + color(red)(3)s)$

By definition, the solubility product constant is equal to

$K_{s p} = [{La}^{3 +}] \cdot {[F^{-}]}^{3}$ $K_(sp) = ["La"^(3+)] * ["F"^(-)]^color(red)(3)$

In this case, you have

$K_{s p} = s \cdot {(0.049 + 3 s)}^{3} (*)$ $K_(sp) = s * (0.049 + color(red)(3)s)^color(red)(3)" " " "color(orange)("(*)")$

This will be equivalent to

$K_{s p} = s \cdot [{0.049}^{3} + {(3 s)}^{3} + 3 \cdot {0.049}^{2} \cdot (3 s) + 3 \cdot 0.049 \cdot {(3 s)}^{2}]$ $K_(sp) = s * [0.049^3 + (3s)^3 + 3 * 0.049^2 * (3s) + 3 * 0.049 * (3s)^2]$

$K_{s p} = s \cdot (0.00011765 + 27 s^{3} + 0.02161 s + 1.323 s^{2})$ $K_(sp) = s * (0.00011765 + 27s^3 + 0.02161s + 1.323s^2)$

$K_{s p} = 27 s^{4} + 1.323 s^{3} + 0.02161 s^{2} + 0.00011765 s$ $K_(sp) = 27s^4 + 1.323s^3 + 0.02161s^2 + 0.00011765s$

This can be rearranged to give

$27 s^{4} + 1.323 s^{3} + 0.02161 s^{2} + 0.00011765 s - K_{s p} = 0$ $27s^4 + 1.323s^3 + 0.02161s^2 + 0.00011765s - K_(sp) = 0$

Since the problem doesn't provide you with a value for the $K_{s p}$ $K_(sp)$ of the salt, I'll pick one

$K_{s p} = 2.0 \cdot 10^{- 19}$ $K_(sp) = 2.0 * 10^(-19)$

Now, this quartic equation will produce four solutions, one positive, one negative, and two complex.

https://www.easycalculation.com/algebra/quartic-equation.php

Since you're looking for $s$ $s$ to be molar solubility, pick the positive solution to get

$s = 1.7 \cdot 10^{- 15}$ $s = 1.7 * 10^(-15)$

Alternatively, you can use the fact that $K_{s p}$ $K_(sp)$ is very, very small when compared with $0.049$ $0.049$ to use the approximation

$0.049 + 3 s \approx 0.049$ $0.049 + 3s ~~ 0.049$

Equation $(*)$ $color(orange)("(*)")$ can thus be written as

$K_{s p} = s \cdot {(0.049)}^{3}$ $K_(sp) = s * (0.049)^3$

This will get you

$s = \frac{K_{s p}}{{0.049}^{3}} = \frac{2.0 \cdot 10^{- 19}}{{0.049}^{3}} = 1.7 \cdot 10^{- 15}$ $s = K_(sp)/0.049^3 = (2.0 * 10^(-19))/(0.049^3) = 1.7 * 10^(-15)$

As you can see, the approximation holds.

This means that in a solution that is $0.049 M$ $"0.049 M"$ in fluoride anions, you can only hope to dissolve $1.7 \cdot 10^{- 15}$ $1.7 * 10^(-15)$ moles of lanthanum trifluoride per liter of solution.

To find the solubility in grams per liter, use the salt's molar mass

$1.8 * 10^(-15) color(red)(cancel(color(black)("moles LaF"_3)))/"L" * "195.9 g"/(1color(red)(cancel(color(black)("mole LaF"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(3.5 * 10^(-13)"g L"^(-1))color(white)(a/a)|)))$

I'll leave the answer rounded to two sig figs.

By comparison, the solubility of lanthanum trifluoride in pure water is $1.8 * 10^(-3)"g L"^(-1)$ , which confirms that the presence of the fluoride anions significantly decreased the salt's solubility.

https://socratic.org/questions/what-s-the-solubility-in-grams-per-liter-of-laf3-in-pure-water

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How would you calculate the solubility of $L a F_{3}$ $LaF_3$ in grams per liter in the following solutions? a) 0.049 M $K F$ $KF$ solution b) 0.070 M $L a C l_{3}$ $LaCl_3$ solution

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How would you calculate the solubility of $L a F_{3}$ $LaF_3$ in grams per liter in the following solutions? a) 0.049 M $K F$ $KF$ solution b) 0.070 M $L a C l_{3}$ $LaCl_3$ solution