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Question #7fcb7

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Answer 1 · 2017-07-06T01:01:15

$2.30 \cdot 10^{- 6}$ $2.30 * 10^(-6)$ ${mol L}^{- 1}$ $"mol L"^(-1)$

Explanation:

The problem wants you to determine the molar solubility of iron(II) hydroxide, which essentially means that you must find the number of moles of iron(II) hydroxide that will dissociate per $1 L$ $"1 L"$ of solution to produce iron(II) cations and hydroxide anions.

Iron(II) hydroxide is considered insoluble in water, which implies that the position of its dissociation equilibrium lies far to the left.

$Fe {(OH)}_{2 (s)} ⇌ {Fe}_{(a q)}^{2 +} + 2 {OH}_{(a q)}^{-}$ $"Fe"("OH")_ (color(red)(2)(s)) rightleftharpoons "Fe"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)$

In other words, only a very, very small number of moles of iron(II) hydroxide will actually dissociate when dissolved in water.

If you take $s$ $s$ to be the molar solubility of the salt, you can say that, at equilibrium, you will have

$[{Fe}^{2 +}] = s \to$ $["Fe"^(2+)] = s ->$ every mole of iron(II) hydroxide that dissociates produces $1$ $1$ mole of iron(II) cations in solution

$[{OH}^{-}] = 2 \cdot s \to$ $["OH"^(-)] = color(red)(2) * s ->$ every mole of iron(II) hydroxide that dissociates produces $2$ $color(red)(2)$ moles of hydroxide anions in solution

Now, the solubility product constant, $K_{s p}$ $K_(sp)$ , is equal to

$K_{s p} = [{Fe}^{2 +}] \cdot {[{OH}^{-}]}^{2}$ $K_(sp) = ["Fe"^(2+)] * ["OH"^(-)]^color(red)(2)$

In your case, this will be equal to

$4.87 \cdot 10^{- 17} = s \cdot {(2 s)}^{2}$ $4.87 * 10^(-17) = s * (color(red)(2)s)^color(red)(2)$

$4.87 \cdot 10^{- 17} = 4 s^{3}$ $4.87 * 10^(-17) = 4s^3$

Rearrange to solve for $s$ $s$

$s = \sqrt[3]{\frac{4.87 \cdot 10^{- 17}}{4}} = 2.30 \cdot 10^{- 6}$ $s = root(3)( (4.87 * 10^(-17))/4) = 2.30 * 10^(-6)$

This means that a saturated solution of iron(II) hydroxide will contain $2.30 \cdot 10^{- 6}$ $2.30 * 10^(-6)$ moles of dissociated salt for every $1 L$ $"1 L"$ of solution, presumably at room temperature.

In other words, the molar solubility of the salt is equal to

$color(darkgreen)(ul(color(black)("molar solubility" = 2.30 * 10^(-6)color(white)(.)"mol L"^(-1))))$

The answer is rounded to three sig figs.

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Question #7fcb7

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