Question #7fcb7

1 Answer
Jul 6, 2017

Answer:

#2.30 * 10^(-6)# #"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"# 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")_ (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# to be the molar solubility of the salt, you can say that, at equilibrium, you will have

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

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

Now, the solubility product constant, #K_(sp)#, is equal to

#K_(sp) = ["Fe"^(2+)] * ["OH"^(-)]^color(red)(2)#

In your case, this will be equal to

#4.87 * 10^(-17) = s * (color(red)(2)s)^color(red)(2)#

#4.87 * 10^(-17) = 4s^3#

Rearrange to solve for #s#

#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 * 10^(-6)# moles of dissociated salt for every #"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.