Question #e8dce

You know that iron(II) oxide-hydroxide is insoluble in water, which implies that a dissociation equilibrium is established in aqueous solution between the undissolved solid and the solvated ions.

#"Fe"("OH")_ (3(s)) rightleftharpoons "Fe"_ ((aq))^(3+) + 3"OH"_( (aq))^(-)#

By definition, the solubility product constant, #K_(sp#, for this solubility equilibrium looks like this

#K_(sp) = ["Fe"^(3+)] * ["OH"^(-)]^3#

Now, in order for precipitation to occur, you need to have

#K_(sp) <= ["Fe"^(3+)] * ["OH"^(-)]^3" "" "color(darkorange)(("*"))#

As you know, an aqueous solution at room temperature has

#color(blue)(ul(color(black)("pOH" = 14 - "pH")))#

Since you know that

#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))#

you can say that

#["OH"^(-)] = 10^(-"pOH")#

which is equivalent to

#["OH"^(-)] = 10^(-(14 - "pH"))#

Plug in the value you have for the #"pH"# of the solution to get

#["OH"^(-)] = 10^(-(14 - 3.48))#

#["OH"^(-)] = 10^(-10.52)#

Next, rearrange equation #color(darkorange)(("*"))# to solve for the minimum concentration of iron(III) cations that will precipitate the solid.

#["Fe"^(3+)] >= K_(sp)/(["OH"^(-)]^3)#

Plug in your values to find

#["Fe"^(3+)] >= (2.79 * 10^(-39))/(10^(-10.52))^3#

#["Fe"^(3+)] >= (2.79 * 10^(-39))/( (10^(-10))^3 * (10^(-0.52))^3)#

#["Fe"^(3+)] >= 2.79 * 10^(1.56) * 10^(-39) * 10^(30) #

Therefore, you can say that you need

#color(darkgreen)(ul(color(black)(["Fe"^(3+)] >= 1.0 * 10^(-7)color(white)(.)"M")))#

The answer is rounded to two sig figs, the number of decimal places you have for the #"pH"# of the solution.