# What concentration of "Fe"^(3+) is required to begin precipitation of "Fe"("OH")_3 if the "pH" of the solution is 3.48? K_(sp) = 2.79 * 10^(-39)

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#### Explanation:

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Nov 20, 2017

$\left[{\text{Fe}}^{3 +}\right] = 1.0 \cdot {10}^{- 9}$ $\text{M}$

#### Explanation:

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.

${\text{Fe"("OH")_ (3(s)) rightleftharpoons "Fe"_ ((aq))^(3+) + 3"OH}}_{\left(a q\right)}^{-}$

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

${K}_{s p} = {\left[{\text{Fe"^(3+)] * ["OH}}^{-}\right]}^{3}$

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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{pOH" = 14 - "pH}}}}$

Since you know that

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

you can say that

$\left[\text{OH"^(-)] = 10^(-"pOH}\right)$

which is equivalent to

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

Plug in the value you have for the $\text{pH}$ of the solution to get

$\left[{\text{OH}}^{-}\right] = {10}^{- \left(14 - 3.48\right)}$

$\left[{\text{OH}}^{-}\right] = {10}^{- 10.52}$

Next, rearrange equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(\text{*}\right)}$ 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

$\left[{\text{Fe}}^{3 +}\right] \ge \frac{2.79 \cdot {10}^{- 39}}{{10}^{- 10.52}} ^ 3$

$\left[{\text{Fe}}^{3 +}\right] \ge \frac{2.79 \cdot {10}^{- 39}}{{\left({10}^{- 10}\right)}^{3} \cdot {\left({10}^{- 0.52}\right)}^{3}}$

$\left[{\text{Fe}}^{3 +}\right] \ge 2.79 \cdot {10}^{1.56} \cdot {10}^{- 39} \cdot {10}^{30}$

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 $\text{pH}$ of the solution.

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