# How would you calculate the solubility of Mn(OH)2 in grams per liter when buffered at ph = 9.5?

Nov 5, 2015

$1.2 \cdot {10}^{- 3} \text{g/L}$

#### Explanation:

First thing first, you need to look up the solubility product constant, ${K}_{s p}$, for manganese(II) hydroxide, "Mn"("OH")_2. which you can find here

http://owl.cengage.com/departments/NYInstitTechBrownHolme2e/appendix/ksp.html

So, the ${K}_{s p}$ of manganese(II) hydroxide is equal to $4.6 \cdot {10}^{- 14}$. As you know, the insoluble manganese(II) hydroxide will not dissociate completely when placed in aqueous solution.

What will happen is an equilibrium rection will take place between the solid manganese(II) hydroxide and the dissolved ions

${\text{Mn"("OH")_text(2(s]) rightleftharpoons "Mn"_text((aq])^(2+) + color(red)(2)"OH}}_{\textrm{\left(a q\right]}}^{-}$

Now, you're trying to determine what the solubility of "Mn"("OH")_2 will be in basic solution. As you know, basic solutions are characterized by an excess of hydroxide ions, ${\text{OH}}^{-}$, when compared with pure water.

More specifically, a solution that has a pH equal to $9$ will contain

$p O H = 14 - p H = 14 - 9.5 = 4.5$

["OH"^(-)] = 10^-"pOH" = 10^(-4.5) = 3.16 * 10^(-5)"M"

You can set up an ICE table to help you find the molar solubility of manganese(II) hydroxide in this solution

${\text{Mn"("OH")_text(2(s]) " "rightleftharpoons" " "Mn"_text((aq])^(2+) " "+" " color(red)(2)"OH}}_{\textrm{\left(a q\right]}}^{-}$

color(purple)("I")" " " "-" " " " " " " " " " " " " " "0" " " " " " " " " "3.16 * 10^(-5)
color(purple)("C")" " " "-" " " " " " " " " " " "(+s)" " " " " " "(+color(red)(2)s)
color(purple)("E")" " " "-" " " " " " " " " " " " " "s" " " " " " "3.16 * 10^(-5) + color(red)(2)s

By definition, the solubility product constant, ${K}_{s p}$, will be equal to

${K}_{s p} = {\left[{\text{Mn"^(2+)] * ["OH}}^{-}\right]}^{\textcolor{red}{2}}$

${K}_{s p} = s \cdot {\left(3.16 \cdot {10}^{- 5} + 2 s\right)}^{2} = 4.6 \cdot {10}^{- 14}$

This is equivalent to

$4 {s}^{3} + 12.64 \cdot {10}^{- 5} {s}^{2} + 9.986 \cdot {10}^{- 10} s - 4.6 \cdot {10}^{- 14} = 0$

This cubic equation will produce three solutions, two negative and one positive. Since $s$ represents molar solubility, the only one that will have any chemical and physical significance will be the positive one

$s = 1.34 \cdot {10}^{- 5}$

SInce this represents molar solubility, you can say that

$s = 1.34 \cdot {10}^{- 5} \text{M}$

To get the solubility in grams per liter, use manganese(II) hydroxide's molar mass

1.34 * 10^(-5)color(red)(cancel(color(black)("moles")))/"L" * "88.953 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)(1.2 * 10^(-3)"g/L")