How would you calculate the solubility of Mn(OH)2 in grams per liter when buffered at ph = 9.5?
1 Answer
Explanation:
First thing first, you need to look up the solubility product constant,
http://owl.cengage.com/departments/NYInstitTechBrownHolme2e/appendix/ksp.html
So, the
What will happen is an equilibrium rection will take place between the solid manganese(II) hydroxide and the dissolved ions
#"Mn"("OH")_text(2(s]) rightleftharpoons "Mn"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-)#
Now, you're trying to determine what the solubility of
More specifically, a solution that has a pH equal to
#pOH = 14 - pH = 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
#"Mn"("OH")_text(2(s]) " "rightleftharpoons" " "Mn"_text((aq])^(2+) " "+" " color(red)(2)"OH"_text((aq])^(-)#
By definition, the solubility product constant,
#K_(sp) = ["Mn"^(2+)] * ["OH"^(-)]^color(red)(2)#
#K_(sp) = s * (3.16 * 10^(-5) + 2s)^2 = 4.6 * 10^(-14)#
This is equivalent to
#4s^3 + 12.64 * 10^(-5)s^2 + 9.986 * 10^(-10)s - 4.6 * 10^(-14) = 0#
This cubic equation will produce three solutions, two negative and one positive. Since
#s = 1.34 * 10^(-5)#
SInce this represents molar solubility, you can say that
#s = 1.34 * 10^(-5)"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")#