Calculate the solubility of mn (oh) 2 in grams per liter when buffered at ph=7.0. assume that buffer capacity is not exhausted

When PH + POH = 14

∴ POH = 14 - 7 = 7

when POH = - ㏒[OH-]

7 = - ㏒ [OH-]

∴[OH-] = 10^-7

by using ICE table:

Mn (OH) 2 (s) ⇄ Mn2 + (aq) + 2OH - (aq)

initial 0 10^-7

change + X + 2X

Equ X (10^-7 + 2X)

when Ksp = [Mn2+][OH-]^2

when Ksp of Mn (OH) 2 = 4.6 x 10^-14

by substitution:

4.6 x 10^-14 = X * (10^-7+2X) ^2 by solving this equation for X

∴ X = 2.3 x 10-5 M

∴ The solubility of Mn (OH) 2 in grams per liter (when the molar mass of Mn (OH) 2 = 88.953 g/mol

= 2.3 x10^-5 moles/L * 88.953 g/mol

= 0.002 g / L