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

Answer///an early form of the atomic theory; dalton atomic theory;