Well this has no relevance whatsoever to Allegiance or my modeling, but basicly this is what I havee been doing when not playing allegiance or working on my models...
Ksp: pH and Common Ion Effect
Just as a review since it is relevant to the example problem:
COMMON STRONG ACIDS AND BASES:
ACIDS: Strong: HCl, HBr, HI, HNO3, H2SO4*diprotic*, HClO4. Most other acids are weak.
BASES: Strong: LiOH, NaOH, KOH, *the following bases are all diprotic*; Ba(OH)2, Sr(OH), Ca(OH)2{^2}. Most other bases are weak.
Ex1: Manganese(II) hydroxide, Mn(OH)2, is sparingly soluble in water: Ksp of Mn(OH)= 4.6E-14 at 25°C. Calculate the solubility of Manganese(II) hydroxide at that temperature in...
a) pure water
b) a solution of pH 11.00
Solutions:
Mn(OH)2(s) ⇌ Mn{^2+} (aq) + OH{^1-} (aq)
Ksp = ([Mn{^2+}] [OH{^1-}]{^2})/1
a) ®ICE table
_____Mn(OH)2(s) ⇌ Mn{^2+} (aq) + OH{^1-} (aq)
I)_____-___________0__________0
C)_____-__________+x_________+2x
E)_____-__________+x_________+2x
4.6E-14 = (x)(2x){^2}
----> (x+0)(2x+0)(2x+0) ---->F.O.I.L.---->
(x)(2x){^2} = 4x{^3}
∛((4.6E-14)/4) = ∛((4x{^3})/4)
2.257...E-5 = x
Final solution Ex1 part a= 2.3E-5 M
b) First convert pH to pOH since we are dealing with the hydroxide ion:
pH= 11.00 ----> 14-11 ----> pOH= 3.00
pOH= 1.00E-3
®ICE table
_____Mn(OH)2(s) ⇌ Mn{^2+} (aq) + OH{^1-} (aq)
I)_____-___________0_________1.00E-3
C)_____-__________+x_________+2x
E)_____-__________+x________1.00E-3 + 2x
4.6E-14 = (x) (1.00E-3 + 2x){^2}
***note 2x because of estimation that 2x<<(1.00E-3) according to 5% rule***
(1.00E-3){^2} = 1.00E-6
4.6E-14 / 1.00E-6 = ((x)(1.00E-6) / 1.00E-6
x = 4.6E-8 M
Check for reasonable answer by substituting the approximate value into (x) (1.00E-3 + 2x){^2}. The of this equation equals our Ksp value, theirfore the approximation is legitimate.
The forum doesnt like the unicode double harpoon for some reason but w/e. Yeah, as you can tell I am pretty bored

, I think I'll go play allegiance for a little while...
edit...nm it does like it... w/e