Calculate the solubility of Mn(OH)2 in grams per liter when buffered at pH = 11.6.?

Refer to: https://www.chm.uri.edu/weuler/chm112/refmater/KspTable.html
Ksp for Mn(OH)₂ = 1.9 × 10⁻¹³
(Ksp values from different sources may be slightly different, and this may lead to slightly different answers.)

pOH at equilibrium = pKw - pH = 14.0 - 11.6 = 2.4
[OH⁻] at equilibrium = 10⁻²˙⁴ mol/L = 3.98 × 10⁻³ mol/M

_______________ Mn(OH)₂(s) _ ⇌ _ Mn²⁺(aq) _ + _ 2OH⁻(aq) ____ Ksp = 1.9 × 10⁻¹³
At eqm (mol/L) : ___________________ s _______ 3.98 × 10⁻³
==
At eqm: Ksp = [Mn²⁺] [OH⁻]²
1.9 × 10⁻¹³ = s (3.98 × 10⁻³)²
s = (1.9 × 10⁻¹³) / (3.98 × 10⁻³)² = 1.2 × 10⁻⁸

Molar mass of Mn(OH)₂ = (54.9 + 16.0×2 + 1.0×2) g/mol = 88.9 g/mol
Solubility of Mn(OH)₂ = (1.2 × 10⁻⁸ mol/L) × (88.9 g/mol) = 1.07 × 10⁻⁶ g/L