The Ksp of tin(II) hydroxide, Sn(OH)2, is 5.45 x 10^-27. Calculate the solubility of this compound in grams per liter

Answer 1

Sn(OH)₂(s) ==> Sn²⁺(aq) + 2OH-(aq)

Let x = [Sn²⁺] and then [OH-] = 2x

Ksp = 5.45x10^-27 = [x][2x]² = 4x³

x³ = 1.36x0^-27

x = 1.11x10^-9 M

Converting this to g/L using the molar mass of Sn(OH)₂ of 152.7 g / mole, we have...

1.11x10^-9 mol/L x 152.7 g / mol = 1.69x10^-7 g/L