Question

The solubility of osmium (IV) oxide is 1.78 grams per liter. What is its solubility product?

Answer 1

osmium (IV) oxide formula is OsO₂

OsO₂ --> Os⁴⁺ + 2O^2-

--------------- s -----   2s ------

K_sp = (s) (2s)²

1 mol OsO₂ gives 1 mol Os⁴⁺ ions and 2 mol O^2- ions

convert solubility 1.78 grams per liter into molar solubility mol/L

moles present in 1.78 g of OsO2 = 1.78g/222.229 g/mol = 0.008009 moles (Molar mass of OsO₂ = 222.229 g/mol)

molar solubility = s = 0.008009 mol/L

K_sp = (s) (2s)² = (0.008009 mol/L) ( 2 x 0.008009 mol/L)² = 0.0000020553 = 2.05 x 10^-6

solubility product (K_sp) = 2.05 x 10^-6

Thank You So Much! Please Rate this answer as you wish.