Question

The solubility of Ba(OH)₂.8H₂O in water at 288 K is 5.6 g per 100 g of water. What is the molality of the hydroxide ions in the saturated solution of barium hydroxide at 288 K? (Atomic masses: Ba = 137, O = 16, H = 1)

Answer 1

Given: Solubility of Ba(OH)₂⋅8H₂O = 5.6 g/100 g water

Temperature = 288 K

Atomic masses: Ba = 137, O = 16, H = 1

We need to find the molality of hydroxide ions (OH⁻):

Molar mass of Ba(OH)₂⋅8H₂O = 137 + 32 + 2 + 144 = 315 g/mol

Moles of Ba(OH)₂⋅8H₂O in 100 g water = `5.6/315` = 0.01778 mol

Ionisation of Ba(OH)₂:

\[\ce{Ba(OH)2 -> Ba^2+ + 2OH-}\]

Each mole gives 2 moles of OH⁻:

Moles of OH⁻ = 0.01778 × 2 = 0.03556 mol

Mass of solvent (water) = 100 g = 0.1 kg

Molality of OH⁻ = `0.03556/0.1` = 0.356 mol/kg