Question

The solubility of `ba(OH)_(2)8H_(2)O` in water at 298 K is 5.6 g per 100 g of water. What is the molality of the hydroxide ions in a saturatd solution of barium hydroxide at 298 K ? (Atomic mass of Ba =137, O=16)

Answer 1

`"Molality of solution(m)"=("Mass of solute"//"Molarmass of solute")/("Mass of solvent in kg")`
Mass of solute, `Ba(HO)_(2)8H_(2)O=5.6g`
Molar mass of solute, `Ba(OH)_(2)8H_(2)O=137+2(16+1)+8xx18=315 g mol^(-1)`
Mass of solvent (water)= 100g=0.1 kg
`Molality (m)=((5.6g)//(315gmol^(-1)))/((0.1kg))=0.18 molkg^(-1)=0.18m`
Calculation of molality of the OH ions
Barium hydroxide ionies inwater as follows:
`Ba(OH)_(2) overset((aq)) Leftrightarrow Ba^(2)+(aq)+2OH(aq)`
This shows that the molality of OH ions is twice the molality of `Ba(OH)_(2)=2xx0.18m=0.36m`