A saturated aqueous dolution of `Mg(OH)_(2)` has a vapour pressure of `759.5 mm` at `373 K`. Calculate the solubility product of `Mg(OH)_(2)`. (Assume molarity equals molality).

Explanation

`p_(s)=759.5 mm, P_(H_(2)O)=760 mm` at `373 K`
`:. DeltaP=p_(s)-P_(H_(2)O)=0.5`
`Deltap//p=0.5//760=x_(2)=6.5xx10^(-4)`
We know,
molality `(m)=("moles of solute")/("Volume of solvent (kg)")`
`:. M=(x_(2)xx1000)/(x_(1)xxMw_(1))=((6.5xx10^(-4))xx1000)/((1-6.5xx10^(-4))xx18)=0.036`
`Mg(OH)_(2) hArr Mg^(2+)+2OH^(ɵ)`
`i=3` (Assuming `100%` ionisation)
Solubility `(S)=m//i`
`S=0.036//3=0.012 M`
`K_(SP)=4S^(3)=4xx(0.012)^(3)=6.8xx10^(-6)`