The solubility of `Mg(OH)_(2)` in pure water is `9.57xx10^(-3)`g `litre^(-1)`. Calculate the pH of its saturated solution. (Assume `100%` ionisation)

Explanation

Solubility of `Mg(OH)_(2)` in pure water
`=9.57xx10^(-3)g//litre`
`=(9.57xx10^(-3))/(58)"mole"//"litre"`
`=1.65xx10^(-3)M`
`Mg(OH)_(2)rarrunderset(S)(Mg^(2+))+underset(2S)(2OH^(-))`
`:. [OH^(-)]=2xx[Mg(OH)_(2)]`
`=2xx1.65xx10^(-4)=3.3xx10^(-4)`
`:. pOH= -log[OH^(-)]=3.48`
`pH=14-3.48=10.52`