The solubility of `Mg(OH)_(2) " is " 8.352 xx 10^(-3) g L^(-1)` at 298 K. Calculate the `K_(sp)` of `Mg(OH)_(2)` at this temerature.

Answer

`1.194 xx 10^(-11) "moll"^(3) L^(-3)`

Explanation

The solubility of Mg `(OH)_(2) =8,352 xx 10^(-3) gL^(-1)` ,
`=((8.352 xx 10^(-3) gL^(-1)))/((58 g " mol ")) = 1.44 xx 10^(-4) " mol"//L^(-1)`
The solubility equilibrium in saturated solution is :
`Mg(OH)_(2) (s) hArr Mg^(2+) (aq) +2OH^(-) (aq)`
`[Mg^(2+)(aq)] =1.44 xx 10^(-4) " mol " L^(-1)`
`[OH^(-)(aq)] =2xx 1.44 xx 10^(-4) =2.88 xx 10^(-4) " mol " L^(-1)`
`K_(sp) " of " Mg (OH)_(2) (s) =[Mg^(2+) (aq)] OH^(-) (aq)]^(2)`
`= (1.44 xx10^(-4) " mol " L^(-1)) xx (2.88 xx 10^(-4) " mol " L^(-1))^(2) =1.194 xx 10^(-11) (" mol " L^(-1))^(3)`