The solubility of `Mg( OH)_2` in pure water is ` 9.57 xx 10 ^(-3) g L^(-1) ` .Calculate its solubility I ` (g L^(-1))` in 0.02 M Mg `(NO_3)_2` solutions.

Explanation

Solubility of `Mg(OH)_(2)`
`=9.57 xx 10^(-3) g L^(-1)`
`=(9.57 xx 10^(-3))/(58)`
`=1.65 xx 10^(-4) mo L^(-1)` (since molecular mass of `Mg(OH)_(2)` is 58) `Mg(OH)_(2)` ionises as
`Mg(OH)_(2)(s) hArr Mg^(2+) (aq) + 2OH^(-) (aq)`
`[Mg^(2+)]= [Mg(OH)_(2)]= 1.65 xx 10^(-4) mol L^(-1)`
`[OH^(-)]= 2 xx [Mg(OH)_(2)] = 2 xx 1.65 xx 10^(-4) mol L^(-1)`
Hence, `K_(sp)= [Mg^(2+)] [OH]^(2)`
`=(1.65 xx 10^(-4)) (2 xx 1.65 xx 10^(-4))^(2)`
`=1.79 xx 10^(-11)`
In 0.02M `Mg(NO_(3))_(2)` solution:
Suppose the solubility of `Mg(OH)_(2)` in 0.02M `Mg(NO_(3))_(2)` solution is x mol `L^(-1)`.
Therefore, `Mg(OH)_(2) (s) hArr underset(x)(Mg^(2+))(aq) + underset(2x)(2OH^(-))(aq)`
`Mg(NO_(3))_(2) (aq) hArr underset(0.02)(Mg^(2+)(aq)) + underset(2 xx 0.02)(2NO_(3)^(-)) (aq)`
`[Mg^(2+)] = x + 0.02`
`[OH^(-)]=2x`
`K_(sp)` of `Mg(OH)_(2)` will remain unchanged. Therefore, `[Mg^(2+)] [OH^(-)]^(2)= K_(sp)`
`(x+ 0.02) (2x)^(2)= 1.79 xx 10^(-11)`
SInce the value of x is much less than 0.02, `x+ 0.02` is approximately equal to 0.02. Therefore, we have
`0.02 xx 4x^(2)= 1.79 xx 10^(-11)`
or `x= ((1.79 xx 10^(-11))/(0.02 xx 4))^(1//2)`
`=1.49 xx 10^(5) mol L^(-1)`
`=1.49 xx 10^(-5) xx 58 g L^(-1)`
`=8.6 xx 10^(-4) g L^(-1)`.
Hence, the solubility, of `Mg(OH)_(2)` in the presence of 0.02M `Mg(NO_(3))_(2)` is `8.6 xx 10^(-4) g L^(-1)`