a. Will a precipitate of`Mg(OH)_(2)` be formed in a `0.001M` solution of `Mg(NO_(3))_(2)` if the `pH` of solution is adjusted to `9.K_(sp)` of `Mg(OH)_(2) = 8.9 xx 10^(-12)`.
b. Calculate `pH` at which `Mg(OH)_(2)` begin to precipitae form a solution containing `0.1M Mg^(2+)` ions. `K_(sp) of Mg(OH)_(2) =1 xx 10^(-11)`.
c. Calculate `[overset(Theta)OH]` of a solution after `100mL` of `0.1M MgC1_(2)` is added to `100mL` of `0.2M NaOH. K_(sp) Mg(OH)_(2) = 1.2 xx 10^(-11)`.

Explanation

a. `pH = 9 :. [H^(o+)] = 10^(-9)M or [OH^(Theta)] = 10^(-5)m`
Now if `Mg(NO_(3))_(2)` is present in a solution of `[OH^(Theta)] = 10^(-5)M`, then
Product of ionic concentration
`= [Mg^(2+)] [OH^(Theta)]^(2)`
`= [0.001][10^(-5)]^(2)`
`10^(-13)` lesser than `K_(sp) of Mg (OH)_(2)` (i.e., `8.9 xx 10^(-12))`
`Mg(OH)_(2)` will not precipitate.
b. When `Mg(OH)_(2)` starts precipitation, then,
`[Mg^(2+)] [OH^(Theta)]^(2) = K_(sp) of Mg (OH)_(2)`
`[0.1] [OH^(Theta)]^(2) = 1 xx 10^(-11)`
`:. [OH^(Theta)] = 10^(-5)M :. pOH =5`
`:. pOH = 14 - pOH`
`pH = 14 - 5 = 9`
c. `{:(,MgCI_(2)+,2NaOhrarr,Mg(OH)_(2)+,2NaCI),("mmoles before reaction",rArr10,20,0,0),("mmoles after reaction",rArr0,0,10,20):}`
Thus, `10mmol` of `Mg(OH)_(2)` are formed. The product of `[Mg^(2)] [OH^(Theta)]^(2)` is, therefore, `[(10)/(200)]xx[(20)/(200)]^(2) = 5 xx 10^(-4)` which is more than `K_(sp)` of `Mg(OH)_(2)`. Now solubility `(S)` of `Mg(OH)_(2)` can be derived by
`K_(sp) = 4S^(3)`
`:. S = 3sqrt((K_(sp))/(4)) = 3sqrt((1.2 xx 10^(-11))/(4)) = 1.4 xx 10^(-4)`
`:. [OH^(Theta)] = 2S =2.8 xx 10^(-4)`.