The `K_(sp)` of `Mg(OH)_(2)` is `8.9 xx 10^(-12)` at `25^(2)C`. the pH of solution is adjusted to `9`. How much `Mg^(2+)` ion will be precipitated as `Mg(OH)_(2)` from a `0.1M MgCl_(2)` solution at `25^(@)C`? Assume that `MgCl_(2)` is completely dissociated.

Options

A. `0.011`
B. `0.89`
C. `0.11`
D. `0.89`

Answer

A

Explanation

`K_(sp)[Mg(OH)_(2)] = 8.9 xx 10^(-2)`
`pH = 9`
`[H^(+)] = 10^(-9)`
`[OH^(-)] = 10^(-5)`
`K_(sp) = [Mg^(2+)][OH^(-)]^(2)`
`8.9 xx 10^(-12) = [Mg^(2+)][10^(-5)]^(2)`
`[Mg^(2+)] = 8.9 xx 10^(-2) = 0.089`
Conc. Of `Mg^(2+)` precipitated `= 0.1 - 0.089`
`= 0.011 "mol"//L.t`