The `K_(sp)` of `Mg(OH)_(2)` is `1xx10^(-12). 0.01M Mg^(2+)` will precipitate at the limiting pH of

Options

A. `8`
B. `9`
C. `10`
D. `12`

Answer

B

Explanation

`{:(Mg(OH)_(2)rarr,Mg^(2+)+,2overset(Theta)OH,),(,S,2S,):}`
Let `S_(1)` is the solubility in `0.1M MgCI_(2)` (common ion `Mg^(2+))`
`:. [Mg^(2+)] = (S_(1) + 0.001) ~~ 0.01`
`[overset(Theta)OH] = 2S_(1)`
`K_(sp) = (0.01) (2S_(1))^(2)`
`S_(1) = ((K_(sp))/(4xx0.01))^(1//2) = ((10^(-12))/(4xx10^(-2)))^(1//2) = 0.5xx10^(-5)M`
`:. [overset(Theta)OH] = 2S_(1) = 2xx 0.5 xx 10^(-5) = 10^(-5)M`
`pOH = 5, pH = 14 - 5 = 9`