Which of the following concentration of `NH_(4)^(+)` will be sufficient to present the precipitation of `Mg(OH)_(2)` form a solution which is 0.01 M `MgCl_(2)`and `0.1 M NH_(3)(aq)`. Given that `K_(sp)Mg(OH)_(2)=2.5xx10^(-11)` and `K_(b)` for `NH_(3) = 2xx10^(-5)`.

Options

A. 0.01 M
B. 0.02 M
C. 0.001 M
D. 0.04 M

Answer

D

Explanation

`K_(SP)=[Mg^(+2)][OH^(-)]^2`
`[OH^(-)]=sqrt((2.5xx10^(-11))/0.01)=5xx10^(-5)`
`K_b=([NH_4^+][OH^(-)])/([NH_3])`
So, `[NH_4^+]=0.04 M`