The formation constant of `Ni(NH_3)_6^(2+)` is `6xx10^8` at `25^@C`.If 50 ml of 2.0 M `NH_3` is added to 50 ml of 0.20 M solution of `Ni^(2+)` , the concentration of `Ni^(2+)` ion will be nearly equal to :

Options

A. `3xx10^(-10)` mole `litre^(-1)`
B. `2xx10^(-10)` mole `litre^(-1)`
C. `2xx10^(-9)` mole `litre^(-1)`
D. `4xx10^(-8)` mole `litre^(-1)`

Answer

D

Explanation

` {:(,Ni^(2+)+,6NH_(3), to,[Ni(NH)_(3)]^(6+),K_(f)=6xx10^(8)),(t=0,0.01 "mole",0.01 "mole",,0):}`
`K_(c) = ([Ni(NH_(3))_(6)^(+6)])/([Ni^(+2)][NH_(3)]^(6))`
`=((0.1))/([Ni^(2+)](0.4)^(6)) =6xx10^(8)`
`[Ni^(2+)]=4xx10^(-8)`