Correct Answer - D
After mixing `[Ag^(+)] = 0.2M, [NH_(3)] = 1M`
Due to very high value of `K_(f), Ag^(+)` is mainly converted into complex.
`Ag^(+)(aq)+2NH_(3)(aq)hArr[Ag(NH_(3))_(2)]^(+)(aq)`
`{:("Initial",0.2,1,),("conc.",,,),("At eq.",x,0.6,=0.2):}`
`[Ag(NH_(3))_(2)]^(+)(aq.) hArr Ag(NH_(3))^(+)(aq.)+NH_(3)(aq.)`
At equilibrium `0.2 - y`
`y = 0.2`
`0.6+y`
`= 0.6`
`(1)/(K_(f_(2))) = (yxx0.6)/(0.2)rArr (0.1)/(10^(4)),y=[Ag(NH_(3))^(+)] rArr 3.333 xx 10^(-5) M`