If 500 ml of 0.4 M AgNO_(3) is mixed with 50 ml of 2 M NH_(3) solution then what is the concentration of [Ag(NH_(3))]^(+) in solution (K_(t), [Ag_(NH_(3))]^(+) = 10^(3), K_(f_(2)) [Ag(NH_(3))_(2)]^(+)= 10^(4))

If 500 ml of 0.4 M AgNO_(3) is mixed with 50 ml of 2 M NH_(3) solution

1.	If 500 ml of 0.4 M AgNO_(3) is mixed with 50 ml of 2 M NH_(3) solution then what is the concentration of [Ag(NH_(3))]^(+) in solution (K_(t), [Ag_(NH_(3))]^(+) = 10^(3), K_(f_(2)) [Ag(NH_(3))_(2)]^(+)= 10^(4))
Answer» `3.33 xx 10^(-7) M` `3.33 xx 10^(-5) M` `3 xx 10^(-4) M` `10^(-7) M` Solution :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))^(+)` `{:(0.2,1,),(x,0.6,02):}` `{:([Ag(NH_(3))_(2)]_((aq))^(+)hArr,[Ag(NH_(3))]_((aq))^(+),+NH_(3(aq))),(0.2-y,y,0.6+y),(0.2,,0.6):}` `(1)/(K_(f_(2))) = (y xx 0.6)/(0.2) = (1)/(10^(4))` `y = [Ag(NH_(3))]^(+) = 3.33 xx 10^(-5) M`