Given: `Ag(NH_(3))_(2)^(+)hArrAg^(+)2NH_(3), K_(C)=6.2xx10^(-8)` and `K_(SP)` of `AgCI=1.8xx10^(-10)` at 298 K. Calculate the concentration of the complex in `1.0M` aqueous ammonia.

Answer

C

Explanation

`[Ag(NH_(3))_(2)]^(o+) (aQ) hArr Ag^(o+) (aq) +2NH_(3)(aq)`
`K_(c) = ([Ag^(o+)][NH_(3)]^(2))/([Ag(NH_(3))_(2)]^(o+))= ([Ag^(o+)][CI^(Theta)][NH_(3)]^(2))/([Ag(NH_(3))_(2)][CI^(Theta)])`
`K_(sp) of AgCI = [Ag^(o+)] [CI^(Theta)] = 1.8 xx 10^(-10)`
`6.2 xx 10^(-8) = (K_(sp) "of" AgCI[NH_(3)]^(2))/([Ag(NH_(3))_(2)]^(o+)[CI^(Theta)])`
`= (1.8 xx 10^(-10) xx [NH_(3)]^(2))/([Ag(NH_(3))_(2)]^(o+)[CI^(Theta)])`
`{:(AgCI,+2NH_(3)hArr,[Ag(NH_(3))_(2)]^(o+)+,CI^(Theta)),(,1,0,0),(,(1-a),a,a):}`
`6.2 xx 10^(-8) = (1.8 xx 10^(-10)xx[NH_(3))]^(2)/([Ag(NH_(3))_(2)]^(o+)[CI^(Theta)])`
`=(1.8 xx 10^(-10)(1-a)^(2))/(axxa)`
Since `a` is very small, it can be neglected. Therefore,
`6.2 xx 10^(-8) = (1.8 xx 10^(-10))/(a^(2))`
or `a = sqrt((1.8xx10^(-10))/(6.2xx10^(-8))) ~~ 0.0538`
` =[Ag(NH_(3))]^(o+)`