A solution is 0.01 M Kl and 0.1 M KCl. If solid `AgNO_(3)` is added to the solution, what is the `[l^(-)]` when AgCl begins to precipitate?
`[K_(SP)(Agl)=1.5xx10^(-16),K_(SP)(AgCl)=1.8xx10^(-10)]`

Options

A. `3.5 xx 10^(-7)`
B. `6.1 xx 10^(-8)`
C. `2.2 xx 10^(-7)`
D. `8.3 xx 10^(-8)`

Answer

D

Explanation

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`