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How many moles of NH(3) must be added to...

How many moles of `NH_(3)` must be added to `1.0L` of `0.75M AgNO_(3)` in order to reduce the `[Ag^(o+)]` to `5.0 xx 10^(-8)M. K_(f) Ag (NH_(3))_(2)^(o+) = 1 xx 10^(8)`.

Text Solution

Verified by Experts

The correct Answer is:
A
`{:(,Ag^(o+)+,2NH_(3)hArr,[Ag(NH_(3))_(2)^(o+)],),("Concentration at",5.0xx10^(-8),?,0.75-(5.0xx10^(-8))~~0.75M,),("equilibrium",,,,):}`
`K_(f) = ([Ag(NH_(3))_(2)^(o+)])/([Ag^(o+)][NH_(3)]^(2))`
`1 xx 10^(8) = (0.75)/((5.0 xx 10^(-8))[NH_(3)]^(2))`
`[NH_(3)]^(2) = 0.15, [NH_(3)] = 0.4M`
From the equation, it is evident,
`1mol` of `Ag(NH_(3))_(2)^(o+)` requires `= 2mol NH_(3)`
`0.75 mol Ag(NH_(3))_(2)^(o+)`requires `= 2xx 0.75 = 1.5mol NH_(3)`.
Total `NH_(3)` required `=(1.5 + 0.4) = 1.9mol = 1.9 M`
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  • 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))

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