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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

`Ag^(+) +2NH_(3) hArr [Ag(NH_(3))_(2)(aq)]^(+1) " " K_(f) = 1//6.8 xx10^(-8)`
As `K_(f) [Ag(NH_(3))_(2)^(+)]=1/(6.8 xx10^(-8) ) = ` very - very large
Hence , almost all `Ag^(+)` ions will be converted to `Ag(NH_(3))_(2)^(+)`
` :. [Ag(NH_(3))_(2)^(+)] approx 0.1 M `
` [Ag^(+)] = 2xx10^(-7)`
` K = ([Ag^(+)][NH_(3)]^(2))/([Ag(NH_(3))_(2)^(+)])`
` rArr 6.8 xx 10^(-8) = (2xx10^(-7)xx[NH_(3)]^(2))/(0.1)`
`[NH_(3)] = 0.184 M`
It is the concentration of free `NH_(3)`
`[NH_(3)] _("total") = [NH_(3)]_("free") + [NH_(3)] _("complexed") = 0.184 + 2xx 0.1 = 0.384 M `
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