1.

10 mol sampleof AgNO_(3)is dissolved in one lit of 1.00 M NH_(3)Is it possibleAgCl(s)form the solution by adding0.010molof NaCl ?

Answer»

Solution :`K_(SP(AgCl)) = 1.8 xx10^(-10), K _(F[Ag(NH_(3))_(2)^(.)])= 1.6 xx10^(7)`
` {:("Ag"^(+),+,2NH_(3),hArr,[Ag(NH_(3))_(2)^(+)]),(0.10 M ,,1.00,,0),(x,,(1-0.20)M,,),(,,=0.80 M,,0.10 M ):}`
It is assumed that all `Ag^(+)` ions have been complexed and only x amount is left
`K_(f)= ([Ag(NH_(3))_(2)])/([Ag^(+)][NH_(3)]^(2)) RARR 1.6 xx10^(7) = (0.10)/(XX(0.80)^(2))`
` :. x = 9.8 xx10^(-9) M = [Ag^(+)] ` undissolved
`[Cl^(-)] = 1.0 xx10^(-2) M `
` :. [Ag^(+)] [Cl^(-)] = 9.8 xx10^(-9) xx1.0 xx10^(2) = 9.8 xx10^(-11) lt 1.8 xx10^(-10) [K_(sp(AgCl))] `
Hence , AgCl (s) will not precipitate


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