InterviewSolution
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InterviewSolution
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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. |
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Answer» Correct Answer - C `[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+)` |
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