InterviewSolution
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InterviewSolution
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If the solution `0.5 M NH_(3)` and stability constant for `A^(+)(NH_(3))_(2)` is `K_(stb)=([Ag(NH_(3))_(2)]^(+))/([Ag^(+)(aq)][NH_(3)]^(2))=6.4xx10^(7),` then find the solubility of AgCl in the above solution `K_(sp)of AgCl=2xx10^(-10)` |
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Answer» `K_(eq)=K_(sp)xxK_(stb)` `implies2xx10^(-10)xx6.4xx10^(7)=([Ag(NH_(3))_(2)][Cl^(-)])/([NH_(3)]^(2))=(SxxS)/((0.5-2S)^(2))` `implies(S)/((0.5-2S))=sqrt(2xx64xx10^(-4))` `impliesS=0.046` |
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