InterviewSolution
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InterviewSolution
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Predict whether a precipitate will be formed or not on mixing 20 mL of 0.001 M NaCI with 80 mL of 0.01 M `AgNO_(3)` solution `(K_(sp) " for " AgCI =1.5 xx 10^(-10))` |
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Answer» Correct Answer - Yes , ionic product `=1.6 xx 10^(-6)` The precipitate of AgCI is to be formed . the solubility equilibrium may be represented as `AgCI (s) hArr Ag^(+) (aq) +CI^(-) (aq)` Now `Ag^(+)` ions are to be provided by `AgNO_(3)` solution while `CI^(-)` ions by NaCI solution as a result of dissociation. `AgNO_(3)overset((aq))(to) Ag^(+) (aq) + NO_(3)^(-) (aq) , NaCI (s) overset((aq))(to) Na^(+) (aq) +CI^(-) (aq)` the total volume of the solution after mixing `=(20 +80) =100 mL` In the solution , the concentration of `Ag^(+)` ions after mixing will be `4//5 (80//100)` while that of `CI^(-)` ions after mixing will be `1//5 (20// 100)` `" Thus "" "[Ag^(+)] " before mixing " =0.01 M , [Ag^(+)] " Aftre mixing " =(0.01 xx 4)/(5) =0.008 M` Similarly, `[CI^(-)]` before mixing =0.001 M `[CI^(-)]` after mixing `=(0.001 xx 1)/(5) =0.0002 M` The ionic product = `Ag^(+)][CI^(-)] = (0.008) xx (0.0002) =8 xx 10^(-3) xx 2xx 10^(-4) =1.6 xx 10^(-6)` `K_(sp)` value of AgCI =`1.5 xx 10^(-10)` (given) As the ionic product is more than the solubility product this means AgCI will get precipitated |
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