InterviewSolution
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InterviewSolution
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The solubility of `Pb(OH)_(2)` in water is `6.7xx10^(-6)M`. Calculate the solubility of `Pb(OH)_(2)` in a buffer solution of pH = 8 |
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Answer» `Pb(OH)_(2) hArr Pb^(2) + 2 OH^(-)` `:. K_(sp)=[Pb^(2+)][OH^(-)]^(2)=s xx (2s)^(2)=4 s^(3) = 4xx(6.7xx10^(-6))^(3) = 1.20 xx 10^(-15)` In a solution with pH = 8, `[H^(+)]=10^(-8) and [OH]^(-) = 10^(-6)` `:. 1.2xx10^(-15)=[Pb^(2+)]xx(10^(-6))^(2) or [Pb^(2+)]=(1.2xx10^(-15))/((10^(-6))^(2))=1.2xx10^(-3)M` |
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