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`K_(sp)` for lead iodate `[Pb(IO_(3))_(2) is 3.2 xx 10^(-14)` at a given temperature. The solubility in `molL^(-1)` will beA. `2.0 xx 10^(-5)`B. `(3.2 xx 10^(-7))^(1//2)`C. `(3.8 xx 10^(-7))`D. `4.0 xx 10^(-6)` |
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Answer» Correct Answer - A `{:([Pb(IO_(3))_(2)]hArr,Pb^(2+)+,2IO_(3)^(Theta),),(,2S,S,):}` `K_(sp) = 4S^(3)`. `S =root3((K_(sp))/(4))root3((3.2xx10^(-14))/(4))=2xx10^(-5)M`. |
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