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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 `K_(sp)=[Pb^(+)]^(2)[IO_(3)^(-)]^(2)=[s(2s)^(2)]=4s^(3)` `s=3 sqrt((K_(sp))/(4))=3sqrt((3.2 xx 10^(-14))/(4))` `3sqrt(8 xx 10^(-15))=2xx10^(-5)mol L^(-1)` |
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