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The solubility of PbI2 is 1.3×10–3 mol.L-1. What is the Ksp for PbI2?
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`PbI_(2(s)) rarr Pb^(2+)_(aq)+2I^-_(aq)`
`Pb^(2+):I^(-) = 1:2`
It is given that the solubility of `PbI_2` is `1.3xx10^(-3)` .
This means `[Pb^(2+)] = 1.3xx10^(-3)`
By molar ratio;
`[I^-] = 2(1.3xx10^(-3))`
`K_(sp) = [Pb^(2+)][I^-]^2`
`K_(sp) = (1.3xx10^(-3))(2xx1.3xx10^(-3))^2`
`K_(sp) = 8.788xx10^(-9)M^3`
So the `K_(sp)` of `PbI_2` is `8.788xx10^(-9)M^3` .
Posted by jeew-m on August 20, 2013 at 4:20 AM (Answer #1)
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