Determine the mass of `PbI_(2)` that will dissolve in (a) `500mL` water (b) `500mL` of `0.01M KI` solution (c) `500mL` of a solution containing `1.33 g Pb(NO_(3))_(2), K_(sp)`of `PbI = 1.4 xx 10^(-8)`.

Explanation

`PbI_(2) hArr Pb^(2+) +2I^(Θ)`,
`K_(sp) = [Pb^(2+)] [I^(Θ)]^(2) = 1.4 xx 10^(-8)`.
a. Let `x - [Pb^(2+)]`, then `[I^(Θ)] = 2x`
`:. 4x^(3) = 1.4 xx 10^(-8), x = 1.5 xx 10^(-3)M`
Number of moles is `500mL H_(2)O`
`= (1.5 xx 10^(-3) xx500)/(1000) 0.75 xx 10^(-3)mol`
Mass of `PbI_(2) = 0.75 xx 10^(-3) xx 461 = 0.35g`
b. `[Pb^(2+)] [0.1]^(2) = 1.4 xx 10^(-8)`
`:. [Pb^(+2)] = 1.4 xx 10^(-6)M`
Mass of `PbI_(2)` is `500mL` of `0.1KI` solution
`= 1.4 xx 10^(-6)M xx 0.5L xx 46g mol^(-1)`
` = 3.2 xx 10^(-4)g = 0.32 mg`
c. Molarity of `Pb(NO_(3))_(2)`
`= (W_(2) xx 1000)/(M_(w_(2)xxV_(Sol))) = (1.33 xx 1000)/(331xx500) = 8.04 xx 10^(-3)M`
If the concentration of `PbI_(2)` is `y`, then `[I^(Θ)] = 2y`
`K_(sp) = [Pb^(2+)] [I^(Θ)]^(2)`
`1.4 xx 10^(-8) = (8.04 xx 10^(-3)) (2y)^(2)`
`:. y = 6.6 xx 10^(-4)M`
Mass of `PbI` in `500mL` of a solution of `1.33g` of `Pb(NO_(3))_(2)`
`= 6.6 xx 10^(-4) xx 0.5L xx 461 = 0.51g`