What mass of `pb^(2+)` ions is left in solution, when `50.0mL` of `0.20M Pb(NO_(3))_(2)` is added to `50.0mL` of `1.5M NAC1`?

Explanation

Assume complete precipitation of the `Pb^(2+)` followed by solution of the equilibrium concentration to be determined.
`(50.0mL) (0.20M) = 10 mmol Pb^(2+)`
`(50.0mL) (1.5M) = 75 mmol C1^(Θ)`
`Pb^(2+) + 2C1^(Θ) hArrPbC1_(2)`
When lead ion concentration can exist in a solution of `C1^(Θ)` containing `(75mmol) -(20 mmol) = 55 mmol` in `100mL? [C1^(Θ)] = 0.55M`
`PbC1_(2) hArr Pb^(2+) + 2C1^(Θ)`
`K_(sp) [Pb^(2+)] [C1^(2+)] [C1^(Θ)]^(2) = 1.7 xx 10^(-4)`
`[Pb^(2+)] = (1.7 xx 10^(-4))/((0.55)^(2)) = 5.6 xx 10^(-4)M`
`(5.6 xx 10^(-4) mol L^(-1) (0.100L) (208 g mol^(-1)) = 12 mg`