`K_(sp)` of `PbC1_(2)` is `10^(-13)`. What will be `[Pb^(2+)]` in a of solution prepared by mixing `100mL` of `0.1MPb(NO_(3))_(2)` of solution `1.0mL 1M HCI` ?

Explanation

`{:(,Pb(NO_(3))_(2)+,2HCIrarr,PbCI_(2)+,2HNO_(3)),(,100xx0.1,1xx1,,),("mmol added",=10,=1,0,0),("mmol left",9.5,0,0.5,1):}`
Since `HCI` is a limiting reagent:
Concentration `= ("mmol")/("Total volume") :. [Pb^(2+)] = (9.5+0.5)/(101)`
Now if `PbC1_(2)` is precipitated, then conrtibution `0.5M` of `[Pb^(2+)]` from `PbC1_(2)` should be left.
To see precipitation, ionic product `gt K_(sp)`
Ionic product `=[Pb^(2+)] [C1^(Θ)]^(2)`
`=[10//101][1//101]^(2) = 9.70 xx 10^(-6)`
Which is greater than `K_(sp)` of `PbC1_(2)` and thus, precipitaiton of `PbC1_(2)` ocurs.
`[Pb^(2+)] = (9.5)/(101) = 9.4 xx 10^(-2)` "mole litre"^(-1)`