`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+2,2HCl to , PbCl_2+ , 2HNO_3),("Millimoles added",100 xx 0.1, 1 xx1 , 0,0),("Millimoles left",9.5,0,0.5, 1):}`
Total `[Pb^(2+)]` in solution `=(9.5+0.5)/101`
Now, if `PbCl_2` is precipitated , then contribution of 0.5 of `[Pb^(2+)]` form `PbCl_2` be left
For precipitation ,
Ionic product > `K_(sp)`
Ionic product = `[Pb^(2+)][Cl^-]^2`
`=(10/101)(1/101)^2=9.70xx10^(-6)`
The ionic product is greater than `K_(sp)`, so that precipitation will occur.
`[Pb^(2+)]=9.5/101=9.4xx10^(-2) "mol L"^(-1)`