Question

If 100.0 mL of 0.0500 F Pb(NO₃)₂ is mixed with 200.0 mL of 0.100 F NaI will a precipitate of PbI₂ from or not?

you will need to justify your answer. Ksp(PbI₂) = 1.4* 10^-8

Answer 1

In 200 mL of 0.100 M NaI :
¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯
n(NaI) = C1 x V1
n(NaI) = 0.150 x 0.300
n(NaI) = 0.045mol of NaI

. . . therefore : n(I(-)) = 0.045 mol

In 100 mL of 0.050 M Pb(NO3)2 :
¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯
n(Pb(NO2)2) = C12 x V2
n(Pb(NO2)2) = 0.050 x 0.100
n(Pb(NO2)2) = 5×10^-3 mol of Pb(NO3)2

. . . therefore : n(Pb(2+)) = 5×10^-3 mol

Concentrations :
¯¯¯¯¯¯¯¯¯¯ ¯¯
[Pb(2+)] = n(n(Pb(2+)) / Vtotal
[Pb(2+)] = 5×10^-3 / ( 0.200 + 0.100 )
[Pb(2+)] = 0.0166 mol/L

[I(-)] = n(I(-)) / Vtotal
[I(-)] = 0.045 / ( 0.200 + 0.100 )
[I(-)] = 0.15 mol/L

[Pb(2+)] x [I(-)] = 0.0166 x 0.15 = 2.50×10^-3

[Pb(2+)] x [I(-)] > Ksp . . . . then there is precipitate . . . !