Question

The molar solubility of PbI2 in,0.2 M of Pb(NO3)2

Answer 1

To determine molar solubility of a slightly soluble salt all you need to consider is whether there is presence of common ion or there is formation of complex ion.

In this case we have common ion, that is Pb²⁺.

The presence of common ion reduces the solubility of a slightly soluble salt. So When we have a solution of lead (II) Iodide and then we and lead (II) nitrate, in this case, the common ion is lead (II) ions and thus the solubility decreases.

As I had mentioned, presence or formation of complex ion normally increases the solubility of a slightly soluble salt.

Answer 2

Answer:

let the solubility of PbI2 be s

then PbI2 <---------> Pb2+ + 2I- ..........(1)

if there were no Pb(NO3)2 then ....at equilibrium ...[Pb2+] = s

and [I-] = 2s

and Ksp = [Pb2+] [I-]^2 = s X (2s)^2 = s X 4s^2 = 4s^3

so s = (Ksp/4 )^1/3

but since Pb(NO3)2 is there and it completely dissociates so ...

Pb(No3)2 --------> Pb2+ + 2NO3-

as [Pb(NO3)2] = 0.2 M

after complete dissociation ...[Pb2+] = 0.2 M

let new soubility of PbI2 be s'

so now conc.of Pb2+ at equilibrium = 0.2 + s'

due to common ion effect equilibrium (1) will shift to left and [Pb2+] ~ 0.2

Ksp = 0.2 X (2s')^2 = 0.2 X 4s^2 = 0.8s'^2

so s'^2 = Ksp/0.8

s' = (Ksp/0.8 )^1/2