Ml of 0.00237 m nai (aq) is combined with 625. ml of 0.00785 m pb(no3)2 (aq). determine if a precipitate will form given that the ksp of pbl2 is 1.40x10-8.

The volume of I- is missing in your question by assuming it = 1L
moles I- =  molarity * volume

              = 0.00237 * 1 L 

              = 0.00237 mol

[I-] = moles / total volume
 
    = 0.00237 / 1.625L
   
    = 0.00146 M

moles Pb2+ = molarity * volume 

                     = 0.00785 * 0.625 L

                    = 0.0049 mol 
  
[Pb2+] = 0.0049 / 1.625L

           =  0.003 M 

when        PbI2(s)  ↔     Pb2+(aq)  +  2I-(aq)

when Q = [Pb2+][I-]^2    and we neglect [PbI2] as it is solid

 ∴ Q = 0.003  * (0.00146)^2

        = 6.4 x 10^-9

by comparing the value of Q with Ksp value  we will found that:

Q < Ksp which mean that more solid will dissolve, and this is an unsaturated solution which has ion concentrations < equilibrium concentrations, so the reaction will go forward until achieving equilibrium.

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