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.

The correct answer should be "electrons orbit the nucleus in circular motions exactly as shown on the classic bohr model."