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 catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer