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.