Chemistry Solubility Math Question 5

Question

A solution contains 0.14 M Pb(NO₃)₂ and 0.02 M AgNO₃. As solid KI is slowly added to the solution, one of the metal iodides begins to precipitate. What is the concentration of Ag⁺(aq) remaining in solution when the second metal iodide begins to precipitate from the solution? K_sp1(AgI)=1.5E-16 and K_sp2(PbI₂)=1.4E-8

Answer 1

According to K_sp1(AgI)=1.5E^-16 and K_sp2(PbI₂)=1.4E^-8, when solid KI is slowly added to the solution, AgI will for precipitate first because its very small Ksp value. One or two KI crystals could have AgI to precipitate.

when KI is added to have PbI₂ begins to precipitate from the solution, [I^-] concentration reach to:

Pb(NO₃)₂ ----> Pb²⁺ + 2 NO₃^- PbI_{2 (s)} ---> Pb²⁺ + 2I^- Ksp = [Pb²⁺] [I^-]²

0.14 M 0.14 M 2(0.14) M 0.14 M x

1.4E^-8 = (0.14) x² x = [I^-] = 3.16 x 10^-4M

When [I^-] concentration reach to 3.16 x 10^-4M in the solution:

AgI (s) ------> Ag⁺ + I^- Ksp = [ Ag⁺] [ I^-] 1.5E^-16 = x (3.16 x 10^-4) x=[Ag⁺] = 4.75 x 10^-13M

x 3.16 x 10^-4

the concentration of Ag⁺(aq) remaining in solution = 0.02 - 4.75 x 10^-13 = 0.02 M almost no change