Question

Consider the following reaction that has reached equilibrium Pbl2 Pb2^2 (aq) + 2 I- (aq)

1). Determine the concentration of Pb2+ and I- ions in the solution at equilibrium

2). Predict the affect of adding potassium iodide, Kl to the solution

3). Predict the affect of adding hydroxide ion, OH-, if the Ksp of Pb(OH)2 = 1.2X10^-15

Just found out this is due in 2 hours!!! Please Help!!! Thank you so much!!!

Answer 1

a) Ksp for PbI2 = 8.5 x 10^-9

let x be the amount of PbI2 present in solution then,

Ksp = [Pb2+][I-]^2

8.5 x 10^-9 = (x)(2x)^2

x^3 = 2.125 x 10^-9

x = 1.29 x 10^-3 M

So the equilibrium concentrations of,

[Pb2+] = 1.29 x 10^-3 M

[I-] = 2.58 x 10^-3 M

2) Following Lechatellier's principle, when we add KI to the solution, the concentration of I- increases (KI is completely soluble). This would shift the equilibrium towards PbI2.

3) Ksp Pb(OH)2 = [Pb2+][OH-]^2

Pb(OH)2 <===> Pb2+ + 2OH-

So as we increase the concentration of OH- in solution by adding NaOH, the equilibrium is disturbed for the reaction and will shift towards left. That is more of Pb(OH)2 would be formed until we have an equilibrium between Pb(H)2 and OH- and reestablish the equilibrium. this is according to the LeChatellier's principle.