3 The Ksp for lead iodide Pbl2 is 14 x 10 8 Calculate the so

3. The Ksp for lead iodide (Pbl2) is 1.4 x 10 8. Calculate the solubility of lead iodide in each of the following. a) Pure water b) 0.10 M Pb(NO3)2 c) 0.010 M Nal

Solution

PbI2(s) --------------> Pb^2+ (aq) + 2I^- (aq)

                                    s                 2s

Ksp    = [Pb^2+][I^-]^2

          = s*(2s)^2

   1.4*10^-8    = 4s^3

s^3                 = 1.4*10^-8/4

   s^3               = 3.5*10^-9

s                     = 1.5*10^-3M

solubility in water = 1.5*10^-3 M

b. 0.1M Pb(NO3)2

PbI2(s) --------------> Pb^2+ (aq) + 2I^- (aq)

                                    s+0.1                 2s

Ksp    = [Pb^2+][I^-]^2

          = (s+0.1)*(2s)^2                          [s+0.1 = 0.1]

   1.4*10^-8    = 4s^2*0.1

s^2                 = 1.4*10^-8/0.4

   s^2               = 3.5*10^-8

s                     = 1.87*10^-4M

solubility in 0.1M Pb(NO3)2 = 1.87*10^-4M

c. 0.01M NaI

PbI2(s) --------------> Pb^2+ (aq) + 2I^- (aq)

                                    s                2s+0.01

Ksp    = [Pb^2+][I^-]^2

          = (s)*(2s+0.01)^2                          [2s+0.01 = 0.01]

   1.4*10^-8    = s*(0.01)^2

s                 = 1.4*10^-8/(0.01)^2

   s               = 0.00014   = 1.4*10^-4M

solubility in 0.01M NaI = 1.4*10^-4M