Question

Solubility Product of Calcium Hydroxide
Post-Lab Questions (40 Points)

1. A. Determine the molar solubility of PbI₂. The Ksp values for PbI^₂ is 8.7 x 10^-9.  

B. Determine the molar solubility of 100 mL of a solution of PbI_² to which 0.01 mole of lead nitrate, Pb(NO₃)₂ has been added. Assume the total volume remains at 100 mL.

C. The molar solubility of cadmium hydroxide, Cd(OH)₂ is 1.842 x 10^-5 M. What is the Ksp value of cadmium hydroxide?

Answer 1

PbI_2 rightarrow Pb^2 + + 2I^-, K_sp = 8.7 times 10^-9 If solubility of PbI_2 = x then, [Pb^2 +] = x, [I^-] = 2x K_sp = [Pb^2 +][I^-]^2 = > 8.7 times 10^-9 = (x) (2x)^2 = > 4x^3 = 8.7 times 10^-9 = > x^3 = 2.175 times 10^-9 = > x = 1.296 times 10^-3 M Hence, molar solubility of PbI_2 = 1.296 times 10^-3 M \r\n Pb(NO_3)_2 = 0.01 mole in 100 ml = > [Pb(NO_3)_2] = 0.01 times 10^3/100 = 0.1 M = > [Pb^2 +] = 0.1M PbI_2 rightarrow Pb^2 + + 2I^- (If solubility of PbI_2 is 'x') K_sp = [Pb^2 +] [I]^2 = > 8.7 times 10^-9 = (0.1) (2x)^2 = > 8.7 times 10^-8 = 4x^2 = > x = 1.475 times 10^-4 M Hence, molar solubility in this case of PbI_2 is 1.475 times 10^-4 M \r\n Cd(OH)_2 rightarrow cd^2 + + 2 OH^- Molar solubility of cd(oH)_2 = 1.842 times 10^-5 m 1.842 times 10^-5 = 1.842 times 10^-5 2 times 1.842 times 10^-5 K_sp = [cd^2 +][OH^-]^2 = (1.842 times 10^-5) (2 times 1.842 times 10^-5)^2 = 2.499 times 10^-14