Question

Calculate the Ksp of Cd(OH)2 if 3.1 mL of 0.01300 M HCl was needed to titrate 1.2 L of saturated Cd(OH)2 solution. (show your work)

Answer 1

Number of moles of HCl, n = Molarity x volume in L

                                      = 0.01300 M x 3.1 mL x10^-3 L/mL

                                     = 4.03x10^-5 mol

2HCl + Cd(OH)2 ----> CdCl2 + 2H2O

From the balanced reaction ,

2 moles of HCl reacts with 1 mole of Cd(OH)2

4.03x10^-5 moles of HCl reacts with ( 4.03x10^-5 /2) = 2.015x10^-5 moles of Cd(OH)2

So Molarity of Cd(OH)2 , M = Number of moles / volume in L

                                        = ( 2.015x10^-5 mol) / 1.2 L

                                        = 1.68x10^-5 M

Cd(OH)2 ----> Cd+ + 2 OH-

So [Cd²⁺] = [Cd(OH)₂] = 1.68x10^-5 M

[OH^-] = 2x [Cd(OH)₂] = 2x1.68x10^-5 M = 3.36x10^-5 M

Solubility product constant , Ksp = [Cd²⁺] [OH^-]2

                                                = (1.68x10^-5 )(3.36x10^-5 )²

                                                = 1.89x10^-14