Question

Will a precipitate form if 100.0 mL of 2.5 × 10−3M Cd(NO3)2 and 75.0 mL of 0.0500 M NaOH are mixed at 25°C? Calculate the concentration of cadmium ion in solution at equilibrium after the two solutions are mixed. (Ksp is 7.2 × 10−15 for Cd(OH)2 at 25°C)

Answer 1

moles of Cd(NO3)2 = 100 x 2.5 x 10^-3 / 1000 = 2.5 x 10^-4

moles of NaOH = 75 x 0.05/1000 = 3.75 x 10^-3

Cd(NO3)2      +     2 NaOH ---------------------> Cd(OH)2 + 2NaNO3

2.5 x 10^-4         3.75 x 10^-3                         2.5 x 10^-4

concentration of Cd(OH)2 = moles / total volume

                                          = 2.5 x 10^-4 / (0.100 + 0.075)

                                          = 1.43 x 10^-3 M

[OH-] concentration = 2 x 1.43 x 10^-3 = 2.86 x 10^-3 M

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

                                                S          2.86 x 10^-3

Ksp = [Cd+2][OH-]^2

7.2 × 10^−15 = S x (2.86 x 10^-3)^2

S = 8.82 x 10^-10 M

[Cd+2] = 8.82 x 10^-10 M