Question

Wastewater resulting from metal processing is polluted with cadmium ions, Cd²⁺, which must be removed from the water. In one method, the water is treated with a solution of concentrated sodium hydroxide. Suppose 1.00 x 10² L of wastewater containing 1.2 x 10^–5 M Cd²⁺ is treated with 1.0 L of 6.0 M NaOH solution.   The K_sp of Cd(OH)₂ is 7.2 x 10^-15

What is the residual concentration of aqueous cadmium ion after treatment?

Answer 1

After treatment, the solid is separated, so the equilibirum concentration of Cd+2 will remain as "residual"

So

Cd(OH)2 = Cd+2 + 2OH-

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

V = 1*10^2 = 100 L

[Cd+2] = 1.2*10^-5 M

M = 6 NaOH and V = 1L

mol of OH = MV = 6*1 = 6 mol of OH-

V total = 100+1 = 101 L

so

[OH-] = 6/101 = 0.005928 M

[Cd+2] = (1.2*10^-5)(100)/101 = 0.0000118

Ksp = 2*10^-14

2*10^-14 = (Cd+2)(0.005928^2)

[Cd+2] = (2*10^-14)/(0.005928^2) = 0.000000000569133 M

residual cd+2 = 0.000000000569133 M