Question

You have a reactor that is releasing Cd^(2+). Your effluent standards state that the concentration of Cd^(2+) leaving the reactor has to be less than 0.01 mg/L. There are two possibilities to control the Cd^(2+) concentration: control the pH or add sulfide ( S^(2-) ). The technician running the reactor reports that adjusting the pH of the reactor to 8.5 or adding enough sulfide such that [S^(2-) ]=0.005 M will meet the effluent goal. Given the following information, will either of these suggestions work?

Cd(OH)2(s) <=> Cd^(2+) + 2OH^(-)

Ks <=> 10-14.27

CdS(s) <=> Cd^(2+) + S^(2-)

Ks <=> 10-27

Answer 1

For the first reaction:

Cd(OH)₂(s) <=> Cd²⁺ + 2OH^-

Ks = 10^-14.27

= 5.37x10^-15

pH = 8.5

pOH = 14 - pH

= 14 - 8.5 = 5.5

[OH-] = 10^-5.5

= 3.16x10^-6

Ks = [Cd²⁺][OH-]²

5.37x10^-15 =  [Cd²⁺][3.16x10^-6]²  

5.37x10^-15 =  [Cd²⁺] * 9.98x10^-12

[Cd²⁺] = 5.37x10^-15 / 9.98x10^-12

= 0.538x10^-3 M

Molar mass of Cd²⁺ = 112.41 g/mole

[Cd²⁺] =  0.538x10^-3 * 112.41

= 0.0604 g/L

[Cd²⁺] = 60.47 mg/L

For second reaction:

CdS(s) <=> Cd²⁺ + S^2-

Ks = 10^-27

[S^2-] = 0.005 M

Ks = [Cd²⁺][S^2-]

10^-27 = [Cd²⁺] * 0.005

[Cd²⁺] = 2.0x10^-25 M

= 2.0x10^-25 * 112.41

= 2.24x10^-23 g/L

[Cd²⁺]= 2.24x10^-20 mg/L

Since [Cd²⁺] is less than 0.01 mg/L in second suggestion. Hence, the second suggestion will work.