You have a reactor that is releasing Cd^(2+). Your effluent standards state that

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

Explanation / Answer

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+ + S2-

Ks = 10-27

[S2-] = 0.005 M

Ks = [Cd2+][S2-]

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.

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.