A) Ignoring any possible side reactions, calculate the molarity of Cd^2+ for a s

Question

A) Ignoring any possible side reactions, calculate the molarity of Cd^2+ for a saturated solution of damium sulfide in pure water.

B) Convert the concentration from part A into ppb Cd

C) The EPA-regulated MCL (maximum contaminant level) for cadmium is 5 ppb— much larger than the concentration you should have found above. However, the chemistry of CdS dissolution is much more complicated than the simple Ksp reaction. Among other issues, the sulfide ion is a relatively strong weak base. Write a balanced chemical equation (including physical states) for the transfer of a proton to the sulfide ion. Where does this proton come from? Does this side reaction increase or decrease the saturated concentration of Cd2+ relative to what you calculated above? Briefly explain your answer, citing Le Châtelier’s principle.

Please show all work!

Explanation / Answer

A)

CdS --> Cd2+ + S2-    : Ksp = 10-27

(Co - S) --> S + S

If S is the solubility of Cd2+ ions in pure water,

S2 = Ksp = 10-27

S = 3.16 x 10-14 M

B)

Atomic mass of Cd2+ = 112.411 g/mol

Mass of Cd2+ in 1 L solution, m = S * 112.411 g

= 3.55 x 10-12 g

Concentration, ppb = m * 109 / 103

= 3.55 x 10-6 ppb

C)

S2- is slightly basic and it can assimilate proton

S2-(aq) + H+(aq) --> HS- (aq)

Proton comes from water. Kw = [H+][OH-]

The above reaction will shift the CdS dissociation in forward direction according Le Chatelier’s principle. Sulphide anions will be replenished by augmentation of CdS dissociation.

Therefore, this side reaction will also increase the saturated concentration of Cd2+ relative to that calculated in part A.

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