Challenge Problem:
(a) The solubility of CdS is normally very low, but it can be increased by lowering the solution pH. Calculate the molar solubility of CdS as a function of pH from pH 11 to pH 1. Find values at every 0.5 pH unit and plot solubility versus pH.
(b) A solution contains 1 × 10-4 M of both Fe2+ and Cd2+. Sulfide ions are slowly added to this solution to precipitate either FeS or CdS. Determine which ion precipitates first and the range of S2- concentration that will allow a clean separation of the two ions.
(c) The analytical concentration of H2S in a solution saturated with H2S(g) is 0.10 M. What pH range is necessary for the clean separation described in part (b)?
(d) If there is no pH control from a buffer, what is the pH of a saturated H2S solution?
(e) Plot the α0 and α1 values for H2S over the pH range of 10 to 1.
(f) A solution contains H2S and NH3. Four Cd2+ complexes form with NH3 in a stepwise fashion: Cd(NH3)2+, Cd(NH3)22+, Cd(NH3)32+, and Cd(NH3)42+. Find the molar solubility of CdS in a solution of 0.1 M NH3.
(g) For the same solution components as in part (f), buffers are prepared with a total concentration of NH3 + NH4Cl = 0.10 M. The pH values are 8.0, 8.5, 9.0, 9.5, 10.0, 10.5, and 11.0. Find the molar solubility of CdS in these solutions.
(h) For the solutions in part (g), how could you determine whether the solubility increase with pH is due to complex formation or to an activity effect?
We need at least 10 more requests to produce the solution.
0 / 10 have requested this problem solution
The more requests, the faster the answer.