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
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To find the molar solubility of CdS in these solutions, we need to determine the pH at which CdS starts to precipitate out of the solution. This is because the solubility of a compound like CdS is affected by the pH of the solution.
To find the molar solubility of CdS, we will make use of the solubility product constant (Ksp) expression for CdS:
CdS ↔ Cd2+ + S2-
The Ksp expression can be written as:
Ksp = [Cd2+][S2-]
Under equilibrium conditions, the concentration of Cd2+ and S2- ions will be equal to the molar solubility of CdS, denoted as "s". Therefore, we can rewrite the Ksp expression as:
Ksp = s^2
Now, let's consider each of the given pH values and determine whether CdS will precipitate out or remain soluble.
1. pH = 8.0:
At pH 8.0, the solution is basic. Since CdS is an insoluble compound, it is likely to precipitate to some extent. We can calculate the concentration of Cd2+ and S2- ions using the given total concentration of NH3 + NH4Cl = 0.10 M and their respective acid-base reactions. Then, substitute these concentrations into the Ksp expression to find the molar solubility of CdS.
Repeat the above steps for each pH value (8.5, 9.0, 9.5, 10.0, 10.5, and 11.0) to determine the molar solubility of CdS in each solution.