Ksp of Ag2S is 4  10-48 at 250C. Calculate its solubility in a pure water and 0.01 M aqueous solution of Ag2S.

The solubility product constant, Ksp, is a measure of the solubility of a compound. In this case, we are given the Ksp of Ag2S and asked to calculate its solubility in pure water and a 0.01 M aqueous solution of Ag2S.

Step 1: Write the balanced chemical equation for the dissolution of Ag2S. Ag2S(s) ⇌ 2Ag+(aq) + S2-(aq)

Step 2: Write the expression for the solubility product constant. Ksp = [Ag+]^2[S2-]

Step 3: Set up the equilibrium concentrations. Since Ag2S is sparingly soluble, we can assume that the initial concentration of Ag+ and S2- is zero. Let s be the solubility of Ag2S in mol/L. At equilibrium, the concentration of Ag+ will be 2s (since there are two Ag+ ions for each Ag2S unit) and the concentration of S2- will be s.

Step 4: Substitute the equilibrium concentrations into the Ksp expression. Ksp = (2s)^2 * s = 4s^3

Step 5: Solve for s. 4s^3 = 4 * 10^-48 s^3 = 10^-48 s = (10^-48)^(1/3) = 10^-16 M

So, the solubility of Ag2S in pure water is 10^-16 M.

For the 0.01 M aqueous solution of Ag2S, the concentration of Ag+ is already 0.01 M due to the Ag2S in the solution. Therefore, the additional solubility of Ag2S will be negligible compared to this concentration. So, the solubility of Ag2S in the 0.01 M solution will be approximately the same as in pure water, 10^-16 M.