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 in a 0.01 M aqueous solution.

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 an ICE (Initial, Change, Equilibrium) table.

For pure water: Initial concentrations are 0 for Ag+ and S2-. The change in Ag+ is +2x (since there are 2 moles of Ag+ for every mole of Ag2S that dissolves) and +x for S2-. At equilibrium, we have 2x for Ag+ and x for S2-.

Step 4: Substitute the equilibrium concentrations into the Ksp expression and solve for x. 4x10^-48 = (2x)^2 * x 4x10^-48 = 4x^3 x^3 = 1x10^-48 x = (1x10^-48)^(1/3) x = 1x10^-16 M

So, the solubility of Ag2S in pure water is 1x10^-16 M.

For the 0.01 M aqueous solution of Ag2S: The initial concentration of Ag+ is 0.01 M. The change in Ag+ is still +2x and +x for S2-. At equilibrium, we have 0.01+2x for Ag+ and x for S2-.

Substitute the equilibrium concentrations into the Ksp expression and solve for x. 4x10^-48 = (0.01+2x)^2 * x This is a cubic equation and may be difficult to solve by hand. You may need to use a calculator or computer software to find the solution. The solubility will be less than in pure water due to the common ion effect.