The solubility of Ag2S is 2.48x10^-15 mol/dm3. Calculate its solubility product.

Write the equation:

Ag2S<<>>2Ag+ + S-- when in equilibrium

ksp= [Ag+]^2 [S--]/[Ag2S]

ksp= (2x)^2 (x)/x

you are given x as 2.48E-15

solve for ksp
=4x^2

To calculate the solubility product (Ksp) of Ag2S, we need to use the solubility value given and understand how it relates to the stoichiometry of the compound.

The chemical equation for the dissolution of Ag2S in water is:

Ag2S ⇌ 2Ag+ + S2-

The equilibrium expression for this reaction is:

Ksp = [Ag+]^2 [S2-]

According to the stoichiometry of the reaction, when Ag2S dissolves, it produces 2 moles of Ag+ ions and 1 mole of S2- ions. Therefore, the concentration of Ag+ ions equals twice the concentration of S2- ions.

Since the solubility of Ag2S is given as 2.48x10^-15 mol/dm3, the concentration of Ag+ ions and S2- ions can be represented as follows:

[Ag+] = 2x [S2-] = 2x (2.48x10^-15 mol/dm3) = 4.96x10^-15 mol/dm3 (approximately)

Now we can substitute these values into the equilibrium expression for Ksp:

Ksp = [Ag+]^2 [S2-] = (4.96x10^-15 mol/dm3)^2 (2.48x10^-15 mol/dm3)
= 6.14x10^-44 mol^3/dm^9 (approximately)

Therefore, the solubility product (Ksp) of Ag2S is approximately 6.14x10^-44 mol^3/dm^9.