Final answer:
The molar solubility of Ag2CO3 in water at 25°C is calculated as 1.3 x 10^-4 M, using the solubility product constant (Ksp) of 8.1 x 10^-12.
Step-by-step explanation:
The molar solubility of silver carbonate (Ag2CO3) in water can be calculated using its solubility product constant (Ksp), which is given as 8.1 × 10-12 at 25°C. If we let x be the molarity of Ag2CO3 that dissolves to form a saturated solution, the dissolution reaction would be:
Ag2CO3 (s) → 2Ag+ (aq) + CO32- (aq)
At equilibrium, the molar concentrations of the ions will be [Ag+] = 2x and [CO32-] = x. Plugging these values into the expression for Ksp, we get:
Ksp = [Ag+]2[CO32-] = (2x)2 × x = 4x3
8.1 × 10-12 = 4x3
Solving for x gives us the molar solubility of Ag2CO3, which is the molarity of a saturated solution:
x = ∓(8.1 × 10-12 / 4) = 1.3 × 10-4 M
Therefore, option D, 1.3 x 10-4 M, is the correct molar solubility of silver carbonate in pure water at 25°C.