Final answer:
The molar solubility of silver(I) carbonate with a Ksp of 6.3 x 10^-9 is 1.3 x 10^-3 M. This value represents the concentration of silver(I) carbonate that can dissolve in a saturated solution at equilibrium.
Step-by-step explanation:
The student is asking about the molar solubility of silver(I) carbonate given its solubility product constant (Ksp) at a specific temperature. To find the molar solubility of silver(I) carbonate, we first need to write down the balanced dissolution equation:
Ag2CO3(s) ⇌ 2Ag+(aq) + CO32-(aq)
If s represents the molar solubility of Ag2CO3, then at equilibrium, the concentration of silver ions will be 2s and the concentration of carbonate ions will be s. The Ksp expression for Ag2CO3 is:
Ksp = [Ag+]2[CO32-]
Substituting the concentrations, we have:
6.3 x 10-9 = (2s)2(s)
6.3 x 10-9 = 4s3
Solving for s:
s = √(6.3 x 10-9 / 4)
s ≈ 1.3 x 10-3 M
This gives us the molar solubility of Ag2CO3. The molar solubility is the concentration of Ag2CO3 in a saturated solution, which can be used to determine how much will dissolve in a certain volume of water.