What is the molar solubility ofAg2CO3 (Ksp = 4 × 10–13) in 0.1 M Na2CO3 solution ?10–610–72 × 10–62 × 10–7

To solve this problem, we need to understand the concept of molar solubility and the solubility product constant (Ksp).

1. Write the balanced chemical equation for the dissolution of Ag2CO3 in water: Ag2CO3(s) ⇌ 2Ag+(aq) + CO3^2-(aq)

2. Write the expression for the solubility product constant (Ksp): Ksp = [Ag+]^2[CO3^2-]

3. In a 0.1 M Na2CO3 solution, the concentration of CO3^2- ions is 0.1 M. Let's assume the molar solubility of Ag2CO3 is 's'. Therefore, the concentration of Ag+ ions is 2s.

4. Substitute these values into the Ksp expression: 4 × 10^-13 = (2s)^2 * 0.1

5. Simplify the equation to solve for 's': 4 × 10^-13 = 4s^2 * 0.1 s^2 = 10^-14 s = sqrt(10^-14) s = 10^-7 M

So, the molar solubility of Ag2CO3 in 0.1 M Na2CO3 solution is 10^-7 M.