Final answer:
To find the Ksp for Ag₂CO₃ using the solubility of 0.035 g/L, convert the solubility to molarity and then use the disassociation equation to find the concentrations of Ag⁺ and CO₃²⁻. The Ksp is calculated as 4x³, resulting in a value of approximately 8.17×10⁻¹ⁱ when rounded to two significant digits.
Step-by-step explanation:
To calculate the solubility product constant (Ksp) for Ag₂CO₃ at 25 °C using the solubility of 0.035 g/L, we first must convert this solubility into molarity. Since the molar mass of Ag₂CO₃ is 275.75 g/mol, we can determine the molarity by dividing the grams per liter by the molar mass:
Solubility in molarity (M) = 0.035 g/L ÷ 275.75 g/mol = 1.27×10⁻´ M
The disassociation of Ag₂CO₃ in water is represented by the equation:
Ag₂CO₃(s) → 2Ag⁺(aq) + CO₃²⁻(aq)
At equilibrium, the concentration of Ag⁺ will be 2x and that of CO₃²⁻ will be x, where x is the molarity of the saturated solution of Ag₂CO₃. Substituting into the Ksp expression we get:
Ksp = [Ag⁺]²[CO₃²⁻] = (2x)²(x) = 4x³
Given that x = 1.27×10⁻´ M, we can calculate Ksp:
Ksp = 4(1.27×10⁻´)³ = 8.17×10⁻¹¹ⁱ
Therefore, the Ksp for Ag₂CO₃ at 25 °C is approximately 8.17×10⁻¹ⁱ when rounded to two significant digits.