Final answer:
The molar solubility of BaCO₃ in a 0.10 M Na₂CO₃ solution is calculated by considering the common ion effect and using the Ksp value of BaCO₃. The final solubility is found to be 8.1 x 10⁻¸ M.
Step-by-step explanation:
To calculate the molar solubility of BaCO₃ in a 0.10 M solution of Na₂CO₃, we need to understand that the presence of a common ion (CO₃²⁻ from Na₂CO₃) will affect the solubility of BaCO₃. The solubility product (Ksp) for BaCO₃ is 8.1 x 10⁻¹ at 25°C. In a solution containing 0.10 M Na₂CO₃, the CO₃²⁻ concentration is essentially fixed at 0.10 M due to the high concentration of the common ion. This suppresses the solubility of BaCO₃ due to the common ion effect.
The solubility equilibrium can be written as:
BaCO₃(s) ⇌ Ba²⁻(aq) + CO₃²⁻(aq)
In the presence of 0.10 M CO₃²⁻, we let the molar solubility of BaCO₃ be 's' where s is much smaller than 0.10 M. Therefore, at equilibrium we have:
[Ba²⁻] = s
[CO₃²⁻] = 0.10 + s ≈ 0.10 M
The Ksp expression for BaCO₃ is:
Ksp = [Ba²⁻][CO₃²⁻] = s(0.10) = 8.1 x 10⁻¹
Solving for s gives us the molar solubility of BaCO₃ in the presence of a common ion:
s = Ksp / [CO₃²⁻]
s = 8.1 x 10⁻¹ / 0.10 = 8.1 x 10⁻¸ M
The molar solubility of BaCO₃ in 0.10 M Na₂CO₃ solution is therefore 8.1 x 10⁻¸ mol/L.