Calculate the solubility of BaCO3 (a) in pure water and (b) in a solution in which [CO32-] = 0.289 M. Solubility in pure water = M Solubility in 0.289 M CO32- = M

the solubility of BaCO₃ in pure  water  and in a solution is 4.472 × 10⁻⁵ M and 6.9204 × 10⁻⁹ M respectively.

Explanation:

To calculate the solubility of BaCO₃ in:

(a) pure water and (b) in a solution in which [CO₃²⁻] = 0.289 M

The (i.e solubility constant ) for BaCO₃= 2.0 × 10⁻⁹

BaCO₃   → Ba²⁺  + CO₃²⁻

ksp = s × s

s² = ksp

s =

s =

s = 4.472 × 10⁻⁵ M

(b) The solubility of BaCO₃ in a solution in which [CO₃²⁻] = 0.289 M

BaCO₃   → Ba²⁺  + CO₃²⁻

ksp = s × s

2.0 × 10⁻⁹ = s × 0.289

s = 2.0 × 10⁻⁹/0.289

s = 6.9204 × 10⁻⁹ M

Thus, the solubility of BaCO₃ in pure  water  and in a solution is 4.472 × 10⁻⁵ M and 6.9204 × 10⁻⁹ M respectively.

the answer is c

tell me if wrong