Answer:
solubility is 1.984x10⁻⁹M
Explanation:
When CaCO₃ is in water, the equilibrium that occurs is:
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Kps = [Ca²⁺] [CO₃²⁻] = 4.96x10⁻⁹
If you have a 0.250M solution of Na₂CO₃, [CO₃²⁻] = 0.250M:
[Ca²⁺] [0.250M] = 4.96x10⁻⁹
Assuming you are adding an amount of CaCO₃:
[X] [0.250 + X] = 4.96x10⁻⁹
<em>Where X is the amoun of CaCO₃ you can add, that means, solubility</em>
X² + 0.250X - 4.96x10⁻⁹ = 0
Solving for X:
X = -0.25M → False answer, there is no negative concentrations.
X = 1.984x10⁻⁹M.
That means, <em>solubility is 1.984x10⁻⁹M</em>
