Question

Calculate the molar solubility of CaCO3 in 0.250M Na2CO3

Kps CaCO3 is 4.96x10-9

Answer 1

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

Where X is the amoun of CaCO₃ you can add, that means, solubility

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, solubility is 1.984x10⁻⁹M

Answer 2

Answer:

$1.984x10^{-8}M$

Explanation:

Hello,

In this case, the equilibrium reaction for the solubility of calcium carbonate is:

$CaCO_3(s) \rightleftharpoons Ca^{2+}(aq)+CO_3^{-2}(aq)$

In such a way, since 0.250 M sodium carbonate solution is the solvent, we assume an initial concentration of carbonate anion to be also 0.250 M since sodium carbonate is completely dissolved, for that reason the equilibrium equation turns out:

$Ksp=[Ca^{2+}][CO_3^{2-}]\\\\4.96x10^{-9}=x*(0.250+x)$

Hence, solving for $x$ we have:

$x=1.984x10^{-8}M$

Which corresponds to the molar solubility if calcium carbonate as well.

Regards.