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Trava [24]
3 years ago
9

Calculate the molar solubility of CaCO3 in 0.250M Na2CO3 Kps CaCO3 is 4.96x10-9

Chemistry
2 answers:
Sveta_85 [38]3 years ago
5 0

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.

Yuliya22 [10]3 years ago
3 0

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>

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