Calculate the molar solubility of CaCO3 in 0.250M Na2CO3

Kps CaCO3 is 4.96x10-9

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

Explanation:

Hello,

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

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:

Hence, solving for we have:

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

Regards.

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.

solute : a solute is a substance that is dissolved in another substance is known as a solvent. it is mixed with a special mixture know as homogeneous mixture.

solvent: a solvent is a substance that dissolved a solute, which results in a solution is known as solvent.