Calcium carbonate, CaCO 3 , has a K sp value of 1.4 ´ 10 - 8 , the solubility of CaCO 3 , in mol/L is?

Question

Calcium carbonate, CaCO 3 , has a K sp value of 1.4 ´ 10 - 8 , the solubility of CaCO 3 , in mol/L is?

Answer 1

I dont think you can balance that equation.

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Answer 2

First write the equation for Ksp:

CaCO3(s) --> Ca^2+(aq) + CO3^2-(aq)

Ksp = [Ca^2+(aq)][CO3^2-(aq)] / [CaCO3(s)]

Then recognize that the concentration of CaCO3(s) remains constant, and is taken as "1".

Realize that when dissolving, the concentration of Ca^2+(aq) is the same as the concentration of CO3^2-(aq): a 1:1 mole ratio from the formula CaCO3(s).

[Ca^2+(aq)] = [CO3^2-(aq)]

Substitute [Ca^2+(aq)] for [CO3^2-(aq)].

Ksp = [Ca^2+(aq)][Ca^2+(aq)] / 1

Ksp = [Ca^2+(aq)]^2 = 1.4 x 10^-8

[Ca^2(aq)] = sq rt 1.4 x 10^-8 = 1.2 x 10^-4 mol/L

Since the concentration of Ca^2+ = the concentration of CO3^2-, this is the concentration of the dissolved CaCO3.

[CaCO3(aq)] = 1.2 x 10^-4 mol/L

This is the solubility of CaCO3.