The solubility or calcium carbonate in pure water is 6.9 x 10^-5 moles per liter. what is its molar solubility in a 0.373 M CaCl2 solution?

Question

i think my answer is wrong, please help.

Answer 1

Hi Jina L.,

You didn't include your answer, so it's a little difficult to judge whether it is right or wrong. I would say, perhaps you are meta-wrong to have left it out?

From the molar solubility, and knowing the respective ions involved, you can write the dissolution reaction:

CaCO3 = Ca(2+) + (CO3)(2-)

So, at saturation in water, each of those ions is at the molar solubility, itself.

Therefore, Ksp(CaCO3) = (6.9*10^-5)^2 M² = [Ca(2+)][CO3(2-)]

Now, when you put the CaCO3 into the 0.373 M CaCl2 solution, the additional Ca(2+) from dissolution of the CaCO3 is negligible, compared with the Ca(2+) already in solution. So you don't need to do an ICE table, at all. You can immediately write the concentration of the Ca(2+) : 0.373 M . So plug that value into the Ksp expression, and solve for [CO3](2-).

Would you like that with a twist of lime (stone)?

--Cheers, --Mr. d.