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lisanoce · May 12, 2009, 02:41 PM

A chemist adds 0.010 g of CaCl2 to 5.0 x 10^2 mL of 0.0015 mol/l sodium carbonate, Na2CO3. Does a precipitate of calcium carbonate form? Include a balanced chemical equation for the formation of the possible precipitate.

This is my balanced chemical equation: CaCl2(aq) + Na2CO3(aq) --> CaCO3(?) + 2NaCl(aq)

I just don't get how to solve this problem .thanks in advance

answer in the back of my book is
Qsp = 2.7 x 10^-7 > Ksp, therfore a precipitate forms

Perito · May 12, 2009, 05:00 PM

A chemist adds 0.010 g of CaCl2 to 5.0 x 10^2 mL of 0.0015 mol/l sodium carbonate, Na2CO3. Does a precipitate of calcium carbonate form? Include a balanced chemical equation for the formation of the possible precipitate.

This is my balanced chemical equation: CaCl2(aq) + Na2CO3(aq) --> CaCO3(?) + 2NaCl(aq)

answer in the back of my book is
Qsp = 2.7 x 10^-7 > Ksp, therfore a precipitate forms

The Qsp is the solubility product for Calcium carbonate -- at least I think that's what it is. I've always used "Ksp" and not "Qsp".

$Ca^{+2} + CO_3^{-2} \rightleftharpoons CaCO_3$

By definition, the Ksp for Calcium carbonate

$K_sp = [Ca^{+2}][CO_3^{-2}]=2.7 \times 10^{-7}$

where the [] represents concentration.

You know how much CaCl2 was added, so you should be able to calculate the concentration of Ca+2 that will be present when it dissolves and ionizes:

$CaCl_2 \rightleftharpoons Ca^{+2} + 2Cl^-$

You are given the concentration of carbonate. The amount, 5.0 x 10^2 mL, is irrelevant to this problem.

$Na_2CO_3 \rightleftharpoons 2 Na^+ + CO_3^{2-}$

$[CO_3^{-2}] = 0.0015$

If the product of the calcium ion concentration and the carbonate concentration exceeds the solubility product, a precipitate will appear.