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.