Will CuCI2+Na2SO4->CuSO4=NaCI2 precipitate and how can you determine this?
By knowing the solubility rules. Here is a simplified set of rules. Memorize them.
http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
To determine if a precipitate will form when CuCl2 reacts with Na2SO4 to form CuSO4 and NaCl, we need to examine the solubility rules and the ionic equation of the reaction.
First, we need to write the balanced ionic equation:
CuCl2 + Na2SO4 → CuSO4 + 2NaCl
According to the solubility rules:
1. Most chloride (Cl-) salts are soluble, except those of silver (Ag+), lead (Pb2+), and mercury (Hg2+).
2. Most sulfate (SO42-) salts are soluble, except those of calcium (Ca2+), strontium (Sr2+), barium (Ba2+), lead (Pb2+), and silver (Ag+).
Based on these solubility rules, we can determine if a precipitate will form in this reaction.
CuCl2 contains the chloride ion (Cl-), which is generally soluble. Na2SO4 contains the sulfate ion (SO42-), which is also generally soluble. CuSO4 contains both the sulfate and copper ions, both of which are generally soluble.
NaCl, on the other hand, contains the chloride ion, which is soluble.
Since all the ions in the reaction (Cu2+, Cl-, Na+, SO42-) are generally soluble, no precipitate will form in this reaction. Both the products, CuSO4 and NaCl, will remain dissolved in the solution.
In summary, based on the solubility rules, we can determine that the reaction CuCl2 + Na2SO4 → CuSO4 + 2NaCl will not produce a precipitate.