Consider the following reaction:
Na2CO3 + NiCl2 �¨ NiCO3 + 2NaCl
Using the solubility rules determine the solubility of all reactants and products. Explain your answer Please.
Which product is the precipitate in this reaction?
Write the complete ionic equation for this reaction:
Write the net ionic equation for this reaction:
You should have the phase; since you don't I will add what I think is appropriate.
Na2CO3(aq) + NiCl2(aq) ==> NiCO3(s) + 2NaCl(aq)
2Na^+(aq) + CO3^2-(aq) + Ni^2+(aq) + 2Cl^-(aq) ==> NiCO3(s) + 2Na^+(aq) + 2Cl^-(aq)
This is the complete ionic equation. To make it a net ionic equation you go through and cancel ions that are the same on each side of the equation What is left is the net ionic equation. For example, there is a Cl^- on both side; cancel them. Etc.
So the net equation is CO3^2-(aq)+Ni^2+(aq)----->NiCO3(s)
Here is a simplified set of solubility rules. You should memorize this table.
http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
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To determine the solubility of the reactants and products in the given reaction, we can use solubility rules. Solubility rules are guidelines that help us predict whether a substance will dissolve in water to form a solution or precipitate out as a solid.
Here are some common solubility rules to consider:
1. All common compounds of Group IA metals (such as Na+) and ammonium (NH4+) are soluble.
2. Most nitrate (NO3-) compounds are soluble.
3. Most chlorides (Cl-) are soluble, except for those of silver (Ag+), lead (Pb2+), and mercury (Hg2+).
4. Most carbonates (CO3^2-), phosphates (PO4^3-), sulfides (S^2-), and hydroxides (OH-) are insoluble, except for those of Group IA metals and ammonium.
Now let's apply these rules to each of the reactants and products in the reaction:
1. Na2CO3 (sodium carbonate) - According to rule #1, all common compounds of Group IA metals (such as Na+) are soluble, so Na2CO3 is soluble.
2. NiCl2 (nickel chloride) - According to rule #3, most chlorides (Cl-) are soluble, except for those of silver (Ag+), lead (Pb2+), and mercury (Hg2+). Therefore, NiCl2 is soluble.
3. NiCO3 (nickel carbonate) - According to rule #4, carbonates (CO3^2-) are generally insoluble except for those of Group IA metals. Since nickel (Ni) is not a Group IA metal, NiCO3 is insoluble.
4. NaCl (sodium chloride) - According to rule #1, NaCl is soluble as it contains a Group IA metal, sodium (Na+).
From the solubility rules, we can conclude that all reactants (Na2CO3 and NiCl2) are soluble, while one of the products (NiCO3) is insoluble.
Next, let's identify the precipitate in this reaction. A precipitate is the solid that forms when two solutions react and a solid product is formed. In this case, the only insoluble product is NiCO3 (nickel carbonate). Therefore, NiCO3 is the precipitate in this reaction.
To write the complete ionic equation for this reaction, we need to break down all the compounds into their ions:
Na2CO3 (aq) + NiCl2 (aq) → NiCO3 (s) + 2NaCl (aq)
Complete ionic equation:
2Na+ (aq) + CO3^2- (aq) + Ni^2+ (aq) + 2Cl- (aq) → NiCO3 (s) + 2Na+ (aq) + 2Cl- (aq)
Finally, let's write the net ionic equation. The net ionic equation only includes the species directly involved in the reaction and eliminates spectator ions (ions that do not participate in any chemical change):
CO3^2- (aq) + Ni^2+ (aq) → NiCO3 (s)
In the net ionic equation, we can see that 2Na+ and 2Cl- ions are present on both sides but do not participate in the actual reaction, so we exclude them.
I hope this explanation helps you understand the solubility of the reactants and products, as well as how to write the complete and net ionic equations for this reaction.