Three solutions are mixed together to form a single solution. One contains 0.2 mol Pb(CH3COO)2, the second contains 0.1 mol Na2S, and the third contains 0.1 mol CaCl2. Write the net ionic equations for the precipitation reaction or reactions that occur.
Pb^2+(aq) + S^=(aq) ==> PbS(s)
Pb^2+(aq) + 2Cl^-(aq) ==> PbCl2(s)
Note: Both reactions will not take place unless there is enough Pb^2+ to ppt BOTH. PbS will ppt first and if there is enough lead ion present, PbCl2 will ppt next.
Sure, let me put on my chemistry clown nose for this one!
To determine the net ionic equation for the precipitation reaction, we first need to identify the possible combinations of ions that could form a precipitate. In this case, we have Pb(CH3COO)2, Na2S, and CaCl2.
When we mix these solutions together, Pb2+ from Pb(CH3COO)2 would react with S2- from Na2S to form PbS, which is insoluble and precipitates out of solution.
The net ionic equation for this reaction can be written as follows:
Pb2+ (aq) + S2- (aq) → PbS (s)
So, when you mix the solutions, you'll get a solid cloud of PbS falling like rain! I hope this precipitation reaction brings a smile to your face!
To determine the net ionic equations for the precipitation reactions, we need to identify the possible combinations that can form precipitates. Let's analyze the three possible combinations:
1. Pb(CH3COO)2 + Na2S:
- The cations are Pb2+ and Na+, and the anions are CH3COO- and S2-.
- The possible reaction is:
Pb2+ + S2- -> PbS ↓
2. Pb(CH3COO)2 + CaCl2:
- The cations are Pb2+ and Ca2+, and the anions are CH3COO- and Cl-.
- The possible reaction is:
Pb2+ + 2Cl- -> PbCl2 ↓
3. Na2S + CaCl2:
- The cations are Na+ and Ca2+, and the anions are S2- and Cl-.
- No precipitation reaction occurs between Na+ and Ca2+.
Therefore, the net ionic equations for the precipitation reactions that occur are:
1. Pb2+ + S2- -> PbS ↓
2. Pb2+ + 2Cl- -> PbCl2 ↓
To write the net ionic equations for the precipitation reactions, we need to determine if any insoluble products will form when the three solutions are mixed together.
First, let's write the chemical formulas for the three compounds involved:
- Lead(II) acetate: Pb(CH3COO)2
- Sodium sulfide: Na2S
- Calcium chloride: CaCl2
Next, let's identify any possible combinations of ions that could form insoluble compounds. We know that most acetates, sulfides, and chlorides are soluble except for a few exceptions.
Upon combining the three solutions:
1. Pb(CH3COO)2 + Na2S
Here, the acetate ions (CH3COO-) and the sulfide ions (S2-) could potentially form an insoluble compound (lead(II) sulfide). To write the net ionic equation, we need to know the states of the compounds (aqueous, solid, etc.). Assuming all compounds are in their aqueous state, the net ionic equation would be:
Pb2+ + S2- → PbS
2. Pb(CH3COO)2 + CaCl2
In this combination, there won't be any insoluble compound formed because both lead(II) acetate (Pb(CH3COO)2) and calcium chloride (CaCl2) are highly soluble in water.
3. Na2S + CaCl2
Again, there won't be any insoluble compound formed in this combination because both sodium sulfide (Na2S) and calcium chloride (CaCl2) are highly soluble in water.
So, the net ionic equation for the precipitation reaction(s) that occur when the three solutions are mixed will be:
Pb2+ + S2- → PbS