Solutions of potassium carbonate, copper(II) chloride, and lead(II) nitrate are each added to a beaker. Will any compound(s) precipitate from solution? Write a balanced net ionic equation for any precipitation reaction that occurs.
I have gotten this so far:
K2CO3(aq) + CuCl2(aq) + Pb(NO3)2(aq) -> ?
But I am not sure how to complete the equation with 3 reactants.
Remember that reactions occur for one of a few reasons.
1. a ppt is formed
2. a gas is formed
3. a slightly ionized substance is formed.
This problem requires that you know the solubility of compounds. Here is a site that gives a simplified set of solubility rules.
http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
Therefore you see that PbCO3 is insoluble which means we can write
K2CO3 + Pb(NO3)2 ==> 2KNO3 + PbCO3(s)
Now you need to turn that into a net ionic equation to answer that part of the question.
This is not the only reaction. Can you find the other one?
So when solving this equation, we only use two of the reactants and not all three?
And since there is more than one reaction, do we use the net ionic equations of both as the answer for this question or just one equation?
So would the other reaction be this:
CuCl2(aq) + Pb(NO3)2(aq) --> Cu(NO3)2(aq) + PbCl2(s)
Thank you so much!
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To determine if any compound(s) will precipitate from solution, we need to examine the solubility of the products formed after the reaction between potassium carbonate (K2CO3), copper(II) chloride (CuCl2), and lead(II) nitrate (Pb(NO3)2).
First, let's write out the ionic compounds as separate ions:
K2CO3(aq) contains K+(aq) and CO3^2-(aq)
CuCl2(aq) contains Cu^2+(aq) and 2Cl^-(aq)
Pb(NO3)2(aq) contains Pb^2+(aq) and 2NO3^-(aq)
Now, we can combine the ions to see if any compounds form:
K+(aq) + Cu^2+(aq) + 2NO3^-(aq) + 2Cl^-(aq) + Pb^2+(aq) + CO3^2-(aq)
To determine if any precipitate forms, we need to reference a solubility table. Based on solubility rules, we find that most nitrates and chlorides are soluble, except for a few exceptions like lead(II) chloride (PbCl2) which is insoluble. On the other hand, most carbonates are insoluble except for alkali metal carbonates, such as potassium carbonate (K2CO3), which are soluble.
Therefore, the balanced net ionic equation for the precipitation reaction can be written as follows:
Pb^2+(aq) + 2Cl^-(aq) -> PbCl2(s)
In this reaction, lead(II) ions (Pb^2+) from lead(II) nitrate (Pb(NO3)2) react with chloride ions (Cl^-) to form a solid precipitate of lead(II) chloride (PbCl2).
It's important to note that the potassium ions (K+) and carbonate ions (CO3^2-) from the potassium carbonate (K2CO3) solution and the copper(II) ions (Cu^2+) from copper(II) chloride (CuCl2) solution do not participate in the precipitation reaction and remain in the solution.
Yes that is another reaction. There is a third one with the formation of CuCO3(s),
And yes you write three net ionic equations.
Pb^2+(aq) + 2Cl^-(aq) ==> PbCl2(s) is one of them.
Pb^2+(aq) + CO3^2-(aq) ==> PbCO3(s) is another one.
Then Cu^2+(aq) + CO3^2-(aq) ==> CuCO3(s) is the third one.