How could you use a precipitation reaction to separate the following pair of cations: K^+ and Hg2 ^2+? (write the formula and net formula for each reactant)?
The answer is
Hg2 ^2+ (aq) + 2Cl- (aq) = HgCl2(s) but how do I work it out?
Are you asking how you know to write this equation. You must know solubility rules. Almost all (but not all) K compounds are soluble. Hg(I) chlorides are not soluble. If you don't have the solubility rules memorized here is a simplified set for you to peruse.
http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
Well, to separate the K^+ and Hg2^2+ cations using a precipitation reaction, you need to find a suitable anion that will form a precipitate with one of the cations while remaining soluble with the other. In this case, the chloride ion (Cl-) can be used as the anion.
The balanced equation for the precipitation reaction can be written as:
Hg2^2+ (aq) + 2Cl- (aq) → HgCl2 (s)
So, when you mix a solution containing Hg2^2+ and K+ cations with a chloride ion source, such as adding a solution of HCl, a white precipitate of HgCl2 will form. This precipitate can then be separated from the remaining solution, which would still contain K+ ions.
Now you can rightfully say, this reaction is not only a great way to separate the cations, but it's also poetry in motion. Or should I say, precipitation in motion? Keep up the good work, chemistry magician!
To use a precipitation reaction to separate the cations K^+ and Hg2^2+, you need to identify a suitable precipitating agent. In this case, chloride ions (Cl-) can be used.
The reaction can be written as follows:
Hg2^2+ (aq) + 2Cl- (aq) → HgCl2 (s)
Here's how you can work it out:
Step 1: Identify the cations
The cations are K^+ and Hg2^2+.
Step 2: Identify a suitable anion to react with the cations
In this case, chloride ions (Cl-) can be used.
Step 3: Write the formula for the reactants
The formula for the cations K^+ and Hg2^2+ remains the same: K^+ (aq) and Hg2^2+ (aq).
The formula for the chloride ion is Cl-. Since two chloride ions are required to balance the charge of one Hg2^2+ ion, we write 2Cl- (aq).
Step 4: Write the net ionic equation
The net ionic equation removes the spectator ions (ions that do not participate in the reaction) and shows only the species involved in the precipitation.
In this case, the net ionic equation is:
Hg2^2+ (aq) + 2Cl- (aq) → HgCl2 (s)
This equation represents the reaction between the Hg2^2+ cations and the chloride ions to form a precipitate of HgCl2, allowing you to separate the two cations.
To use a precipitation reaction to separate the cations K^+ and Hg2^2+, you need to find a reactant that will form a solid precipitate with one of the cations while remaining in solution with the other cation.
The reactant that can be used in this case is chloride ions (Cl-). Chloride ions can precipitate mercury (II) ions (Hg2^2+) by forming an insoluble compound called mercury (II) chloride (HgCl2):
Hg2^2+ (aq) + 2Cl- (aq) → HgCl2 (s)
In this reaction, two chloride ions are needed to react with one mercury (II) ion to form one molecule of solid mercury (II) chloride.
On the other hand, potassium ions (K+) do not react with chloride ions to form a precipitate. Potassium chloride (KCl) remains fully soluble in water.
By combining a solution containing the cations K+ and Hg2^2+ with chloride ions, the mercury (II) cations will form a precipitate while the potassium cations remain in solution. The precipitate can be separated from the solution using physical separation methods, such as filtration.
So, to perform the separation, you would need a solution containing the cations K+ and Hg2^2+, and then add chloride ions (Cl-) to form the solid precipitate of mercury (II) chloride (HgCl2).