If I added sodium carbonate to magnesium nitrate, would it be insoluble & a precipitate forms?
If I added sodium carbonate to calcium nitrate, would it be insoluble & precipitate forms?
Adding sodium carbonate to strontium nitrate: insoluble, precipitate forms?
adding sodium carbonate to barium nitrate--> insoluble, precipitate forms?
All of these look ok to me.
Just as a side note, the formation of ppts and their color and what they will dissolve in, is the basis for qualitative analysis. For example, if I added dilute HCl to an aqueous solution of an unknown and I obtained a white ppt, I would immediately thing of Hg2Cl2, AgCl, AgBr, AgI, and PbCl2.
If it dissolves in hot water it's Pb. If it dissolves in NH3 it's one of the chlorides of Ag. And if it turns black when NH3 is added to test for the Ag halides, it is Hg(+1--not +2).
Ok, thanks!
When sodium carbonate is added to magnesium nitrate, the reaction that takes place is as follows:
Mg(NO3)2(aq) + Na2CO3(aq) → MgCO3(s) + 2NaNO3(aq)
In this reaction, a precipitate of magnesium carbonate (MgCO3) is formed. Magnesium carbonate is insoluble in water, so it will appear as a solid (precipitate) in the solution.
Now, let's consider adding sodium carbonate to calcium nitrate:
Ca(NO3)2(aq) + Na2CO3(aq) → No precipitate forms
When sodium carbonate is added to calcium nitrate, no precipitate is formed because calcium carbonate (CaCO3) is soluble in water. Therefore, it dissolves and remains in the solution.
Next, let's explore adding sodium carbonate to strontium nitrate:
Sr(NO3)2(aq) + Na2CO3(aq) → SrCO3(s) + 2NaNO3(aq)
In this case, a precipitate of strontium carbonate (SrCO3) is formed. Strontium carbonate is insoluble in water, so it will appear as a solid (precipitate) in the solution.
Finally, let's examine adding sodium carbonate to barium nitrate:
Ba(NO3)2(aq) + Na2CO3(aq) → BaCO3(s) + 2NaNO3(aq)
When sodium carbonate is added to barium nitrate, a precipitate of barium carbonate (BaCO3) is formed. Barium carbonate is insoluble in water, so it will appear as a solid (precipitate) in the solution.
To summarize:
- Adding sodium carbonate to magnesium nitrate results in the formation of an insoluble precipitate (magnesium carbonate).
- Adding sodium carbonate to calcium nitrate does not form a precipitate.
- Adding sodium carbonate to strontium nitrate results in the formation of an insoluble precipitate (strontium carbonate).
- Adding sodium carbonate to barium nitrate results in the formation of an insoluble precipitate (barium carbonate).
To determine if a reaction between two compounds will produce a precipitate, you need to consider the solubility rules of the compounds involved. Solubility rules can help us predict whether a compound will be soluble or insoluble in water.
Here are the solubility rules for the compounds mentioned in your question:
1. Sodium carbonate (Na2CO3) is generally soluble in water. Therefore, it will dissociate into sodium ions (Na+) and carbonate ions (CO32-) when added to water.
2. Magnesium nitrate (Mg(NO3)2) is also soluble in water. It will dissociate into magnesium ions (Mg2+) and nitrate ions (NO3-) when dissolved in water.
Based on the solubility rules, both sodium carbonate and magnesium nitrate will dissociate completely in water, meaning they will not form an insoluble precipitate when mixed together. The resulting solution will contain sodium ions, carbonate ions, magnesium ions, and nitrate ions.
The same reasoning applies to calcium nitrate (Ca(NO3)2) and strontium nitrate (Sr(NO3)2). Both compounds are soluble in water, and therefore, when sodium carbonate is added, no precipitate will form.
However, barium nitrate (Ba(NO3)2) is an exception. According to the solubility rules, it is insoluble in water, and when sodium carbonate is added, a precipitation reaction will occur. This reaction will result in the formation of an insoluble precipitate called barium carbonate (BaCO3), along with sodium nitrate (Na(NO3)) in the aqueous phase.
It's important to note that while these predictions are based on general solubility rules, there can sometimes be exceptions or complications. It's always best to consult references or conduct experimental tests for specific cases.