X is a solution containing both sodium chloride and sodium sulfate. Three of the following solutions will produce a white precipitate on addition of X. Which one remains clear?
A) Aqueous ammonia
C) Lead(ll) nitrate
B) Barium chloride
D) Silver nitrate
Could someone explain to me how to get the answer, thks!
To determine which solution will produce a white precipitate when added to solution X, we need to understand the reactions that occur between the reagents.
1) Aqueous ammonia (option A):
When ammonia is added to a solution containing sodium chloride and sodium sulfate, it reacts with the sodium chloride to form a white precipitate of silver chloride. However, it does not react with sodium sulfate.
The reaction is as follows:
2NH3 + NaCl -> NH4Cl + AgCl
2) Lead(II) nitrate (option C):
When lead(II) nitrate is added to a solution containing sodium chloride and sodium sulfate, it reacts with both sodium chloride and sodium sulfate to form white precipitates.
The reactions are as follows:
Pb(NO3)2 + 2NaCl -> PbCl2 + 2NaNO3
Pb(NO3)2 + Na2SO4 -> PbSO4 + 2NaNO3
3) Barium chloride (option B):
When barium chloride is added to a solution containing sodium chloride and sodium sulfate, it reacts with both sodium chloride and sodium sulfate to form white precipitates.
The reactions are as follows:
BaCl2 + 2NaCl -> BaCl2 + 2NaCl
BaCl2 + Na2SO4 -> BaSO4 + 2NaCl
4) Silver nitrate (option D):
When silver nitrate is added to a solution containing sodium chloride and sodium sulfate, it reacts with both sodium chloride and sodium sulfate to form white precipitates.
The reactions are as follows:
AgNO3 + NaCl -> AgCl + NaNO3
AgNO3 + Na2SO4 -> Ag2SO4 + 2NaNO3
Based on the reactions, it can be concluded that option A (aqueous ammonia) will remain clear when added to solution X, as it only reacts with sodium chloride to form a white precipitate of silver chloride. Therefore, option A is the correct answer. Options C, B, and D will all produce white precipitates upon addition to solution X.
To determine which solution will not produce a white precipitate on addition of X, you need to understand the concept of solubility rules. Solubility rules help us predict whether a compound will dissolve in water or form a precipitate when mixed with another compound.
Here are the general solubility rules that will be helpful in this case:
1) Most nitrates (NO3-) are soluble.
2) Most chloride (Cl-) salts are soluble, except for silver chloride (AgCl), lead chloride (PbCl2), and mercury(I) chloride (Hg2Cl2).
3) Most sulfate (SO42-) salts are soluble, except for lead sulfate (PbSO4), barium sulfate (BaSO4), and calcium sulfate (CaSO4).
Based on these rules, let's evaluate each option:
A) Aqueous ammonia (NH3): Aqueous ammonia does not contain any of the cations that are known to form insoluble chlorides or sulfates. Therefore, it will not produce a white precipitate on addition of X.
B) Barium chloride (BaCl2): As per the solubility rules, barium chloride will form a white precipitate when mixed with sulfate ions. Since X contains sodium sulfate, when barium chloride is added to X, it will form barium sulfate (BaSO4), which is insoluble.
C) Lead(II) nitrate (Pb(NO3)2): Lead(II) nitrate does not contain chloride ions, so it will not form a white precipitate on addition of X.
D) Silver nitrate (AgNO3): As per the solubility rules, silver nitrate will form a white precipitate when mixed with chloride ions. Since X contains sodium chloride, when silver nitrate is added to X, it will form silver chloride (AgCl), which is insoluble.
Therefore, the solution that remains clear on addition of X is option A) Aqueous ammonia.