A mixture of two salts is used to prepare a solution S, which gives the following results:
\(\text{White precipitate(s) only} \underset{\text{Room temperature}}{\overset{\text{Dilute NaOH(aq)}}{\longleftarrow}} \text{S (aq solution of the salts)}\underset{\text{Room temperature}}{\overset{\text{Dilute HCl(aq)}}{\longrightarrow}} \text{White precipitate(s) only}\)
The correct option(s) for the salt mixture is(are)
\(\text{White precipitate(s) only} \underset{\text{Room temperature}}{\overset{\text{Dilute NaOH(aq)}}{\longleftarrow}} \text{S (aq solution of the salts)}\underset{\text{Room temperature}}{\overset{\text{Dilute HCl(aq)}}{\longrightarrow}} \text{White precipitate(s) only}\)
The correct option(s) for the salt mixture is(are)
Pb(NO3)2 and Zn(NO3)2
Pb(NO3)2 and Bi(NO3)2
AgNO3 and Bi(NO3)3
Pb(NO3)2 and Hg(NO3)2
The Correct Option is A, B, C
Solution and Explanation
The given experiment involves a solution "S" containing two salts, which gives the following results:
- Room temperature + Dilute NaOH(aq) → White precipitate(s) only.
- Room temperature + Dilute HCl(aq) → White precipitate(s) only.
This suggests that the solution contains salts that react with dilute NaOH and dilute HCl to form white precipitates. We need to identify salts that fit this behavior.
Step 2: Reaction with Dilute NaOH(aq)
The formation of a white precipitate in the presence of dilute NaOH suggests that one or more of the salts in the solution could be lead(II) (Pb2+), bismuth(III) (Bi3+), or silver (Ag+) ions. These ions are known to form white precipitates when reacted with NaOH:
- Lead(II) ions (Pb2+) form lead(II) hydroxide Pb(OH)2, which is white and insoluble in water.
- Bismuth(III) ions (Bi3+) form bismuth(III) hydroxide Bi(OH)3, which is white and insoluble in water.
- Silver ions (Ag+) form silver(I) hydroxide AgOH, which is white and insoluble in water.
Therefore, NaOH can produce a white precipitate of these hydroxides.
Step 3: Reaction with Dilute HCl(aq)
The formation of a white precipitate upon adding dilute HCl suggests that the solution also contains ions that form white precipitates with chloride ions:
- Lead(II) chloride PbCl2 is white and forms when Pb2+ reacts with chloride ions.
- Silver chloride AgCl is white and forms when Ag+ reacts with chloride ions.
- Bismuth(III) chloride BiCl3 is also white and forms when Bi3+ reacts with chloride ions.
Therefore, the presence of chloride ions in dilute HCl results in the formation of these white precipitates.
Step 4: Identifying the correct mixture
Based on the reactions, we need to identify which two salts can lead to the observed behavior of forming white precipitates with both NaOH and HCl.
- Option A:
Pb(NO3)2 and Zn(NO3)2
- Pb(NO3)2 reacts with NaOH to form a white precipitate of Pb(OH)2 and with HCl to form a white precipitate of PbCl2. However, Zn(NO3)2 does not react with NaOH to form a white precipitate, so this mixture is incorrect.
- Option B:
Pb(NO3)2 and Bi(NO3)2
- Pb(NO3)2 reacts with NaOH to form Pb(OH)2 and with HCl to form PbCl2.
- Bi(NO3)2 reacts with NaOH to form Bi(OH)3 and with HCl to form BiCl3, both of which are white precipitates.
- Therefore, this mixture fits the observed behavior, and is correct.
- Option C:
AgNO3 and Bi(NO3)3
- AgNO3 reacts with NaOH to form AgOH (white precipitate) and with HCl to form AgCl (white precipitate).
- Bi(NO3)3 reacts similarly to form Bi(OH)3 (white precipitate) with NaOH and BiCl3 (white precipitate) with HCl.
- Therefore, this mixture also fits the behavior, and is correct.
Step 5: Conclusion
The correct options are:
A Pb(NO3)2 and Zn(NO3)2
B Pb(NO3)2 and Bi(NO3)2
C AgNO3 and Bi(NO3)3
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