To determine which hydrogen halides react with aqueous silver nitrate (AgNO3) to give a precipitate that dissolves in aqueous sodium thiosulfate (Na2S2O3), we analyze the behavior of each hydrogen halide (HX):
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Reaction with Aqueous AgNO3:
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HF (Option B):
HF(aq)+AgNO3(aq)→AgF(aq)+HNO3(aq)
Silver fluoride (AgF) is highly soluble in water due to the high hydration energy of the fluoride ion (F−). Therefore, no precipitate is formed.
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HCl (Option A):
HCl(aq)+AgNO3(aq)→AgCl(s)↓+HNO3(aq)
A white precipitate of silver chloride (AgCl) is formed.
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HBr (Option C):
HBr(aq)+AgNO3(aq)→AgBr(s)↓+HNO3(aq)
A pale yellow precipitate of silver bromide (AgBr) is formed.
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HI (Option D):
HI(aq)+AgNO3(aq)→AgI(s)↓+HNO3(aq)
A yellow precipitate of silver iodide (AgI) is formed.
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Dissolution of Precipitates in Aqueous Na2S2O3:
The silver halide precipitates (AgCl, AgBr, and AgI) react with sodium thiosulfate to form a soluble complex, sodium dithiosulphatoargentate(I):
AgX(s)+2Na2S2O3(aq)→Na3[Ag(S2O3)2](aq)+NaX(aq)(where X=Cl,Br,I)
Because the formation constant (Kf) of the complex [Ag(S2O3)2]3− is extremely high, even the least soluble silver halide, AgI, dissolves in aqueous Na2S2O3.
Thus, HCl, HBr, and HI form precipitates with AgNO3(aq) that subsequently dissolve in Na2S2O3(aq).
Correct Options:
(A), (C), and (D)