Reaction of hydrogen halides with $AgNO_3(aq.)$

Aqueous $AgNO_3$ gives precipitates with halide ions except fluoride:

$Ag^+ + X^- \rightarrow AgX(s)$ where $X^- = Cl^-, Br^-, I^-$.

Silver fluoride, $AgF$, is soluble in water, so no precipitate is obtained with $F^-$.

Check each hydrogen halide

• A: HCl

  $HCl \rightarrow H^+ + Cl^-$ $Ag^+ + Cl^- \rightarrow AgCl(s)$ A white precipitate of $AgCl$ forms.

• B: HF

  $HF \rightleftharpoons H^+ + F^-$ Even if $F^-$ is present, $AgF$ is soluble, so no precipitate forms.

• C: HBr

  $HBr \rightarrow H^+ + Br^-$ $Ag^+ + Br^- \rightarrow AgBr(s)$ A pale yellow precipitate of $AgBr$ forms.

• D: HI

  $HI \rightarrow H^+ + I^-$ $Ag^+ + I^- \rightarrow AgI(s)$ A yellow precipitate of $AgI$ forms.

Dissolution in $Na_2S_2O_3(aq.)$

Silver halides dissolve in sodium thiosulfate due to complex formation:

$AgX(s) + 2S_2O_3^{2-} \rightarrow [Ag(S_2O_3)_2]^{3-} + X^-$

Thus, $AgCl$, $AgBr$, and $AgI$ dissolve in $Na_2S_2O_3(aq.)$.

Conclusion

The hydrogen halides that give a precipitate with $AgNO_3(aq.)$ and that precipitate dissolves in $Na_2S_2O_3(aq.)$ are:

$\boxed{\text{HCl, HBr, HI}}$

Comparison with stored answer

Stored correct answer: A, C, D

Our derived answer is also A, C, D, so it agrees.