When a salt is treated with sodium hydroxide | vedclass

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Question

(B) When an ammonium salt is treated with sodium hydroxide $(NaOH)$,it releases ammonia gas $(NH_3)$: $NH_4^{+} + OH^{-} \longrightarrow NH_3 \uparrow

Vedclass · Accepted Answer

$X=NH_3$ and $Y=K_2HgI_4+KOH$

Vedclass · Answer

(B) When an ammonium salt is treated with sodium hydroxide $(NaOH)$,it releases ammonia gas $(NH_3)$: $NH_4^{+} + OH^{-} \longrightarrow NH_3 \uparrow + H_2O$ Ammonia gas $(NH_3)$ reacts with Nessler's reagent $(K_2[HgI_4] + KOH)$ to form a brown-coloured precipitate known as the iodide of Millon's base: $2[HgI_4]^{2-} + NH_3 + 3OH^{-} \longrightarrow HgO \cdot Hg(NH_2)I + 7I^{-} + 2H_2O$ Thus,$X = NH_3$ and $Y = K_2HgI_4 + KOH$.

When a salt is treated with sodium hydroxide solution,it gives gas $X$. On passing gas $X$ through reagent $Y$,a brown-coloured precipitate is formed. $X$ and $Y$ respectively,are

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