In qualitative analysis,in order to detect II | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $H_2S$ is a weak acid that dissociates as $H_2S \rightleftharpoons 2H^+ + S^{2-}$.When $HCl$ is added,the concentration of $H^+$ ions increases si

Vedclass · Accepted Answer

Decrease the dissociation of $H_2S$

Vedclass · Answer

(C) $H_2S$ is a weak acid that dissociates as $H_2S \rightleftharpoons 2H^+ + S^{2-}$.When $HCl$ is added,the concentration of $H^+$ ions increases significantly.According to Le Chatelier's principle,this common ion effect suppresses the ionization of $H_2S$,thereby decreasing the concentration of $S^{2-}$ ions.This ensures that only the solubility product $(K_{sp})$ of $II$ group metal sulfides is exceeded,allowing for their selective precipitation.

In qualitative analysis,in order to detect $II$ group basic radicals,$H_2S$ gas is passed in the presence of dilute $HCl$ to:

Similar Questions

If $Pb^{2+}, Cu^{2+}, Al^{3+}, Cd^{2+}, Zn^{2+}$ radicals are present in an aqueous solution,and it reacts with $H^{+}/H_2S$,a precipitate is produced. The solution $(X)$ obtained after the removal of the precipitate is treated with excess $NaOH$. What is the result?

Which of the following match is $INCORRECT$ :

$X_2S_n + H_2O \to X(OH)_n \downarrow + Y \uparrow (Gas) \xrightarrow{Pb(CH_3COO)_2} Z \downarrow (Black \ ppt.)$
Then $(X)$ cation can not be

The $V^{-}A$ group precipitate was dissolved in $HNO_3$ and treated with an excess of $NH_4OH$. It gives a white precipitate because of:

Which compound gives a black color with $NH_4OH$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module