Wet Test for Acid Radical Questions in English

(C) $AgNO_3 + HCl \longrightarrow AgCl \downarrow$ (White precipitate)
$AgNO_3 + HBr \longrightarrow AgBr \downarrow$ (Pale yellow precipitate)
$AgNO_3 + HI \longrightarrow AgI \downarrow$ (Yellow precipitate)
$HF$ does not form a precipitate with $AgNO_3$ because $AgF$ is soluble in water.
$AgCl$,$AgBr$,and $AgI$ dissolve in $Na_2S_2O_3$ (sodium thiosulfate) due to the formation of the soluble complex $Na_3[Ag(S_2O_3)_2]$.

(C) The reaction between sodium nitroprusside and sulphide ions is a standard test for the detection of sulphide ions in a salt sample.
$Na_2S + Na_2[Fe(CN)_5NO] \rightarrow Na_4[Fe(CN)_5NOS]$
This reaction produces a characteristic violet or purple coloured complex,$Na_4[Fe(CN)_5NOS]$,which confirms the presence of the sulphide ion $(S^{2-})$.

(D) The brown ring test is used to detect the presence of nitrate ions $(NO_3^{-})$.
In this test,freshly prepared ferrous sulfate $(FeSO_4)$ solution is added to the nitrate solution,followed by the addition of concentrated sulfuric acid $(H_2SO_4)$ along the sides of the test tube.
$Fe^{+2}$ ions act as a reducing agent to reduce $NO_3^{-}$ to $NO$ gas.
Reaction: $NO_3^{-} + 3Fe^{+2} + 4H^{+} \rightarrow NO + 3Fe^{+3} + 2H_2O$.
Then,$NO$ reacts with the remaining $Fe^{+2}$ to form the brown complex: $[Fe(H_2O)_5(NO)]SO_4$.
Since $Fe^{+2}$ is required for the reduction of $NO_3^{-}$ and the formation of the complex,$Fe^{+3}$ cannot be used.
Thus,both $(A)$ and $(R)$ are correct,and $(R)$ explains why $Fe^{+3}$ is not used (because $Fe^{+2}$ is the necessary reductant).

(B) When a salt containing iodide ions $(I^{-})$ is treated with concentrated $H_2SO_4$,it produces hydrogen iodide $(HI)$.
$HI$ is a strong reducing agent and is further oxidized by concentrated $H_2SO_4$ to produce iodine $(I_2)$ vapors,which are violet in color.
The reaction is: $2I^{-} + 2H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O + SO_4^{2-}$.
Thus,the presence of violet vapors indicates the presence of $I^{-}$ ions.

(B) $P$. $NaBr + \text{Conc. } H_2SO_4 \rightarrow NaHSO_4 + HBr$. $HBr$ is oxidized by $\text{conc. } H_2SO_4$ to $Br_2$ (Brown colour) and $SO_2$. $Br_2$ is diamagnetic,but $NO_2$ is paramagnetic. Wait,$Br_2$ is brown. $P$ gives $B$.
$Q$. $Na_2S + \text{dil. } HCl \rightarrow 2NaCl + H_2S$. $H_2S$ has a rotten egg smell. $Q$ gives $C$.
$R$. $NaNO_2 + \text{dil. } HCl \rightarrow NaCl + HNO_2$. $HNO_2$ decomposes to $NO$ (colourless gas,paramagnetic). $R$ gives $A, D$.
$S$. $NaNO_3 + \text{Conc. } H_2SO_4 \rightarrow NaHSO_4 + HNO_3$. $HNO_3$ decomposes to $NO_2$ (brown colour,paramagnetic). $S$ gives $B, D$.
Thus,the correct matching is $P-B; Q-C; R-A, D; S-B, D$.

(C) The gas $X$ with a rotten egg smell is hydrogen sulfide $(H_2S)$.
$H_2S$ reacts with metal ions to form metal sulfides.
$1$. $H_2S + Zn^{2+} \rightarrow ZnS (\text{white ppt}) + 2H^+$.
$2$. $H_2S + Cd^{2+} \rightarrow CdS (\text{yellow ppt}) + 2H^+$.
$3$. $H_2S + Ni^{2+} \rightarrow NiS (\text{black ppt}) + 2H^+$.
$4$. $H_2S + Cu^{2+} \rightarrow CuS (\text{black ppt}) + 2H^+$.
Option $C$ states that $Ni^{2+}$ gives a pink precipitate,which is incorrect because $NiS$ is black.

(D) The reaction of halides with $AgNO_3$ in the presence of dilute $HNO_3$ is as follows:
$1$. $Cl^{-}$ reacts with $Ag^{+}$ to form $AgCl$ (white precipitate),which is soluble in $NH_4OH$ due to the formation of $[Ag(NH_3)_2]Cl$. Thus,statement $D$ is correct.
$2$. $Br^{-}$ reacts with $Ag^{+}$ to form $AgBr$ (pale yellow precipitate),which is partially soluble in $NH_4OH$. Thus,statement $B$ is correct.
$3$. $I^{-}$ reacts with $Ag^{+}$ to form $AgI$ (yellow precipitate),which is insoluble in $NH_4OH$. Thus,statement $C$ and $E$ are correct.
Comparing with the given options,$B$ and $C$ are correct statements.

(NONE) $(P) NaBr + \text{conc. } H_2SO_4 \rightarrow NaHSO_4 + HBr$. $HBr$ is oxidized to $Br_2$ (brown colour) and $SO_2$ (colourless gas,paramagnetic). Thus,$P-B, D$.
$(Q) Na_2S + \text{dil. } HCl \rightarrow 2NaCl + H_2S$. $H_2S$ is a colourless gas with a rotten egg smell. Thus,$Q-A, C$.
$(R) NaNO_2 + \text{dil. } HCl \rightarrow NaCl + HNO_2$. $HNO_2$ decomposes to $NO$ (colourless gas,paramagnetic). Thus,$R-A, D$.
$(S) NaNO_3 + \text{conc. } H_2SO_4 \rightarrow NaHSO_4 + HNO_3$. $HNO_3$ decomposes to $NO_2$ (brown colour,paramagnetic). Thus,$S-B, D$.
The correct matching is $P-B, D; Q-A, C; R-A, D; S-B, D$.

(C) The brown ring test is a common laboratory test used to detect the presence of the nitrate ion $(NO_3^-)$ in a solution.
When a freshly prepared ferrous sulfate $(FeSO_4)$ solution is added to the nitrate solution,followed by the careful addition of concentrated sulfuric acid $(H_2SO_4)$ along the sides of the test tube,a brown ring is formed at the junction of the two layers.
This brown ring is due to the formation of the complex $[Fe(H_2O)_5(NO)]SO_4$.

(B) Gas $B$ turns the colour of acidified $K_{2}Cr_{2}O_{7}$ green,thus it is $SO_{2}$. $SO_{2}$ is obtained along with a yellow precipitate of sulfur when thiosulfate is treated with dilute acids. Thus,$A$ is $Na_{2}S_{2}O_{3}$,$B$ is $SO_{2}$,and $C$ is $Cr_{2}(SO_{4})_{3}$.
The reactions are as follows:
$Na_{2}S_{2}O_{3} + 2HCl \rightarrow 2NaCl + H_{2}O + SO_{2} + S \text{ (yellow precipitate)}$
$K_{2}Cr_{2}O_{7} + 3SO_{2} + H_{2}SO_{4} \rightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} \text{ (green solution)} + H_{2}O$

(D) The chromyl chloride test is used for the detection of $Cl^-$ ions.
In this test,$4NaCl + K_2Cr_2O_7 + 3H_2SO_4 \longrightarrow K_2SO_4 + 2Na_2SO_4 + 2CrO_2Cl_2 \uparrow + 3H_2O$.
The $CrO_2Cl_2$ (chromyl chloride) formed appears as reddish-brown vapours.
When these vapours are passed into $NaOH$ solution,it forms sodium chromate: $CrO_2Cl_2 + 4NaOH \longrightarrow 2NaCl + Na_2CrO_4 + 2H_2O$.
This solution,when treated with lead acetate,gives a yellow precipitate of lead chromate: $Na_2CrO_4 + (CH_3COO)_2Pb \longrightarrow 2CH_3COONa + PbCrO_4 \downarrow$.
Chlorine gas is not liberated in this test. Therefore,the statement 'liberation of chlorine' is incorrect.

(A) The salt $(A)$ reacts with $BaCl_{2}$ to form a white precipitate. Sulfite salts $(SO_{3}^{2-})$ form $BaSO_{3}$ (white precipitate) with $BaCl_{2}$.
$BaSO_{3(s)} + 2HCl_{(aq)} \longrightarrow BaCl_{2(aq)} + H_{2}O_{(l)} + SO_{2(g)}$.
The gas $(B)$ is $SO_{2}$,which is a reducing agent and decolourises acidified $KMnO_{4}$ solution by reducing $Mn^{7+}$ to $Mn^{2+}$.
Therefore,$(A)$ is $Na_{2}SO_{3}$ and $(B)$ is $SO_{2}$.

(A) When a salt containing chloride $(Cl^-)$ is treated with concentrated sulphuric acid $(H_2SO_4)$,it undergoes a displacement reaction to produce hydrogen chloride gas $(HCl)$.
The reaction is: $NaCl(s) + H_2SO_4(conc.) \rightarrow NaHSO_4(s) + HCl(g) \uparrow$.
$HCl$ is a colourless gas with a pungent odour.

(B) The reaction of $Ba(NO_3)_2$ with $H_2SO_4$ is a double displacement reaction that results in the formation of a white precipitate of $BaSO_4$.
Because $BaSO_4$ is highly insoluble,the reaction proceeds in the forward direction even with dilute $H_2SO_4$.
The chemical equation is: $Ba(NO_3)_2 + H_2SO_4 \rightarrow BaSO_4(s) + 2HNO_3$.

(A) The brown ring test is a standard laboratory test used to detect the presence of the nitrate ion $(NO_3^-)$ in a solution.
In this test,freshly prepared ferrous sulfate $(FeSO_4)$ solution is added to the aqueous solution containing nitrate ions,followed by the careful addition of concentrated sulfuric acid $(H_2SO_4)$ along the sides of the test tube.
$A$ brown ring is formed at the interface of the two layers due to the formation of the nitroso-ferrous sulfate complex.
The chemical reactions involved are:
$2HNO_3 + 3H_2SO_4 + 6FeSO_4 \rightarrow 3Fe_2(SO_4)_3 + 2NO + 4H_2O$
$[Fe(H_2O)_6]SO_4 + NO \rightarrow [Fe(H_2O)_5(NO)]SO_4 + H_2O$

(A) $SO_{3}^{2-}$ and $SO_{4}^{2-}$ ions react with $BaCl_{2}$ to form white precipitates of $BaSO_{3}$ and $BaSO_{4}$ respectively.
$Ba^{2+} + SO_{3}^{2-} \longrightarrow BaSO_{3} (s)$ (White precipitate)
$Ba^{2+} + SO_{4}^{2-} \longrightarrow BaSO_{4} (s)$ (White precipitate)
$BaSO_{3}$ is a salt of a weak acid $(H_{2}SO_{3})$,so it dissolves in dilute $HCl$ due to the formation of $SO_{2}$ gas.
$BaSO_{3} + 2HCl \longrightarrow BaCl_{2} + H_{2}O + SO_{2} \uparrow$
$BaSO_{4}$ is a salt of a strong acid $(H_{2}SO_{4})$ and is insoluble in dilute $HCl$.

(A) The reaction of lead nitrate with a chloride salt produces lead$(II)$ chloride,which is a white precipitate.
$Pb^{2+} (aq) + 2Cl^{-} (aq) \rightarrow PbCl_2 (s) \text{ (white precipitate)}$
Lead$(II)$ chloride $(PbCl_2)$ is sparingly soluble in cold water but dissolves significantly in hot water.
Upon cooling,the solubility decreases,and the lead$(II)$ chloride reprecipitates as needle-like crystals.
Therefore,the acid radical present is the chloride ion $(Cl^{-})$.

(A) According to the question,$[X] +$ dil. $H_{2}SO_{4} \longrightarrow [Y]$ (a colourless,suffocating gas).
$[Y] + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} \longrightarrow$ Green colouration of the solution.
Sulphites $(SO_{3}^{2-})$ react with dil. $H_{2}SO_{4}$ to release sulphur dioxide $(SO_{2})$ gas,which is colourless and has a suffocating odour.
The reaction is: $SO_{3}^{2-} + 2H^{+} \longrightarrow SO_{2} + H_{2}O$.
$SO_{2}$ acts as a reducing agent and reduces acidified potassium dichromate $(K_{2}Cr_{2}O_{7})$ from $Cr(VI)$ (orange) to $Cr(III)$ (green) as $Cr_{2}(SO_{4})_{3}$.
The reaction is: $K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 3SO_{2} \longrightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + H_{2}O$.

(D) When $BaCl_{2}$ is added to an aqueous solution containing $CO_{3}^{2-}$,$SO_{3}^{2-}$,or $SO_{4}^{2-}$ ions,white precipitates are formed in all three cases due to the formation of $BaCO_{3}$,$BaSO_{3}$,and $BaSO_{4}$ respectively.
Therefore,the presence of a white precipitate does not uniquely identify any one of these anions,as all three anions react with $Ba^{2+}$ to produce a white precipitate.

(C) The sodium salt is sodium thiosulphate $(Na_2S_2O_3 \cdot 5H_2O)$,which is a colourless efflorescent salt.
When dilute acid is added to it,it undergoes disproportionation to form sulphur (white precipitate) and sulphur dioxide gas $(SO_2)$:
$Na_2S_2O_{3(s)} + 2H^+_{(aq)} \rightarrow S_{(s)} + SO_{2(g)} + H_2O_{(\ell)} + 2Na^+_{(aq)}$
$SO_2$ gas turns acidified potassium dichromate solution green due to the reduction of $Cr(VI)$ to $Cr(III)$:
$K_2Cr_2O_{7(\text{aq})} + 3SO_{2(\text{g})} + H_2SO_{4(\text{aq})} \rightarrow K_2SO_{4(\text{aq})} + Cr_2(SO_4)_{3(\text{aq})} + H_2O_{(\ell)}$
Thus,the correct option is $C$.

(B) The Griess-Ilosvay test is a standard analytical method used for the detection of nitrite ions $(NO_2^-)$.
In this test,sulphanilic acid reacts with nitrous acid $(HNO_2)$,which is generated from the nitrite ion in an acidic medium,to form a diazonium salt.
This diazonium salt then undergoes a coupling reaction with $\alpha$-naphthylamine to produce a red-colored azo-dye.
Both statements accurately describe the chemical principles of this test.
Therefore,both Statement $I$ and Statement $II$ are true.

(C) The smell of vinegar is a characteristic property of acetic acid $(\text{CH}_3\text{COOH})$.
When dilute $\text{H}_2\text{SO}_4$ is added to a salt containing the acetate anion $(\text{CH}_3\text{COO}^-)$,the following reaction occurs:
$2\text{CH}_3\text{COO}^- + \text{H}_2\text{SO}_4 \rightarrow 2\text{CH}_3\text{COOH} + \text{SO}_4^{2-}$.
Acetic acid is liberated as colourless vapours,which are acidic in nature and turn blue litmus paper red.