Wet Test for Acid Radical Questions in English

(A) The sodium salt is sodium bicarbonate $(NaHCO_3)$.
When $NaHCO_3$ reacts with $MgCl_2$,it forms magnesium bicarbonate,which is soluble in water at room temperature: $2NaHCO_3 + MgCl_2 \to Mg(HCO_3)_2 + 2NaCl$.
Upon heating,magnesium bicarbonate decomposes to form magnesium carbonate,which is a white precipitate: $Mg(HCO_3)_2 \xrightarrow{\Delta} MgCO_3 \downarrow + H_2O + CO_2 \uparrow$.
Therefore,the anion is $HCO_3^-$.

(A) When a salt containing iodide $(I^-)$ is heated with concentrated $H_2SO_4$,it produces hydrogen iodide $(HI)$.
$KI + H_2SO_4 \rightarrow KHSO_4 + HI$
Concentrated $H_2SO_4$ acts as a strong oxidizing agent and oxidizes $HI$ to iodine $(I_2)$ gas,which appears as violet vapours.
$2HI + H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O$
Therefore,the salt is an iodide.

(B) When sodium sulphite $(Na_2SO_3)$ reacts with dilute $HCl$,it liberates sulphur dioxide $(SO_2)$ gas:
$Na_2SO_3 + 2HCl \to 2NaCl + H_2O + SO_2 \uparrow$
$SO_2$ gas acts as a reducing agent and reduces acidified potassium dichromate $(K_2Cr_2O_7)$ to chromium sulphate $(Cr_2(SO_4)_3)$,which is green in color:
$K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \to K_2SO_4 + Cr_2(SO_4)_3 \text{ (green)} + H_2O$
Therefore,the gas turns acidified potassium dichromate paper green.

(D) The reaction between sodium nitroprusside and sulphide ions in an alkaline medium is a standard test for the detection of sulphide ions.
The chemical reaction is: $Na_2[Fe(CN)_5NO] + S^{2-} \to [Fe(CN)_5NOS]^{4-}$.
The complex formed is the violet/purple coloured ion $[Fe(CN)_5NOS]^{4-}$,which is known as the thio-nitroprusside ion.
Thus,the correct option is $D$.

(C) The chromyl chloride test is a qualitative test used to detect the presence of chloride ions $(Cl^-)$ in a sample.
In this test,the salt containing chloride ions is heated with solid potassium dichromate $(K_2Cr_2O_7)$ and concentrated sulfuric acid $(H_2SO_4)$.
The chemical reaction is:
$4 NaCl + K_2Cr_2O_7 + 6 H_2SO_4 \rightarrow 2 KHSO_4 + 2 CrO_2Cl_2 + 4 NaHSO_4 + 3 H_2O$
The formation of deep red vapors of chromyl chloride $(CrO_2Cl_2)$ confirms the presence of chloride ions.

(A) $AgF$ is highly soluble in water due to its high hydration energy compared to its lattice energy. Therefore,it does not form a precipitate with $AgNO_3$ solution. In contrast,$AgBr$,$Ag_2CO_3$,and $Ag_3PO_4$ are insoluble or sparingly soluble in water and form precipitates.

(D) When chromyl chloride $(CrO_2Cl_2)$ vapours are passed into $NaOH$ solution,sodium chromate $(Na_2CrO_4)$ is formed.
$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$
When this solution is acidified with acetic acid $(CH_3COOH)$ and treated with lead acetate $((CH_3COO)_2Pb)$,a yellow precipitate of lead chromate $(PbCrO_4)$ is formed.
$Na_2CrO_4 + (CH_3COO)_2Pb \rightarrow PbCrO_4 \downarrow (\text{yellow ppt.}) + 2CH_3COONa$

(B) When a salt containing iodide $(I^{-})$ is treated with concentrated $H_2SO_4$,it produces hydrogen iodide $(HI)$ gas.
$KI + H_2SO_4 \to KHSO_4 + HI$
This $HI$ is then oxidized by concentrated $H_2SO_4$ to produce iodine $(I_2)$ vapours,which are violet in color.
$2HI + H_2SO_4 \to I_2 + 2H_2O + SO_2$
Therefore,the salt contains $I^{-}$ ions.

(B) The reaction involves the displacement of halide ions by chlorine water.
Chlorine $(Cl_2)$ is a stronger oxidizing agent than iodine $(I_2)$.
When $Cl_2$ water is added to a solution containing iodide $(I^-)$ ions,$Cl_2$ oxidizes $I^-$ to $I_2$.
The reaction is: $2I^- + Cl_2 \to I_2 + 2Cl^-$.
The liberated iodine $(I_2)$ dissolves in the chloroform $(CHCl_3)$ layer,imparting a characteristic violet color to it.
Therefore,the salt contains $I^-$ ions.

(D) The salt contains the $SO_4^{2-}$ anion.
When a salt containing the $SO_4^{2-}$ group is treated with $H_2SO_4$ (dilute or concentrated),no reaction occurs because the common ion effect prevents the reaction,and $H_2SO_4$ is the acid corresponding to the sulfate radical.
Nitrate,sulphide,and oxalate salts would react with $H_2SO_4$ to evolve gases like $NO_2$,$H_2S$,or $CO/CO_2$ respectively.

(C) The phosphate radical $(PO_4^{3-})$ reacts with ammonium molybdate $(NH_4)_2MoO_4$ in the presence of concentrated nitric acid $(HNO_3)$ to form a canary yellow precipitate of ammonium phosphomolybdate.
The chemical reaction is:
$H_3PO_4 + 12(NH_4)_2MoO_4 + 21HNO_3 \rightarrow (NH_4)_3PO_4 \cdot 12MoO_3 + 21NH_4NO_3 + 12H_2O$
The resulting precipitate,$(NH_4)_3PO_4 \cdot 12MoO_3$,is known as canary yellow precipitate.

(A) When halide ions like $Cl^{-}$,$Br^{-}$,and $I^{-}$ are treated with $AgNO_3$ solution,they form silver halides.
$AgCl$ is a white precipitate that is soluble in $NH_4OH$ due to the formation of a soluble complex $[Ag(NH_3)_2]Cl$.
$AgBr$ is a pale yellow precipitate that is sparingly soluble in $NH_4OH$.
$AgI$ is a yellow precipitate that is insoluble in $NH_4OH$.
Since the precipitate is white and dissolves in $NH_4OH$,the radical present is $Cl^{-}$.
Chemical reaction:
$AgNO_3 + NaCl \rightarrow AgCl \downarrow (\text{white}) + NaNO_3$
$AgCl + 2NH_4OH \rightarrow [Ag(NH_3)_2]Cl + 2H_2O$

(A) The correct answer is $A$.
When a mixture containing bromide ions $(Br^-)$ is heated with concentrated $H_2SO_4$ and $MnO_2$,bromine gas $(Br_2)$ is evolved,which appears as brown fumes.
The chemical reaction is:
$2KBr + MnO_2 + 2H_2SO_4 \to Br_2(g) + 2KHSO_4 + MnSO_4 + 2H_2O$
Thus,the brown fumes are due to the presence of $Br^-$ ions.

(C) The gas liberated is $SO_2$ (sulphur dioxide).
$(i)$ $SO_2$ reacts with baryta water $(Ba(OH)_2)$ to form $BaSO_3$,which causes turbidity: $Ba(OH)_2 + SO_2 \to BaSO_3 \downarrow + H_2O$.
$(ii)$ $SO_2$ reduces acidified potassium dichromate $(K_2Cr_2O_7)$ to chromium$(III)$ sulphate,which is green in colour: $K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \to K_2SO_4 + Cr_2(SO_4)_3 + H_2O$.
These tests are characteristic of the sulphite ion $(SO_3^{2-})$.

(D) The white precipitate formed in the test for sulphate radical (usually $BaSO_4$) is insoluble in common mineral acids like $HCl$,$HNO_3$,and $H_2SO_4$.
Therefore,the correct answer is $(D)$.

(D) When dilute $H_2SO_4$ is added to an inorganic mixture in the cold:
$1$. Sulphite $(SO_3^{2-})$ evolves $SO_2$ gas,which has a suffocating,burning sulphur odour.
$2$. Acetate $(CH_3COO^-)$ evolves acetic acid,which has a characteristic vinegar-like odour.
$3$. Nitrite $(NO_2^-)$ evolves $NO_2$ gas,which is reddish-brown in colour.
$4$. Carbonate $(CO_3^{2-})$ evolves $CO_2$ gas,which is colourless and odourless.
Therefore,the correct option is $(D)$.

(C) The Chromyl chloride test is a specific qualitative test used to detect the presence of $Cl^-$ (chloride) ions in a salt sample.
When a chloride salt is heated with potassium dichromate $(K_2Cr_2O_7)$ and concentrated sulphuric acid $(H_2SO_4)$,deep red vapours of chromyl chloride $(CrO_2Cl_2)$ are evolved.
These vapours,when passed into a solution of sodium hydroxide $(NaOH)$,give a yellow solution of sodium chromate $(Na_2CrO_4)$,confirming the presence of chloride ions.

(B) To remove sulphate ions $(SO_4^{2-})$ from a solution containing both sulphate and chloride ions,we need a reagent that forms an insoluble precipitate with sulphate ions but remains soluble or does not react with chloride ions.
$Ba^{2+}$ ions react with $SO_4^{2-}$ ions to form $BaSO_4$,which is a white precipitate.
$Ba(OH)_2$ provides $Ba^{2+}$ ions in the solution.
$BaSO_4$ is insoluble in water,effectively removing the sulphate ions from the solution.
$BaCl_2$ would also work,but among the given options,$Ba(OH)_2$ is the correct choice.

(A) solution of a salt in dilute sulphuric acid reacts with iodide ions in the presence of starch to produce a deep blue colour,which confirms the presence of nitrite ions $(NO_2^{-})$.
The chemical reaction is as follows:
$2NO_2^{-} + 2I^{-} + 4H^{+} \rightarrow 2NO + I_2 + 2H_2O$
The liberated iodine $(I_2)$ forms a complex with starch,resulting in a characteristic deep blue colour.

(A) The brown ring test is a common chemical 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.
The brown ring is due to the formation of the complex $[Fe(H_2O)_5NO]SO_4$.

(B) The correct answer is $(B)$.
Most nitrate salts $(NO_3^{-})$ are highly soluble in water and do not form precipitates with common cations.
In contrast,$Cl^{-}$ forms a precipitate with $Ag^{+}$,$CO_3^{2-}$ forms precipitates with many metal ions like $Ca^{2+}$ or $Ba^{2+}$,and $SO_4^{2-}$ forms a precipitate with $Ba^{2+}$.

(A) $Na_2CO_3$ is used to identify acid radicals by the evolution of $CO_2$ gas due to the reaction with stronger acids.
$CO_3^{2-}$ ions cannot be identified using $Na_2CO_3$ because there is no reaction between the reagent and the analyte containing the same ion.
$SO_3^{2-}$,$S^{2-}$,and $SO_4^{2-}$ salts of stronger acids react with $Na_2CO_3$ to evolve gases or form precipitates,allowing for their identification.

(A) The reaction of a salt $[X]$ containing sulfite ion $(SO_3^{2-})$ with dilute $H_2SO_4$ produces sulfur dioxide gas $(SO_2)$,which is a colourless gas with a pungent,irritating smell.
$SO_3^{2-} + 2H^+ \to H_2O + SO_2 \uparrow$.
Sulfur dioxide $(SO_2)$ acts as a reducing agent and turns acidified potassium dichromate $(K_2Cr_2O_7)$ solution green due to the formation of chromium$(III)$ sulfate.
$K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \to K_2SO_4 + Cr_2(SO_4)_3 + H_2O$.
Thus,$[X]$ is $SO_3^{2-}$ and $[Y]$ is $SO_2$.

(A) $Pb^{2+}$ ions react with $Cl^-$ ions to form a white precipitate of $PbCl_2$,whereas $Ba^{2+}$ ions do not form a precipitate with $NaCl$ because $BaCl_2$ is highly soluble in water.
$Pb(NO_3)_2(aq) + 2NaCl(aq) \to PbCl_2(s) + 2NaNO_3(aq)$
$Ba(NO_3)_2(aq) + 2NaCl(aq) \to \text{No reaction}$
Therefore,$NaCl$ is the correct answer.

(A) The reaction involves the oxidation of iodide ions $(I^-)$ to iodine $(I_2)$ by chlorine water $(Cl_2)$.
$2I^- + Cl_2 \rightarrow I_2 + 2Cl^-$
$I_2$ dissolves in chloroform to give a violet colour.
On adding excess chlorine water,$I_2$ is further oxidized to iodate $(IO_3^-)$,which is colourless.
$I_2 + 5Cl_2 + 6H_2O \rightarrow 2HIO_3 + 10HCl$
Thus,the disappearance of the violet colour confirms the presence of iodide.

(B) The correct answer is $B$.
When a mixture containing a sulphide $(S^{2-})$ is heated with dilute $H_2SO_4$,it releases hydrogen sulphide gas $(H_2S)$.
$H_2S$ gas reacts with lead acetate paper to form lead sulphide $(PbS)$,which is black in color.
The reaction is: $Pb(CH_3COO)_2 + H_2S \rightarrow PbS \text{ (black ppt.)} + 2CH_3COOH$.

(D) The brown ring test is a common test for the detection of nitrate ions $(NO_3^-)$.
When a freshly prepared ferrous sulfate $(FeSO_4)$ solution is added to an 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 junction of the two layers.
The brown ring is due to the formation of the nitroso-ferrous sulfate complex,which is represented as $[Fe(H_2O)_5NO]SO_4$.

(D) $AgNO_3$ reacts with $K_2CrO_4$ to form silver chromate $(Ag_2CrO_4)$,which is a red-colored precipitate.
The balanced chemical equation is:
$2AgNO_3(aq) + K_2CrO_4(aq) \to Ag_2CrO_4(s) \text{ (red ppt.)} + 2KNO_3(aq)$.

(C) The $Chromyl$ $Chloride$ test is a specific qualitative test used to detect the presence of $Cl^-$ (chloride) ions in an inorganic salt.
When a chloride salt is heated with potassium dichromate $(K_2Cr_2O_7)$ and concentrated sulphuric acid $(H_2SO_4)$,reddish-brown vapours of chromyl chloride $(CrO_2Cl_2)$ are evolved.
These vapours,when passed into sodium hydroxide $(NaOH)$ solution,turn the solution yellow due to the formation of sodium chromate $(Na_2CrO_4)$.

(D) The test for sulfate ions $(SO_4^{2-})$ involves adding barium chloride $(BaCl_2)$ solution to the test solution,which results in the formation of a white precipitate of barium sulfate $(BaSO_4)$.
$Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4 (s) \text{ (white precipitate)}$
Barium sulfate $(BaSO_4)$ is highly insoluble in water and common mineral acids like $HCl$,$H_2SO_4$,and $HNO_3$ at room temperature.
Therefore,it does not dissolve in any of the given solutions.

(A) $1$. The reaction of $[X]$ with dilute $H_2SO_4$ produces a colorless gas $[Y]$ with a pungent smell,which is characteristic of $SO_2$ gas. Thus,$[X]$ is a sulfite salt $(SO_3^{2-})$.
$2$. $SO_2$ gas acts as a reducing agent. When it reacts with acidified potassium dichromate $(K_2Cr_2O_7 + H_2SO_4)$,the orange dichromate ion $(Cr_2O_7^{2-})$ is reduced to green chromium$(III)$ ion $(Cr^{3+})$.
$3$. The reaction is: $K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$.
$4$. The formation of the green solution confirms the presence of $SO_2$.

(A) The reaction of $AgNO_3$ with chloride ions $(Cl^{-})$ produces a white precipitate of silver chloride $(AgCl)$:
$Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s) \text{ (white precipitate)}$.
This $AgCl$ precipitate is soluble in excess ammonium hydroxide $(NH_4OH)$ due to the formation of a soluble complex,diamminesilver$(I)$ chloride:
$AgCl(s) + 2NH_4OH(aq) \rightarrow [Ag(NH_3)_2]Cl(aq) + 2H_2O(l)$.

(C) To remove $SO_4^{2-}$ and $Cl^{-}$ ions,we need a reagent that forms insoluble precipitates with both.
Lead$(II)$ nitrate,$Pb(NO_3)_2$,reacts with $SO_4^{2-}$ to form white precipitate of lead$(II)$ sulfate $(PbSO_4)$ and with $Cl^{-}$ to form white precipitate of lead$(II)$ chloride $(PbCl_2)$.
Therefore,$Pb(NO_3)_2$ is the correct reagent.

(A) When a mixture containing bromide ions $(Br^{-})$ is heated with $MnO_2$ and concentrated $H_2SO_4$,the bromide ions are oxidized to bromine gas $(Br_2)$,which is reddish-brown in color.
The reaction is: $2NaBr + MnO_2 + 2H_2SO_4 \rightarrow Na_2SO_4 + MnSO_4 + 2H_2O + Br_2(g)$.
Therefore,the presence of $Br^{-}$ ions leads to the evolution of a brown gas.

(A) When sodium sulfide $(Na_2S)$ reacts with dilute hydrochloric acid $(HCl)$,hydrogen sulfide gas $(H_2S)$ is evolved.
$Na_2S + 2HCl \rightarrow 2NaCl + H_2S \uparrow$
Hydrogen sulfide gas $(H_2S)$ is known for its characteristic rotten egg smell and its ability to turn lead acetate paper black due to the formation of lead sulfide $(PbS)$:
$Pb(CH_3COO)_2 + H_2S \rightarrow PbS \downarrow (\text{black}) + 2CH_3COOH$

(A) The reaction of nitrite ions $(NO_2^-)$ with iodide ions $(I^-)$ in the presence of dilute $H_2SO_4$ produces iodine $(I_2)$.
$2NO_2^- + 2I^- + 4H^+ \rightarrow 2NO + I_2 + 2H_2O$
Iodine $(I_2)$ reacts with starch to form a characteristic blue-colored complex.
Therefore,the blue color indicates the presence of nitrite ions $(NO_2^-)$.

(A) The reaction of $AgNO_3$ with halide ions and other anions produces silver salts.
Silver fluoride $(AgF)$ is highly soluble in water,unlike other silver halides $(AgCl, AgBr, AgI)$,which are insoluble and form precipitates.
Therefore,$F^{-}$ ions do not form a precipitate with $AgNO_3$.

(C) When phosphate ions $(PO_4^{3-})$ react with ammonium molybdate $((NH_4)_2MoO_4)$ in the presence of nitric acid $(HNO_3)$,they form a canary yellow precipitate of ammonium phosphomolybdate $((NH_4)_3PO_4 \cdot 12MoO_3)$.

(A) $1$. $Pb(NO_3)_2$ reacts with $NaCl$ to form $PbCl_2$ (white precipitate),while $Ba(NO_3)_2$ reacts with $NaCl$ to form $BaCl_2$ (soluble in water).
$2$. $Pb(NO_3)_2$ reacts with sodium acetate to form lead acetate (soluble),while $Ba(NO_3)_2$ also forms soluble barium acetate.
$3$. Sodium nitrate does not form a precipitate with either.
$4$. Sodium hydrogen phosphate $(Na_2HPO_4)$ forms precipitates with both $Pb^{2+}$ and $Ba^{2+}$ ions.
$5$. Therefore,$NaCl$ is the correct answer as it selectively precipitates $Pb^{2+}$ as $PbCl_2$.

(B) The solubility rules state that most nitrate salts $(NO_3^{-})$ are soluble in water.
Unlike $Cl^{-}$,$CO_3^{2-}$,and $SO_4^{2-}$,which form insoluble precipitates with various metal cations (e.g.,$Ag^{+}$,$Ba^{2+}$,$Ca^{2+}$),nitrate ions do not form common insoluble precipitates with most metal cations.
Therefore,$NO_3^{-}$ cannot be easily removed from an aqueous solution by precipitation.

(B) When chlorine water ($Cl_2$ water) is added to a solution containing iodide ions $(I^-)$,the iodide ions are oxidized to iodine $(I_2)$.
$2I^- + Cl_2 \rightarrow I_2 + 2Cl^-$
When chloroform $(CHCl_3)$ is added to this mixture,the liberated iodine $(I_2)$ dissolves in the chloroform layer.
Since iodine is soluble in organic solvents like chloroform,it imparts a characteristic violet color to the organic layer.

(B) When an iodide salt $(I^{-})$ reacts with concentrated $H_2SO_4$,it produces hydrogen iodide $(HI)$.
$2I^{-} + 2H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O + SO_4^{2-}$.
The iodine $(I_2)$ produced is in the form of violet vapors.

(C) The brown ring test for $NO_2^-$ and $NO_3^-$ ions involves the reaction of $Fe^{2+}$ ions with $NO$ (formed by the reduction of nitrate/nitrite) in the presence of water. The resulting complex is pentaaquanitrosyliron$(II)$ ion,which has the formula $[Fe(H_2O)_5(NO)]^{2+}$. This complex is responsible for the characteristic brown ring.

(A) The brown ring test is a common laboratory test used to detect the presence of the nitrate ion $(NO_3^-)$.
When concentrated $H_2SO_4$ is added to a mixture of the nitrate solution and freshly prepared $FeSO_4$ solution,a brown ring is formed at the junction of the two liquids.
The brown ring is due to the formation of the complex $[Fe(H_2O)_5(NO)]SO_4$.

(D) $Na_2CO_3$ is used in the qualitative analysis of inorganic salts,particularly for the preparation of a sodium carbonate extract to identify anions.
It is useful for identifying anions like $SO_3^{2-}$,$S^{2-}$,and $CO_3^{2-}$ because these ions form insoluble precipitates with various metal cations in the presence of carbonate.
However,$Na_2CO_3$ is $NOT$ useful for the identification of the $SO_4^{2-}$ ion because the sulfate ion does not form a characteristic precipitate with the carbonate reagent that allows for its specific identification in this context.
Therefore,the correct option is $D$.

(C) $Ba^{2+}$ ions react with $SO_4^{2-}$ ions to form an insoluble white precipitate of $BaSO_4$.
By adding a reagent containing $Ba^{2+}$ ions (like $BaCl_2$ or $Ba(OH)_2$),$SO_4^{2-}$ can be precipitated out,leaving $Cl^-$ ions in the solution.
Thus,$Ba(OH)_2$ is a suitable reagent for this separation.

(C) The reaction is: $SO_4^{2-} + BaCl_2 \to BaSO_4 \downarrow + 2Cl^-$.
$BaSO_4$ is a heavy white precipitate that is insoluble in $dil. HCl$.

(A) When a sodium bicarbonate $(NaHCO_3)$ solution is added to a magnesium salt solution (like $MgCl_2$),it forms magnesium bicarbonate:
$MgCl_2 + 2NaHCO_3 \to Mg(HCO_3)_2 + 2NaCl$
Upon heating,magnesium bicarbonate decomposes to form a white precipitate of magnesium carbonate:
$Mg(HCO_3)_2 \xrightarrow{\Delta} MgCO_3 \downarrow + H_2O + CO_2$
Therefore,the anion is bicarbonate $(HCO_3^-)$.

(D) The salt is a white sodium salt that is neutral to litmus,suggesting it is a salt of a strong acid and a strong base,such as $NaCl$.
When $AgNO_3$ is added to a solution containing $Cl^-$ ions,a white precipitate of $AgCl$ is formed.
$Ag^+_{(aq)} + Cl^-_{(aq)} \to AgCl_{(s)} \text{ (white precipitate)}$.
This $AgCl$ precipitate is insoluble in dilute $HNO_3$.
Therefore,the anion is $Cl^-$.

(C) When neutral ferric chloride $(FeCl_3)$ is added to an aqueous solution of acetate ions $(CH_3COO^-)$,a blood-red colour is produced due to the formation of a complex compound known as ferric acetate,represented as $[Fe(OH)_2(CH_3COO)]_3$ or often simplified as $[Fe(CH_3COO)_3]$.
The reaction is: $3CH_3COONa + FeCl_3 \to [Fe(OH)_2(CH_3COO)]_3 + 3NaCl$.