
An iron salt A reacts with \[{\text{NaOH}}\] to form a green precipitate. Another iron salt B reacts with \[{\text{NaOH}}\] to form a brown precipitate. Identify the iron salts, A and B along with their colours and write the reactions involved.
Answer
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Hint: We know that the green precipitate is of iron (II) hydroxide and the brown precipitate is of iron (III) oxide. Now we have to find out which iron salts react with sodium hydroxide to form iron (II) hydroxide and iron (III) hydroxide.
Complete step by step answer:
An iron salt A reacts with \[{\text{NaOH}}\] to form a green precipitate. The green precipitate is of iron (II) hydroxide. The molecular formula for iron (II) hydroxide is ${\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{2}}}$.
The iron (II) hydroxide is formed when iron (II) sulphate reacts with sodium hydroxide. The molecular formula for iron (II) sulphate is ${\text{FeS}}{{\text{O}}_{\text{4}}}$.
The iron (II) sulphate on reaction with \[{\text{NaOH}}\] forms iron (II) hydroxide which is green in colour. The by-product of the reaction is sodium sulphate $\left( {{\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}} \right)$. The reaction is as follows:
${\text{FeS}}{{\text{O}}_{\text{4}}} + {\text{NaOH}} \to {\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{2}}} + {\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}$
Thus, the iron salt A is iron (II) sulphate $\left( {{\text{FeS}}{{\text{O}}_{\text{4}}}} \right)$. The colour of iron (II) sulphate is white.
An iron salt B reacts with \[{\text{NaOH}}\] to form a green precipitate. The brown precipitate is of iron (III) oxide. The molecular formula for iron (II) hydroxide is ${\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{3}}}$.
The iron (III) hydroxide is formed when iron (III) sulphate reacts with sodium hydroxide. The molecular formula for iron (III) sulphate is ${\text{F}}{{\text{e}}_2}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_3}$.
The iron (III) sulphate on reaction with \[{\text{NaOH}}\] forms iron (III) hydroxide which is brown in colour. The by-product of the reaction is sodium sulphate $\left( {{\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}} \right)$. The reaction is as follows:
${\text{F}}{{\text{e}}_2}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_3} + {\text{NaOH}} \to {\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{3}}} + {\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}$
Thus, the iron salt B is iron (III) sulphate $\left( {{\text{F}}{{\text{e}}_2}{{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)}_3}} \right)$. The colour of iron (III) sulphate is white.
Note:
Iron (II) hydroxide is also known as ferrous hydroxide or green rust. Iron (II) hydroxide is used in nickel-iron batteries. Iron (III) hydroxide is also known as ferric hydroxide. Iron (III) hydroxide is used in cosmetics and tattoo inks.
Complete step by step answer:
An iron salt A reacts with \[{\text{NaOH}}\] to form a green precipitate. The green precipitate is of iron (II) hydroxide. The molecular formula for iron (II) hydroxide is ${\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{2}}}$.
The iron (II) hydroxide is formed when iron (II) sulphate reacts with sodium hydroxide. The molecular formula for iron (II) sulphate is ${\text{FeS}}{{\text{O}}_{\text{4}}}$.
The iron (II) sulphate on reaction with \[{\text{NaOH}}\] forms iron (II) hydroxide which is green in colour. The by-product of the reaction is sodium sulphate $\left( {{\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}} \right)$. The reaction is as follows:
${\text{FeS}}{{\text{O}}_{\text{4}}} + {\text{NaOH}} \to {\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{2}}} + {\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}$
Thus, the iron salt A is iron (II) sulphate $\left( {{\text{FeS}}{{\text{O}}_{\text{4}}}} \right)$. The colour of iron (II) sulphate is white.
An iron salt B reacts with \[{\text{NaOH}}\] to form a green precipitate. The brown precipitate is of iron (III) oxide. The molecular formula for iron (II) hydroxide is ${\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{3}}}$.
The iron (III) hydroxide is formed when iron (III) sulphate reacts with sodium hydroxide. The molecular formula for iron (III) sulphate is ${\text{F}}{{\text{e}}_2}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_3}$.
The iron (III) sulphate on reaction with \[{\text{NaOH}}\] forms iron (III) hydroxide which is brown in colour. The by-product of the reaction is sodium sulphate $\left( {{\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}} \right)$. The reaction is as follows:
${\text{F}}{{\text{e}}_2}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_3} + {\text{NaOH}} \to {\text{Fe}}{\left( {{\text{OH}}} \right)_{\text{3}}} + {\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}$
Thus, the iron salt B is iron (III) sulphate $\left( {{\text{F}}{{\text{e}}_2}{{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)}_3}} \right)$. The colour of iron (III) sulphate is white.
Note:
Iron (II) hydroxide is also known as ferrous hydroxide or green rust. Iron (II) hydroxide is used in nickel-iron batteries. Iron (III) hydroxide is also known as ferric hydroxide. Iron (III) hydroxide is used in cosmetics and tattoo inks.
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