
In Kjeldahl’s method of estimation of nitrogen, ${\rm{CuS}}{{\rm{O}}_{\rm{4}}}$ act as:
A. Oxidizing agent
B. Reducing agent
C. Catalytic agent
D. Hydrolysis agent
Answer
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Hint: Before deciding copper sulphate acts as oxidizing, reducing, catalytic or hydrolysis agent, we must know about the Kjeldahl’s method. This method is used to determine the amount of nitrogen in an organic compound, that is, quantitative analysis.
Complete step-by-step answer:
A catalytic reagent is the reagent used in chemical reaction to speed up a reaction. Catalytic reagent does not get changed during the course of chemical reaction.
In Kjeldahl’s method, an organic compound is heated with concentrated sulphuric acid $\left( {{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)$in presence of potassium sulphate $\left( {{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)$ and little copper sulphate $\left( {{\rm{CuS}}{{\rm{O}}_{\rm{4}}}} \right)$for conversion of all nitrogen in the particular compound to the ammonium sulphate ${\left( {{\rm{N}}{{\rm{H}}_4}} \right)_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$. In this reaction, copper sulphate fastens the reaction. So, copper sulphate acts as a catalytic reagent in Kjeldahl’s method.
Hence, the correct answer is C.
Additional Information:
Kjeldahl’s method takes place in three steps.
1) First step is digestion. In this step, nitrogenous organic compound reacts with sulphuric acid, a catalyst $\left( {{\rm{CuS}}{{\rm{O}}_{\rm{4}}}} \right)$ and potassium sulphate $\left( {{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)$ which increases boiling point of the sulphuric acid. This reaction produces ammonium sulphate.
Now, we write the equation for the above reaction is,
${\text{Nitrogenous compound}} + {{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} + {\rm{CuS}}{{\rm{O}}_{\rm{4}}} + {{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} \to {\left( {{\rm{N}}{{\rm{H}}_4}} \right)_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} + {{\rm{H}}_{\rm{2}}}{\rm{O}} + {\rm{C}}{{\rm{O}}_{\rm{2}}} + {\rm{Other}}\;{\rm{by}}\;{\rm{products}}$
2) Second step is distillation. In this step, excess base is added to the digestion mixture (step 1) to convert ammonium ions $\left( {{\rm{N}}{{\rm{H}}_{\rm{4}}}^ + } \right)$ to ammonia $\left( {{\rm{N}}{{\rm{H}}_3}} \right)$, followed by boiling and condensing ammonia gas in a receiving solution.
3) Third step is titration. In this step, the amount of ammonia in the receiving solution is measured.
Note: The other method of estimation of nitrogen in an organic compound is Duma’s method. In this method, organic compounds of known weight are strongly heated with dry cupric oxide. Carbon and hydrogen of the compound is oxidized to carbon dioxide and water vapour. If nitrogen is present in the compound, nitrogen is converted to nitrogen gas and its volume is determined at STP.
Complete step-by-step answer:
A catalytic reagent is the reagent used in chemical reaction to speed up a reaction. Catalytic reagent does not get changed during the course of chemical reaction.
In Kjeldahl’s method, an organic compound is heated with concentrated sulphuric acid $\left( {{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)$in presence of potassium sulphate $\left( {{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)$ and little copper sulphate $\left( {{\rm{CuS}}{{\rm{O}}_{\rm{4}}}} \right)$for conversion of all nitrogen in the particular compound to the ammonium sulphate ${\left( {{\rm{N}}{{\rm{H}}_4}} \right)_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$. In this reaction, copper sulphate fastens the reaction. So, copper sulphate acts as a catalytic reagent in Kjeldahl’s method.
Hence, the correct answer is C.
Additional Information:
Kjeldahl’s method takes place in three steps.
1) First step is digestion. In this step, nitrogenous organic compound reacts with sulphuric acid, a catalyst $\left( {{\rm{CuS}}{{\rm{O}}_{\rm{4}}}} \right)$ and potassium sulphate $\left( {{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)$ which increases boiling point of the sulphuric acid. This reaction produces ammonium sulphate.
Now, we write the equation for the above reaction is,
${\text{Nitrogenous compound}} + {{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} + {\rm{CuS}}{{\rm{O}}_{\rm{4}}} + {{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} \to {\left( {{\rm{N}}{{\rm{H}}_4}} \right)_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} + {{\rm{H}}_{\rm{2}}}{\rm{O}} + {\rm{C}}{{\rm{O}}_{\rm{2}}} + {\rm{Other}}\;{\rm{by}}\;{\rm{products}}$
2) Second step is distillation. In this step, excess base is added to the digestion mixture (step 1) to convert ammonium ions $\left( {{\rm{N}}{{\rm{H}}_{\rm{4}}}^ + } \right)$ to ammonia $\left( {{\rm{N}}{{\rm{H}}_3}} \right)$, followed by boiling and condensing ammonia gas in a receiving solution.
3) Third step is titration. In this step, the amount of ammonia in the receiving solution is measured.
Note: The other method of estimation of nitrogen in an organic compound is Duma’s method. In this method, organic compounds of known weight are strongly heated with dry cupric oxide. Carbon and hydrogen of the compound is oxidized to carbon dioxide and water vapour. If nitrogen is present in the compound, nitrogen is converted to nitrogen gas and its volume is determined at STP.
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