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The equivalent volume of \[{\text{C}}{{\text{O}}_{\text{2}}}\] in the following reaction is
\[{\text{2NaHC}}{{\text{O}}_{\text{3}}} \to {\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O + C}}{{\text{O}}_{\text{2}}}\]
A) 22.4 liters
B) 11.2 liters
C) 5.6 liters
D) 22 liters

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Last updated date: 26th Feb 2024
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IVSAT 2024
Answer
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Hint: At STP volume of one mole of any gas is 22.4 L. Equivalent volume is the ratio of standard volume to factor n. Factor n indicates the number of replaceable \[{{\text{H}}^{\text{ + }}}\]ions.

Formula Used: \[{\text{Equivalent volume = }}\dfrac{{{\text{Standard volume}}}}{{{\text{n - factor}}}}\]

Complete answer:
The chemical reaction given to us is
\[{\text{2NaHC}}{{\text{O}}_{\text{3}}} \to {\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O + C}}{{\text{O}}_{\text{2}}}\]
From the balanced chemical reaction, we can say that in this reaction 2 moles of sodium bicarbonate
(\[{\text{NaHC}}{{\text{O}}_{\text{3}}}\]) dissociates into 1 mole of sodium carbonate (\[{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\]), 1 mole of water (\[{{\text{H}}_{\text{2}}}{\text{O}}\]) and 1 mole of carbon dioxide gas (\[{\text{C}}{{\text{O}}_{\text{2}}}\]).
To calculate the equivalent volume of \[{\text{C}}{{\text{O}}_{\text{2}}}\] gas we need to determine the standard volume of \[{\text{C}}{{\text{O}}_{\text{2}}}\] gas and n-factor.
We know that At the STP volume of one mole of any gas is 22.4L. So the standard volume of \[{\text{C}}{{\text{O}}_{\text{2}}}\] gas is 22.4L.
To determine the n-factor indicates the number of replaceable \[{{\text{H}}^{\text{ + }}}\]ions of acid. In this reaction, we can see that 2 moles of \[{\text{NaHC}}{{\text{O}}_{\text{3}}}\]contain 2 moles of replaceable \[{{\text{H}}^{\text{ + }}}\]ions. Hence the value of the n-factor is 2.
Now, using the standard volume 22.4L and n-factor 2 we can calculate the equivalent volume of \[{\text{C}}{{\text{O}}_{\text{2}}}\] gas as follows:
\[{\text{Equivalent volume = }}\dfrac{{{\text{Standard volume}}}}{{{\text{n - factor}}}}\]
\[{\text{Equivalent volume = }}\dfrac{{{\text{22}}{\text{.4L}}}}{{\text{2}}} = {\text{11}}{\text{.2L}}\]
Thus, the equivalent volume of \[{\text{C}}{{\text{O}}_{\text{2}}}\] gas for the given reaction is 11.2 L.

Hence, the correct answer is option (B) 11.2 liters.

Note: Acid is the species that donate \[{{\text{H}}^{\text{ + }}}\] ions in an aqueous solution. Basicity of the acid indicates a number of replaceable \[{{\text{H}}^{\text{ + }}}\] ions present in acid. While n-factor indicates an actual number of \[{{\text{H}}^{\text{ + }}}\] ions replaced during the reaction. So in the case of diprotic, triprotic acids n-factor might be different from the basicity of acids.
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