Question

Observe the following reaction: \[ I_2+10HNO_3 \rightarrow 2HIO_3+10NO_2+4H_2O \] The equivalent weight of \(HNO_3\) is (molar mass of \(HNO_3=M\))

• A. \(M\)
• B. \(\dfrac{M}{4}\)
• C. \(\dfrac{M}{2}\)
• D. \(\dfrac{M}{5}\)

Hint:

For redox reactions, \[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{Change in oxidation number per molecule}} \] Always calculate the \(n\)-factor from the change in oxidation state of the oxidizing or reducing species involved in the reaction.

Answer 1

The Correct Option is A

Step 1: Determine the oxidation state of nitrogen in \(HNO_3\).
In nitric acid, \[ HNO_3 \] Let the oxidation state of nitrogen be \(x\). \[ (+1)+x+3(-2)=0 \] \[ 1+x-6=0 \] \[ x=+5 \] Thus, nitrogen is in the \(+5\) oxidation state.

Step 2: Determine the oxidation state of nitrogen in \(NO_2\).
For \(NO_2\), \[ x+2(-2)=0 \] \[ x-4=0 \] \[ x=+4 \] Thus, nitrogen changes from \[ +5 \rightarrow +4 \] and gains one electron per nitrogen atom.

Step 3: Calculate the \(n\)-factor of \(HNO_3\).
The change in oxidation number of nitrogen is \[ (+5)-(+4)=1 \] Therefore, one mole of \(HNO_3\) accepts one mole of electrons. Hence, \[ n\text{-factor}=1 \]

Step 4: Use the formula for equivalent weight.
Equivalent weight is given by \[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n\text{-factor}} \] Substituting, \[ \text{Equivalent Weight} = \frac{M}{1} \] \[ =M \]

Step 5: Final conclusion.
Therefore, the equivalent weight of \(HNO_3\) in the given reaction is \[ \boxed{M} \] which corresponds to option (1).