The nitrate ion can be converted into ammonium ion. The equivalent mass of `NO_(3)^(-)` ion in this reaction would be :

Options

A. 6.20 g
B. 7.75 g
C. 10.5 g
D. 21.0 g

Answer

To find the equivalent mass of the nitrate ion (\(NO_3^-\)) when it is converted into ammonium ion (\(NH_4^+\)), we need to follow these steps:

### Step 1: Write the Reaction
The conversion of the nitrate ion to the ammonium ion can be represented as:
\[ NO_3^- \rightarrow NH_4^+ \]

### Step 2: Determine the Oxidation States
Next, we need to find the oxidation states of nitrogen in both ions.

- For \(NO_3^-\):
- Let the oxidation state of nitrogen be \(X\).
- The oxidation state of oxygen is \(-2\).
- The equation for the charge balance is:
\[ X + 3(-2) = -1 \implies X - 6 = -1 \implies X = +5 \]

- For \(NH_4^+\):
- Let the oxidation state of nitrogen be \(Y\).
- The oxidation state of hydrogen is \(+1\).
- The equation for the charge balance is:
\[ Y + 4(+1) = +1 \implies Y + 4 = +1 \implies Y = -3 \]

### Step 3: Calculate the Change in Oxidation State
Now, we can calculate the change in oxidation state as follows:
\[ \text{Change in oxidation state} = \text{Final state} - \text{Initial state} = (-3) - (+5) = -8 \]
This indicates that the nitrogen in \(NO_3^-\) is reduced, gaining 8 electrons.

### Step 4: Determine the n-factor
In redox reactions, the n-factor is defined as the number of moles of electrons gained or lost per mole of substance. Here, the n-factor for \(NO_3^-\) is 8, since it gains 8 electrons.

### Step 5: Calculate the Molar Mass of \(NO_3^-\)
The molar mass of \(NO_3^-\) can be calculated as follows:
- Atomic mass of nitrogen (N) = 14 g/mol
- Atomic mass of oxygen (O) = 16 g/mol
- Molar mass of \(NO_3^-\) = \(14 + 3 \times 16 = 14 + 48 = 62 \, \text{g/mol}\)

### Step 6: Calculate the Equivalent Mass
The equivalent mass is calculated using the formula:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{62}{8} = 7.75 \, \text{g} \]

### Final Answer
The equivalent mass of the nitrate ion (\(NO_3^-\)) in this reaction is \(7.75 \, \text{g}\).

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