In the ionic equation `2K^(+)BrO_(3)^(-)+12H^(+)+10e^(-)rarrBr_(2)+6H_(2)O+2K^(+)`, the equivalent weight of `KBrO_(3)` will be :
(where M = molecular weight of `KBrO_(3)`)

Options

A. M/5
B. M/2
C. M/6
D. M/4

Answer

To find the equivalent weight of \( KBrO_3 \) from the given ionic equation, we will follow these steps:

### Step 1: Understand the Reaction
The given ionic equation is:
\[ 2K^+ + 2BrO_3^- + 12H^+ + 10e^- \rightarrow Br_2 + 6H_2O + 2K^+ \]

### Step 2: Identify the Species Involved
In this reaction, \( KBrO_3 \) is being reduced to bromine gas \( Br_2 \). We need to determine the change in oxidation states to find the valency factor.

### Step 3: Determine Oxidation States
- The oxidation state of bromine in \( BrO_3^- \):
- Oxygen has an oxidation state of \(-2\).
- Let the oxidation state of bromine be \( x \).
- The equation for the oxidation state is:
\[ x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5 \]
- The oxidation state of bromine in \( Br_2 \) is \( 0 \).

### Step 4: Calculate the Change in Oxidation State
- The change in oxidation state for bromine is:
\[ +5 \text{ (in } BrO_3^-) \rightarrow 0 \text{ (in } Br_2\text{)} \]
- Thus, the change in oxidation state is:
\[ 5 - 0 = 5 \]

### Step 5: Calculate the Valency Factor
In redox reactions, the valency factor is equal to the net change in oxidation state. Here, the valency factor is \( 5 \).

### Step 6: Calculate the Equivalent Weight
The equivalent weight of a substance is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Valency Factor}} \]
Let \( M \) be the molecular weight of \( KBrO_3 \). Therefore, the equivalent weight is:
\[ \text{Equivalent Weight of } KBrO_3 = \frac{M}{5} \]

### Final Answer
The equivalent weight of \( KBrO_3 \) is \( \frac{M}{5} \).

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