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 KBrO₃ from the given ionic equation, we will follow these steps:

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

### Step 2: Identify the Change in Oxidation States
In the ionic equation, we need to determine the change in oxidation state of bromine (Br) in KBrO₃:
- In KBrO₃, the oxidation state of Br is +5.
- In Br₂, the oxidation state of Br is 0.

### Step 3: Calculate the Change in Oxidation Number
The change in oxidation number for Br is:
\[ \text{Change in oxidation number} = 0 - (+5) = -5 \]
Since we are interested in the absolute value for equivalent weight calculation, we take:
\[ \text{Change in oxidation number} = +5 \]

### Step 4: Determine the Number of Electrons Transferred
From the equation, we see that 10 electrons (e⁻) are involved in the reduction of bromine from +5 to 0.

### Step 5: Calculate the Equivalent Weight
The formula for equivalent weight is given by:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{Change in oxidation number}} \]

Let the molecular weight of KBrO₃ be \( M \). Therefore, the equivalent weight of KBrO₃ can be calculated as:
\[ \text{Equivalent weight of KBrO₃} = \frac{M}{5} \]

### Step 6: Conclusion
Thus, the equivalent weight of KBrO₃ is:
\[ \frac{M}{5} \]

### Final Answer
The equivalent weight of KBrO₃ is \( \frac{M}{5} \).

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