If the molecular weight of `Ba(MnO_(4))_(2)` is M , then the equivalent weight of `Ba(MnO_(4))_(2)` in acidic medium is

Options

A. `M/5`
B. `M/10`
C. `M/3`
D. M

Answer

To find the equivalent weight of `Ba(MnO₄)₂` in acidic medium, we can follow these steps:

### Step 1: Determine the molecular weight of `Ba(MnO₄)₂`
The molecular weight (M) of `Ba(MnO₄)₂` is given in the question.

### Step 2: Identify the oxidation states of manganese in `MnO₄⁻`
In the permanganate ion (`MnO₄⁻`), the oxidation state of manganese (Mn) can be calculated as follows:
- Let the oxidation state of Mn be x.
- The total charge of the ion is -1.
- The contribution from 4 oxygen atoms (each with an oxidation state of -2) is -8.

So, we can set up the equation:
\[ x + 4(-2) = -1 \]
\[ x - 8 = -1 \]
\[ x = +7 \]

### Step 3: Determine the change in oxidation state of manganese
In acidic medium, `MnO₄⁻` is reduced to `Mn²⁺`. The change in oxidation state is:
- From +7 (in `MnO₄⁻`) to +2 (in `Mn²⁺`).

### Step 4: Calculate the number of electrons gained per manganese ion
The number of electrons gained by one manganese ion is:
\[ 7 - 2 = 5 \text{ electrons} \]

### Step 5: Calculate the total number of electrons gained by `Ba(MnO₄)₂`
Since there are 2 manganese ions in `Ba(MnO₄)₂`, the total number of electrons gained is:
\[ 2 \times 5 = 10 \text{ electrons} \]

### Step 6: Calculate the equivalent weight
The equivalent weight (EW) can be calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Number of electrons gained}} \]
Substituting the values:
\[ \text{Equivalent Weight} = \frac{M}{10} \]

### Conclusion
Thus, the equivalent weight of `Ba(MnO₄)₂` in acidic medium is:
\[ \frac{M}{10} \]

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