In a titration, `H_(2)O_(2)` is oxidised to `O_(2)` by `MnO_(4)^(-)`. 24 mL of 0.1M `H_(2)O_(2)` requires 16 mL of 0.1M `MnO_(4)^(-)` solution. Hence `MnO_(4)^(-)` changes to :

Explanation

To solve the problem, we need to determine what happens to the permanganate ion (MnO4^-) during the titration with hydrogen peroxide (H2O2). Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the reaction and the oxidation states
In the reaction, hydrogen peroxide (H2O2) is oxidized to oxygen (O2) by permanganate ion (MnO4^-). We need to find the oxidation states of the relevant species.

- For H2O2:
- The oxidation state of O in H2O2 is -1 (since there are 2 H atoms contributing +2).
- In O2, the oxidation state of O is 0.

- For MnO4^-:
- The oxidation state of Mn can be calculated as follows:
- Let the oxidation state of Mn be x.
- The equation is: x + 4(-2) = -1 (since the overall charge is -1).
- This simplifies to: x - 8 = -1, thus x = +7.

### Step 2: Calculate the N-factor for H2O2
The N-factor is defined as the number of electrons exchanged per molecule during the reaction.

- The change in oxidation state for oxygen in H2O2 is from -1 to 0.
- Since there are 2 oxygen atoms in H2O2, the total number of electrons exchanged is:
- Change in oxidation state = 0 - (-1) = 1 electron per O atom.
- Therefore, N-factor for H2O2 = 2 (because there are 2 O atoms).

### Step 3: Calculate the equivalence of H2O2
Using the formula for equivalence:
\[ \text{Equivalence} = \text{N-factor} \times \text{Molarity} \times \text{Volume (in L)} \]

For H2O2:
- Molarity = 0.1 M
- Volume = 24 mL = 0.024 L
- N-factor = 2

Calculating equivalence:
\[ \text{Equivalence of H2O2} = 2 \times 0.1 \times 0.024 = 0.0048 \text{ equivalents} \]

### Step 4: Calculate the equivalence of MnO4^-
Now, we set the equivalence of H2O2 equal to the equivalence of MnO4^-.

Let the N-factor for MnO4^- be n. The molarity and volume for MnO4^- are given as:
- Molarity = 0.1 M
- Volume = 16 mL = 0.016 L

Using the equivalence formula:
\[ \text{Equivalence of MnO4^-} = n \times 0.1 \times 0.016 \]

Setting the equivalences equal:
\[ 0.0048 = n \times 0.1 \times 0.016 \]

### Step 5: Solve for N-factor of MnO4^-
Rearranging the equation to solve for n:
\[ n = \frac{0.0048}{0.1 \times 0.016} = \frac{0.0048}{0.0016} = 3 \]

### Step 6: Determine the change in oxidation state of Mn
Since the N-factor for MnO4^- is 3, it means that Mn changes from +7 to +4 (since it is reduced).

### Conclusion
Thus, the permanganate ion (MnO4^-) changes to manganese(IV) oxide (MnO2) during the titration.

### Final Answer
**MnO4^- changes to MnO2.**

---