The addition of `NaH_(2)PO_(4)` to `0.1M H_(3)PO_(4)` will cuase

Options

A. No change in `pH` value
B. Increases in its `pH` value
C. Decrease in its `pH` value
D. Change in `pH` but cannot be predicted

Explanation

To solve the question regarding the addition of `NaH2PO4` to `0.1M H3PO4`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components**:
- `NaH2PO4` is the sodium salt of the weak acid `H2PO4^-`, which acts as a weak base in this context.
- `H3PO4` is phosphoric acid, a weak acid.

2. **Understand the Reaction**:
- When `NaH2PO4` (which can donate `H2PO4^-` ions) is added to `H3PO4`, a weak acid, it creates a mixture of a weak acid and its conjugate base. This is a classic scenario for forming a buffer solution.

3. **Buffer Solution Formation**:
- The combination of a weak acid (H3PO4) and its conjugate base (H2PO4^-) results in the formation of an acidic buffer. Buffers resist changes in pH when small amounts of acid or base are added.

4. **Effect on pH**:
- In this case, since we are adding a weak base (from `NaH2PO4`) to a weak acid (`H3PO4`), the overall pH of the solution will increase. This is because the weak base will neutralize some of the acid, leading to a decrease in the concentration of free hydrogen ions (H^+) in the solution.

5. **Conclusion**:
- Therefore, the addition of `NaH2PO4` to `0.1M H3PO4` will cause an increase in the pH of the solution due to the formation of an acidic buffer.

### Final Answer:
The addition of `NaH2PO4` to `0.1M H3PO4` will cause an increase in pH, resulting in the formation of an acidic buffer.

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