What will be the pH of a solution formed by mixing 10 ml 0.1 M `NaH_(2)PO_(4)` and 15 mL 0.1 M `Na_(2)HPO_(4)`?
`["Given: for "H_(3)PO_(4)pK_(a_(1))=2.12, pK_(a_(2))=7.2]`

Options

A. `7.0`
B. `6.9`
C. `7.4`
D. `7.5`

Answer

To find the pH of the solution formed by mixing 10 mL of 0.1 M NaH₂PO₄ and 15 mL of 0.1 M Na₂HPO₄, we can use the Henderson-Hasselbalch equation, which is suitable for buffer solutions.

### Step-by-Step Solution:

1. **Identify the Components**:
- NaH₂PO₄ is the weak acid (H₂PO₄⁻).
- Na₂HPO₄ is the conjugate base (HPO₄²⁻).

2. **Calculate the Moles of Each Component**:
- Moles of NaH₂PO₄ = Volume (L) × Concentration (M) = 0.010 L × 0.1 M = 0.001 moles.
- Moles of Na₂HPO₄ = Volume (L) × Concentration (M) = 0.015 L × 0.1 M = 0.0015 moles.

3. **Convert Moles to Milliequivalents**:
- Since both NaH₂PO₄ and Na₂HPO₄ are salts of phosphoric acid, we can directly use the moles as milliequivalents (mEq) for this calculation:
- Milliequivalents of NaH₂PO₄ = 0.001 moles × 1000 = 1 mEq.
- Milliequivalents of Na₂HPO₄ = 0.0015 moles × 1000 = 1.5 mEq.

4. **Use the Henderson-Hasselbalch Equation**:
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]
Here, we can use pK₂ of phosphoric acid (given as 7.2) since we are dealing with the second dissociation:
\[ \text{pH} = 7.2 + \log\left(\frac{1.5}{1}\right) \]

5. **Calculate the Logarithm**:
\[ \log\left(\frac{1.5}{1}\right) = \log(1.5) \approx 0.1761 \]

6. **Final Calculation of pH**:
\[ \text{pH} = 7.2 + 0.1761 \approx 7.3761 \]
Rounding this to one decimal place gives us:
\[ \text{pH} \approx 7.4 \]

### Conclusion:
The pH of the solution formed by mixing 10 mL of 0.1 M NaH₂PO₄ and 15 mL of 0.1 M Na₂HPO₄ is approximately 7.4.