pH of 0.1 M `Na_2HPO_4` and 0.2 M `NaH_2PO_4` are respectively : (`pK_a` for `H_3PO_4` are 2.2,7.2,12.0)

Options

A. 4.7,9.6
B. 9.6,4.7
C. 4.7,5.6
D. 5.6,4.7

Answer

To find the pH of the solutions of `Na2HPO4` and `NaH2PO4`, we can use the Henderson-Hasselbalch equation, which is applicable for buffer solutions.

### Step-by-Step Solution:

1. **Identify the relevant acid-base pairs:**
- `Na2HPO4` contains the anion `HPO4^2-`, which is the conjugate base of `H2PO4^-`.
- `NaH2PO4` contains the anion `H2PO4^-`, which is the conjugate base of `H3PO4`.

2. **Determine the pKa values:**
- The dissociation of phosphoric acid (`H3PO4`) occurs in three steps:
1. `H3PO4 ⇌ H2PO4^- + H^+` (pKa1 = 2.2)
2. `H2PO4^- ⇌ HPO4^2- + H^+` (pKa2 = 7.2)
3. `HPO4^2- ⇌ PO4^3- + H^+` (pKa3 = 12.0)

3. **Calculate the pH of `Na2HPO4`:**
- For `Na2HPO4`, we use the second dissociation (pKa2) and the third dissociation (pKa3):
\[ \text{pH} = \frac{\text{pKa2} + \text{pKa3}}{2} \]
Substituting the values:
\[ \text{pH} = \frac{7.2 + 12.0}{2} = \frac{19.2}{2} = 9.6 \]

4. **Calculate the pH of `NaH2PO4`:**
- For `NaH2PO4`, we use the first dissociation (pKa1) and the second dissociation (pKa2):
\[ \text{pH} = \frac{\text{pKa1} + \text{pKa2}}{2} \]
Substituting the values:
\[ \text{pH} = \frac{2.2 + 7.2}{2} = \frac{9.4}{2} = 4.7 \]

5. **Final Results:**
- The pH of `0.1 M Na2HPO4` is **9.6**.
- The pH of `0.2 M NaH2PO4` is **4.7**.

### Summary:
- The pH of `0.1 M Na2HPO4` is **9.6**.
- The pH of `0.2 M NaH2PO4` is **4.7**.