A buffer solution 0.04 M in `Na_(2)HPO_(4)` and 0.02 in `Na_(3)PO_(4)` is prepared. The electrolytic oxidation of 1.0 milli-mole of the organic compound RNHOH is carried out in 100 mL of the buffer. The reaction is `RNHOH+H_(2)OrarrRNO_(2)+4H^(+)+4e^(-)` The approximate pH of solution after the oxidation is complete is :
`[Given : for H_(3)PO_(4),pK_(a1)=2.2,pK_(a2)=7.20,pK_(a3)=12]`
(a)`6.90`
(b)`7.20`
(c)`7.5`
(d)None of these

Options

A. `6.90`
B. `7.20`
C. `7.5`
D. None of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial concentrations of the buffer components
We have two components in the buffer solution:
- Sodium hydrogen phosphate (Na₂HPO₄) with a concentration of 0.04 M
- Sodium phosphate (Na₃PO₄) with a concentration of 0.02 M

Since the volume of the buffer solution is 100 mL, we can calculate the millimoles of each component:
- For Na₂HPO₄:
\[ \text{Millimoles} = \text{Concentration (M)} \times \text{Volume (L)} = 0.04 \, \text{mol/L} \times 0.1 \, \text{L} = 4 \, \text{mmol} \]
- For Na₃PO₄:
\[ \text{Millimoles} = 0.02 \, \text{mol/L} \times 0.1 \, \text{L} = 2 \, \text{mmol} \]

### Step 2: Understand the reaction and its impact on the buffer
The reaction given is:
\[ \text{RNHOH} + \text{H}_2\text{O} \rightarrow \text{RNO}_2 + 4\text{H}^+ + 4e^- \]
From this reaction, we see that 1.0 mmol of RNHOH produces 4 mmol of H⁺ ions.

### Step 3: Calculate the final concentrations after the reaction
Initially, we have:
- H⁺ ions from the reaction: 4 mmol
- H₂PO₄⁻ from Na₂HPO₄: 4 mmol
- PO₄³⁻ from Na₃PO₄: 2 mmol

After the reaction, the total amount of H⁺ ions will be:
\[ \text{Total H}^+ = 4 \, \text{mmol (from the reaction)} + 0 \, \text{mmol (initially)} = 4 \, \text{mmol} \]

### Step 4: Determine the concentrations of the buffer components after the reaction
The total volume of the solution remains approximately 100 mL (assuming the volume change is negligible). Thus, the concentrations will be:
- Concentration of H₂PO₄⁻ (acting as the weak acid):
\[ [\text{H}_2\text{PO}_4^-] = \frac{4 \, \text{mmol}}{0.1 \, \text{L}} = 0.04 \, \text{M} \]
- Concentration of PO₄³⁻ (acting as the weak base):
\[ [\text{PO}_4^{3-}] = \frac{2 \, \text{mmol}}{0.1 \, \text{L}} = 0.02 \, \text{M} \]

### Step 5: Use the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]
In our case:
- The base is PO₄³⁻ and the acid is H₂PO₄⁻.
- We will use pK₂ (7.20) for the equilibrium between H₂PO₄⁻ and HPO₄²⁻.

Substituting the values into the equation:
\[ \text{pH} = 7.20 + \log\left(\frac{0.02}{0.04}\right) \]
Calculating the logarithm:
\[ \log\left(\frac{0.02}{0.04}\right) = \log(0.5) \approx -0.301 \]
Thus, the pH becomes:
\[ \text{pH} = 7.20 - 0.301 \approx 6.90 \]

### Conclusion
The approximate pH of the solution after the oxidation is complete is **6.90**.