Choose the incorrect match :

Options

A. `K[PtCl_(3)(C_(2)H_(4))]` : Zeise's salt
B. `Fe(CO)_(5)` : Trigonal bipyramidal
C. `[Co(NH_(3))_(6)]^(3+)` : Inner orbital complex
D. `[Cr(NH_(3))_(6)]^(3+)` : Diamagnetic

Answer

To solve the question about the pH of a buffer solution formed by mixing NaH2PO4 and Na2HPO4, we will follow these steps:

### Step 1: Identify the components of the buffer solution
We have two components:
- **NaH2PO4** (which acts as the weak acid)
- **Na2HPO4** (which acts as the salt of the weak acid)

### Step 2: Determine the volumes and concentrations
We are given:
- Volume of NaH2PO4 = 10 ml
- Concentration of NaH2PO4 = 0.1 M
- Volume of Na2HPO4 = 15 ml
- Concentration of Na2HPO4 = 0.1 M

### Step 3: Calculate the milli equivalents of each component
To find the milli equivalents, we use the formula:
\[ \text{milli equivalents} = \text{volume (L)} \times \text{concentration (mol/L)} \times 1000 \]

- For **NaH2PO4**:
\[ \text{milli equivalents of NaH2PO4} = 0.01 \, \text{L} \times 0.1 \, \text{mol/L} \times 1000 = 1 \, \text{meq} \]

- For **Na2HPO4**:
\[ \text{milli equivalents of Na2HPO4} = 0.015 \, \text{L} \times 0.1 \, \text{mol/L} \times 1000 = 1.5 \, \text{meq} \]

### Step 4: Use the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation for a buffer solution is given by:
\[ \text{pH} = \text{pKa} + \log\left(\frac{\text{milli equivalents of base}}{\text{milli equivalents of acid}}\right) \]

Here, the base is Na2HPO4 and the acid is NaH2PO4. The pKa for the phosphate buffer system (H2PO4^- to HPO4^2-) is approximately 7.2.

### Step 5: Substitute the values into the equation
\[ \text{pH} = 7.2 + \log\left(\frac{1.5}{1}\right) \]
\[ \text{pH} = 7.2 + \log(1.5) \]

Using the logarithmic value:
\[ \log(1.5) \approx 0.176 \]

So,
\[ \text{pH} = 7.2 + 0.176 \approx 7.376 \]

### Step 6: Round the pH value
Rounding this value gives us approximately:
\[ \text{pH} \approx 7.4 \]

### Step 7: Conclusion
The closest answer to our calculated pH is 7.5, which corresponds to option D.

### Final Answer:
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 **7.5** (Option D).

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