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 follow these steps:

### Step 1: Identify the components
- **NaH₂PO₄** is the salt of the weak acid H₂PO₄⁻ (acting as the weak acid).
- **Na₂HPO₄** is the salt of the weak base HPO₄²⁻ (acting as the conjugate base).

### Step 2: Determine the moles of each component
- For **NaH₂PO₄**:
\[ \text{Moles of NaH₂PO₄} = \text{Volume (L)} \times \text{Molarity (M)} = 0.010 \, \text{L} \times 0.1 \, \text{M} = 0.001 \, \text{mol} = 1 \, \text{mmol} \]

- For **Na₂HPO₄**:
\[ \text{Moles of Na₂HPO₄} = \text{Volume (L)} \times \text{Molarity (M)} = 0.015 \, \text{L} \times 0.1 \, \text{M} = 0.0015 \, \text{mol} = 1.5 \, \text{mmol} \]

### Step 3: Calculate the total volume of the solution
\[ \text{Total Volume} = 10 \, \text{mL} + 15 \, \text{mL} = 25 \, \text{mL} \]

### Step 4: Calculate the concentrations of the acid and the conjugate base
- Concentration of H₂PO₄⁻ (acid):
\[ [\text{H₂PO₄⁻}] = \frac{\text{moles}}{\text{total volume (L)}} = \frac{0.001 \, \text{mol}}{0.025 \, \text{L}} = 0.04 \, \text{M} \]

- Concentration of HPO₄²⁻ (conjugate base):
\[ [\text{HPO₄²⁻}] = \frac{\text{moles}}{\text{total volume (L)}} = \frac{0.0015 \, \text{mol}}{0.025 \, \text{L}} = 0.06 \, \text{M} \]

### Step 5: Use the Henderson-Hasselbalch equation
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]
Given:
- \( \text{pK}_a = 7.2 \) (for H₂PO₄⁻)
- \( [\text{Base}] = [\text{HPO₄²⁻}] = 0.06 \, \text{M} \)
- \( [\text{Acid}] = [\text{H₂PO₄⁻}] = 0.04 \, \text{M} \)

Substituting the values:
\[ \text{pH} = 7.2 + \log\left(\frac{0.06}{0.04}\right) \]

### Step 6: Calculate the log term
\[ \frac{0.06}{0.04} = 1.5 \quad \Rightarrow \quad \log(1.5) \approx 0.176 \]

### Step 7: Final calculation of pH
\[ \text{pH} = 7.2 + 0.176 = 7.376 \]

Thus, the pH of the solution is approximately **7.38**.

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