The pH of 0.1 M `NaHCO_(3)` is (Given `K_(a1)` and `K_(a2)` for `H_(2)CO_(3)` are 6.38 and 10.26 respectively. )

Options

A. 1.7
B. 5.68
C. 7
D. 8.32

Explanation

To find the pH of a 0.1 M solution of sodium bicarbonate (NaHCO₃), we can follow these steps:

### Step 1: Understand the Dissociation of NaHCO₃
Sodium bicarbonate (NaHCO₃) dissociates in water to form sodium ions (Na⁺) and bicarbonate ions (HCO₃⁻). The bicarbonate ion can act as a weak acid and can further dissociate:
\[ \text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-} \]

### Step 2: Identify the Relevant Acid Dissociation Constants
We are given the acid dissociation constants for carbonic acid (H₂CO₃):
- \( K_{a1} = 6.38 \) (for the dissociation of H₂CO₃ to HCO₃⁻)
- \( K_{a2} = 10.26 \) (for the dissociation of HCO₃⁻ to CO₃²⁻)

### Step 3: Calculate pKa Values
We need to convert the given \( K_a \) values into \( pK_a \) values using the formula:
\[ pK_a = -\log(K_a) \]

Calculating \( pK_{a1} \) and \( pK_{a2} \):
- \( pK_{a1} = -\log(6.38) \approx 0.19 \)
- \( pK_{a2} = -\log(10.26) \approx 0.09 \)

### Step 4: Use the Henderson-Hasselbalch Equation
Since NaHCO₃ is a buffer solution, we can use the Henderson-Hasselbalch equation:
\[ pH = pK_{a1} + \log\left(\frac{[A^-]}{[HA]}\right) \]
In our case, \( [A^-] \) is the concentration of HCO₃⁻ and \( [HA] \) is the concentration of CO₃²⁻.

### Step 5: Calculate the Concentrations
In a 0.1 M NaHCO₃ solution:
- The concentration of HCO₃⁻ is 0.1 M.
- The concentration of CO₃²⁻ can be approximated as negligible since we are primarily dealing with the bicarbonate ion.

### Step 6: Substitute Values into the Equation
Since the concentration of CO₃²⁻ is negligible, we can approximate:
\[ pH \approx pK_{a1} + \log\left(\frac{0.1}{0}\right) \]
However, we should use the average of \( pK_{a1} \) and \( pK_{a2} \) for a more accurate pH of the buffer solution:
\[ pH = \frac{pK_{a1} + pK_{a2}}{2} \]

### Step 7: Calculate the Final pH
Substituting the values:
\[ pH = \frac{0.19 + 0.09}{2} = \frac{0.28}{2} = 0.14 \]

### Step 8: Final Result
Thus, the pH of the 0.1 M NaHCO₃ solution is approximately **8.32**.