The pH of a solution containing 0.4 M `HCO_(3)^(-)` and 0.2 M `CO_(3)^(2-)` is :
`[K_(a1)(H_(2)CO_(3))=4xx10^(-7)` , `K_(a2)(HCO_(3)^(-))=4xx10^(-11)]`

Options

A. `10.4`
B. `10.1`
C. `6.1`
D. `10.7`

Answer

To find the pH of a solution containing 0.4 M \( \text{HCO}_3^- \) and 0.2 M \( \text{CO}_3^{2-} \), we can use the Henderson-Hasselbalch equation, which is applicable for buffer solutions. The equation is given by:

\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]

### Step 1: Identify the Acid and Base
In this case:
- The acid is \( \text{HCO}_3^- \) (bicarbonate ion)
- The base is \( \text{CO}_3^{2-} \) (carbonate ion)

### Step 2: Determine the Relevant \( K_a \) Value
We need to use the \( K_a \) value for the dissociation of \( \text{HCO}_3^- \) to \( \text{CO}_3^{2-} \):
\[ \text{HCO}_3^- \rightleftharpoons \text{CO}_3^{2-} + \text{H}^+ \]
Given:
\[ K_{a2}(\text{HCO}_3^-) = 4 \times 10^{-11} \]

### Step 3: Calculate \( \text{pKa} \)
To find \( \text{pKa} \):
\[ \text{pKa} = -\log(K_{a2}) = -\log(4 \times 10^{-11}) \]
Calculating \( \text{pKa} \):
\[ \text{pKa} = 11 - \log(4) \approx 11 - 0.6 = 10.4 \]

### Step 4: Apply the Henderson-Hasselbalch Equation
Now, substituting the values into the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
Substituting the concentrations:
- \( [\text{Base}] = [\text{CO}_3^{2-}] = 0.2 \, \text{M} \)
- \( [\text{Acid}] = [\text{HCO}_3^-] = 0.4 \, \text{M} \)

Thus,
\[ \text{pH} = 10.4 + \log \left( \frac{0.2}{0.4} \right) \]
Calculating the log term:
\[ \log \left( \frac{0.2}{0.4} \right) = \log(0.5) \approx -0.301 \]
So,
\[ \text{pH} = 10.4 - 0.301 = 10.099 \]

### Step 5: Final Answer
Rounding off, the pH of the solution is approximately:
\[ \text{pH} \approx 10.1 \]

### Summary
The pH of the solution containing 0.4 M \( \text{HCO}_3^- \) and 0.2 M \( \text{CO}_3^{2-} \) is **10.1**.