Calculate the concentration of `CO_(2)` in a soft drink that is bottled with a partial pressure of `CO_(2)` of 4 atm over the liquid at `25^(@)C`. The Henry's law constant for `CO_(2)` in water at `25^(@)C` is `3.1 xx 10^(-2) "mol/litre-atm"`.

Answer

To calculate the concentration of \( CO_2 \) in a soft drink that is bottled with a partial pressure of \( CO_2 \) of 4 atm at \( 25^\circ C \), we will use Henry's Law.

### Step-by-Step Solution:

1. **Understand Henry's Law**:
Henry's Law states that the concentration of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. The mathematical expression of Henry's Law is:
\[ P = k_H \cdot C \]
where:
- \( P \) is the partial pressure of the gas (in atm),
- \( k_H \) is Henry's Law constant (in mol/L·atm),
- \( C \) is the concentration of the gas in the liquid (in mol/L).

2. **Identify Given Values**:
- Partial pressure of \( CO_2 \), \( P = 4 \, \text{atm} \)
- Henry's Law constant for \( CO_2 \) in water at \( 25^\circ C \), \( k_H = 3.1 \times 10^{-2} \, \text{mol/L·atm} \)

3. **Rearrange Henry's Law Equation**:
To find the concentration \( C \), we rearrange the equation:
\[ C = \frac{P}{k_H} \]

4. **Substitute the Values**:
Now, substitute the known values into the rearranged equation:
\[ C = \frac{4 \, \text{atm}}{3.1 \times 10^{-2} \, \text{mol/L·atm}} \]

5. **Calculate the Concentration**:
Performing the calculation:
\[ C = \frac{4}{3.1 \times 10^{-2}} = \frac{4}{0.031} \approx 129.03 \, \text{mol/L} \]

6. **Final Result**:
The concentration of \( CO_2 \) in the soft drink is approximately:
\[ C \approx 129.03 \, \text{mol/L} \]