For pure water `(pH=7), K_(w)` at 298 is `10^(-14)`. On adding some acid to it, its pH changes to 3. The value of `K_(w)` for this acidicfied water will be

Answer

To solve the problem, we need to understand the relationship between pH, the concentration of hydrogen ions \([H^+]\), and the ion product of water \(K_w\).

### Step-by-Step Solution:

1. **Understanding the pH Scale**:
- The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration:
\[ pH = -\log[H^+] \]

2. **Calculating \([H^+]\) for pH = 3**:
- Given that the pH of the acidified water is 3, we can calculate the concentration of hydrogen ions:
\[ [H^+] = 10^{-pH} = 10^{-3} \, \text{mol/L} \]

3. **Using the Ion Product of Water \(K_w\)**:
- The ion product of water at any temperature is given by:
\[ K_w = [H^+][OH^-] \]
- At 298 K, the value of \(K_w\) is given as \(10^{-14}\).

4. **Finding \([OH^-]\)**:
- Since we know \(K_w\) and \([H^+]\), we can find \([OH^-]\):
\[ K_w = [H^+][OH^-] \]
\[ 10^{-14} = (10^{-3})[OH^-] \]
- Rearranging the equation to find \([OH^-]\):
\[ [OH^-] = \frac{10^{-14}}{10^{-3}} = 10^{-11} \, \text{mol/L} \]

5. **Conclusion about \(K_w\)**:
- The value of \(K_w\) is dependent on temperature, not on the concentrations of ions in the solution. Since the temperature remains constant at 298 K, the value of \(K_w\) for the acidified water remains the same:
\[ K_w = 10^{-14} \]

### Final Answer:
The value of \(K_w\) for the acidified water is \(10^{-14}\).

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