The pH of pure water at `90^@`C will be

Options

A. less than 7
B. more than 7
C. equal to 7
D. cannot be predicted

Answer

To determine the pH of pure water at \(90^\circ C\), we need to understand how temperature affects the dissociation of water and the value of \(K_w\) (the ion product of water).

### Step-by-Step Solution:

1. **Understanding the Ionization of Water**:
Water (\(H_2O\)) dissociates into hydrogen ions (\(H^+\)) and hydroxide ions (\(OH^-\)):
\[ H_2O \rightleftharpoons H^+ + OH^- \]

2. **Ion Product of Water (\(K_w\))**:
At \(25^\circ C\), the value of \(K_w\) is \(1.0 \times 10^{-14}\). However, as the temperature increases, \(K_w\) also increases. At \(90^\circ C\), \(K_w\) is approximately \(7.5 \times 10^{-14}\).

3. **Finding the Concentration of Ions**:
In pure water at equilibrium, the concentrations of \(H^+\) and \(OH^-\) are equal. Therefore, we can express \(K_w\) as:
\[ K_w = [H^+][OH^-] = [H^+]^2 \]
Thus, at \(90^\circ C\):
\[ [H^+]^2 = 7.5 \times 10^{-14} \]
To find \([H^+]\), we take the square root:
\[ [H^+] = \sqrt{7.5 \times 10^{-14}} \approx 8.66 \times 10^{-7} \, \text{M} \]

4. **Calculating pH**:
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \([H^+]\):
\[ \text{pH} = -\log(8.66 \times 10^{-7}) \approx 6.06 \]

5. **Conclusion**:
Therefore, the pH of pure water at \(90^\circ C\) is approximately \(6.06\), which is less than \(7\).

### Final Answer:
The pH of pure water at \(90^\circ C\) is approximately \(6.06\).