Pure water ionises as
`2H_2O (?) Leftrightarrow H_3O^+ (a)+OH^¯" (aq) At "25^@C` the pH of pure water is approximately 7.0 At `37^@C` its pH is -

Options

A. More than 7.0
B. Less than 7.0
C. Equal to 7.0
D. None of these

Answer

To solve the question regarding the pH of pure water at 37°C, we will follow these steps:

### Step 1: Understand the Ionization of Water
Pure water ionizes according to the equation:
\[ 2H_2O \rightleftharpoons H_3O^+ (aq) + OH^- (aq) \]

At equilibrium, the concentrations of hydronium ions \([H_3O^+]\) and hydroxide ions \([OH^-]\) are equal in pure water.

### Step 2: Know the Ionic Product of Water
At 25°C, the ionic product of water (\(K_w\)) is:
\[ K_w = [H_3O^+][OH^-] = 10^{-14} \]

Since \([H_3O^+] = [OH^-]\) in pure water, we can set:
\[ [H_3O^+] = [OH^-] = x \]
Thus,
\[ x^2 = 10^{-14} \]
\[ x = 10^{-7} \]
This gives us:
\[ [H_3O^+] = 10^{-7} \]

### Step 3: Calculate the pH at 25°C
The pH is calculated using the formula:
\[ \text{pH} = -\log[H_3O^+] \]
Substituting the value of \([H_3O^+]\):
\[ \text{pH} = -\log(10^{-7}) = 7 \]

### Step 4: Consider the Effect of Temperature on \(K_w\)
When the temperature increases (from 25°C to 37°C), the ionization of water increases, which means that \(K_w\) also increases. For example, at 37°C, \(K_w\) is approximately \(10^{-13}\).

### Step 5: Set Up the Equation for 37°C
At 37°C, we can set:
\[ K_w = [H_3O^+][OH^-] = 10^{-13} \]
Again, since \([H_3O^+] = [OH^-]\) in pure water, we have:
\[ x^2 = 10^{-13} \]
Thus,
\[ x = 10^{-6.5} \]
This means:
\[ [H_3O^+] = 10^{-6.5} \]

### Step 6: Calculate the pH at 37°C
Now, we can calculate the pH:
\[ \text{pH} = -\log[H_3O^+] \]
Substituting the value of \([H_3O^+]\):
\[ \text{pH} = -\log(10^{-6.5}) = 6.5 \]

### Step 7: Conclusion
Since the pH at 37°C is less than 7, we conclude that:
- The pH of pure water at 37°C is less than 7.

### Final Answer
The pH of pure water at 37°C is less than 7.

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