pH is the negative logarithm of `H^(+)`
` pH=log[H^(+)]`
`HCllt H^(+)+Cl^(-)`
`H_(2)OhArrH^(+)+OH^(-)`
` K_(W)=[H^(+)][OH^(-)]`
`K_(W)` depend on the temperatue. With rise in temperature `K_(W)` increases.
At 298 K, pH of pure water = 7
At 373 K, pH of the pure water is

Options

A. 7
B. `gt7`
C. `lt7`
D. Cannot be stated

Answer

To solve the question regarding the pH of pure water at 373 K, we will follow these steps:

### Step 1: Understand the relationship between pH and [H⁺]
The pH is defined as the negative logarithm of the hydrogen ion concentration:
\[ \text{pH} = -\log[H^+] \]

### Step 2: Recall the value of \( K_w \)
At 298 K, the value of \( K_w \) (the ion product of water) is \( 1.0 \times 10^{-14} \). This means that at this temperature:
\[ [H^+][OH^-] = 1.0 \times 10^{-14} \]
In pure water, the concentration of hydrogen ions \( [H^+] \) is equal to the concentration of hydroxide ions \( [OH^-] \), so:
\[ [H^+] = [OH^-] = \sqrt{K_w} = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \, \text{M} \]
Thus, the pH at 298 K is:
\[ \text{pH} = -\log(1.0 \times 10^{-7}) = 7 \]

### Step 3: Analyze the effect of temperature on \( K_w \)
The problem states that \( K_w \) increases with temperature. At 373 K, \( K_w \) is approximately \( 2.0 \times 10^{-14} \) (this value can vary slightly depending on the source).

### Step 4: Calculate the new concentration of \( [H^+] \)
Using the new value of \( K_w \):
\[ K_w = [H^+][OH^-] = 2.0 \times 10^{-14} \]
In pure water, again, \( [H^+] = [OH^-] \). Therefore:
\[ [H^+]^2 = 2.0 \times 10^{-14} \]
Taking the square root:
\[ [H^+] = \sqrt{2.0 \times 10^{-14}} \approx 1.41 \times 10^{-7} \, \text{M} \]

### Step 5: Calculate the pH at 373 K
Now, we can find the pH at 373 K:
\[ \text{pH} = -\log(1.41 \times 10^{-7}) \]
Calculating this gives:
\[ \text{pH} \approx 6.85 \]

### Conclusion
Thus, the pH of pure water at 373 K is approximately 6.85, which is less than 7.