Question

The value of ionic product of water at 25°C is:

• A. \(10^{-4}\)
• B. \(10^{-40}\)
• C. \(10^{-14}\)
• D. \(10^{-24}\)

Hint:

K\(_w\) = \(10^{-14}\) at 25°C.
pH = -log[H\(^+\)].
Neutral pH = 7 at 25°C.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The ionic product of water is a constant.
It is the product of H\(^+\) and OH\(^-\) concentrations.

Step 2: Key Formula or Approach:
K\(_w\) = [H\(^+\)][OH\(^-\)] = \(10^{-14}\) at 25°C.

Step 3: Detailed Explanation:
At 25°C, the concentration of H\(^+\) and OH\(^-\) in pure water is \(10^{-7}\) M each.
Thus, K\(_w\) = \(10^{-7}\) \(\times\) \(10^{-7}\) = \(10^{-14}\).
This is a fundamental constant in acid-base chemistry.

Step 4: Final Answer:
The ionic product of water at 25°C is \(10^{-14}\).
Hence, the correct option is (C).