Ionic Product Of Water
At 25°C, the ionic product of water (Kw) is 1.0 × 10⁻¹⁴. The pH of pure water at this temperature is:
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About This Question
- Subject
- chemistry
- Chapter
- equilibrium
- Topic
- ionic product of water
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
7
The Ionic Product of Water (Kw) is the product of concentrations of H+ and OH− ions. Kw=[H+][OH−]=1.0×10−14 at 25∘C.
- In pure water (neutral), [H+]=[OH−].
- Let [H+]=x. Then x2=1.0×10−14.
- x=[H+]=10−14=10−7 M.
- pH=−log[H+]=−log(10−7)=7.
- This is why a pH of 7 defines a neutral solution at 25∘C.
This easy difficulty chemistry question is from the chapter equilibrium, covering the topic of ionic product of water. It appeared in the 2025 exam.
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