Ionic Product Of Water
Easychemistry
At 25°C, the ionic product of water (Kw) is 1.0 × 10⁻¹⁴. The pH of pure water at this temperature is:
Select the correct option:
Solution
Incorrect! 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.
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About This Question
- Subject
- chemistry
- Chapter
- equilibrium
- Topic
- ionic product of water
- Difficulty
- Easy
- Year
- 2025
This easy difficulty chemistry question is from the chapter equilibrium, covering the topic of ionic product of water. It appeared in the 2025 exam. Practice this and similar questions to strengthen your understanding of equilibrium concepts.
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