Ionic Product Of Water
At 25°C, the ionic product of water K_w = 1.0 \times 10^{-14}. What is the pOH of a solution whose [H^+] = 10^{-3} M?
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About This Question
- Subject
- chemistry
- Chapter
- equilibrium
- Topic
- ionic product of water
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
11
The ionic product of water is K_w = [H^+][OH^-] = 1.0 \times 10^{-14} at 25°C. This leads to the relationship pH + pOH = 14 at 25°C, where pH = -log[H^+] and pOH = -log[OH^-]. Given [H^+] = 10^{-3} M, pH = -log(10^{-3}) = 3. Therefore pOH = 14 - pH = 14 - 3 = 11. Alternatively, [OH^-] = K_w/[H^+] = (10^{-14})/(10^{-3}) = 10^{-11} M, so pOH = -log(10^{-11}) = 11. Option 3 is the pH of the solution, not the pOH; this is the most common error where students confuse pH with pOH. Option 14 is the sum pH + pOH, not pOH alone. Option 7 would be the pOH only if the solution were neutral (equal [H^+] and [OH^-]), which it is not since [H^+] = 10^{-3} makes it acidic. This K_w application is from NCERT Equilibrium and ionic equilibrium sections, and the pH-pOH relationship is a JEE staple. Plausibility check: an acidic solution (pH = 3) must have a basic pOH (pOH > 7), and 11 satisfies this.
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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