The pH of a 0.001M aqueous solution of NaOH is:
Correct answer: C. 11
- A. 6
- B. 13
- C. 11
- D. 12
Explanation
NaOH is a strong base, so [OH⁻] = 0.001 M = 10⁻³ M. The pOH is -log[OH⁻] = -log(10⁻³) = 3. The pH is calculated as pH = 14 - pOH = 14 - 3 = 11.
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