Difficulty: Easy
Correct Answer: 10^-11 M
Explanation:
Introduction / Context:In water at 25 °C, the autoionization equilibrium defines the ion product Kw = [H+][OH−] = 1.0 × 10^-14. Using pH and pOH relationships enables rapid interconversion of hydrogen- and hydroxide-ion concentrations for acid–base calculations in environmental, biochemical, and process applications.
Given Data / Assumptions:
Concept / Approach:Use the relations pH = −log10[H+] and pOH = −log10[OH−], with pH + pOH = 14 at 25 °C. From pH we first get [H+], then compute pOH and [OH−].
Step-by-Step Solution:
Given pH = 3 ⇒ [H+] = 10^-3 M.pOH = 14 − pH = 14 − 3 = 11.Therefore, [OH−] = 10^-pOH = 10^-11 M.Check: [H+][OH−] = 10^-3 × 10^-11 = 10^-14 = Kw at 25 °C.Verification / Alternative check:Directly from Kw: [OH−] = Kw/[H+] = 1.0 × 10^-14 / 10^-3 = 10^-11 M, identical to the pOH method.
Why Other Options Are Wrong:
Common Pitfalls:Forgetting the temperature dependence of Kw; at temperatures other than 25 °C, pH + pOH ≠ 14 exactly, though the qualitative approach remains valid.
Final Answer:10^-11 M
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