What is the pH of a 0.010 M aqueous solution of sodium hydroxide (NaOH) at 25 °C (where Kw = 1.0 x 10^-14)?
Answer and explanation
Correct answer: B. 12.0
NaOH is a strong base that dissociates completely: [OH-] = 0.010 M = 1.0 x 10^-2 M. pOH = -log10[OH-] = -log10(1.0 x 10^-2) = 2.0. Since pH + pOH = 14.0 at 25 °C, pH = 14.0 - 2.0 = 12.0.
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