常温下，将 0.04 g 氢氧化钠固体溶于水配成 1 L 溶液，则该溶液的 pH 为多少？（NaOH 摩尔质量为 40 g/mol，忽略体积变化，25℃时 Kw=1.0×10^-14）
Show the answer and explanation
Correct answer
A. 11
Principle or equation
c(OH-) = n(NaOH)/V，pOH = -lg c(OH-)，pH = 14 - pOH。
Why this answer is correct
n(NaOH) = 0.04 g / 40 g/mol = 0.001 mol；c(OH-) = 0.001 mol / 1 L = 0.001 mol/L；pOH = 3；pH = 14 - 3 = 11。
Example
若 0.08 g NaOH 配成 1 L 溶液，则 c(OH-) = 0.002 mol/L，pOH≈2.3，pH≈11.7。
This published item includes a stored explanation and passed the platform’s publication workflow. It is independent preparation material, not a claim of an official or recalled examination question.