在 25℃时，将 0.80 g 氢氧化钠固体溶于水，配成 200 mL 溶液。取该溶液 25.0 mL，加水稀释至 500 mL。稀释后溶液的 pH 值最接近多少？（已知 NaOH 的摩尔质量为 40 g/mol，忽略体积变化，25℃时 Kw=1.0×10^-14）
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Correct answer
B. 12
Principle or equation
c(NaOH) = n/V，稀释时溶质物质的量不变，pH = 14 - pOH，pOH = -lg[OH-]。
Why this answer is correct
原溶液浓度 c1 = (0.80 g / 40 g/mol) / 0.200 L = 0.10 mol/L。取 25.0 mL 稀释至 500 mL，稀释后浓度 c2 = c1 × 25.0/500 = 0.10 × 0.05 = 0.0050 mol/L。NaOH 完全电离，[OH-] = 0.0050 mol/L，pOH = -lg(0.0050) ≈ 2.3，pH = 14 - 2.3 = 11.7，最接近 12。
Example
若将 0.1 mol/L NaOH 稀释 10 倍，浓度变为 0.01 mol/L，pH 约为 12。
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