在25℃时，将0.30 g 醋酸（CH3COOH，摩尔质量60 g/mol）溶于水配成500 mL溶液。已知醋酸的电离度约为1.0%，忽略水的电离，该溶液的pH最接近多少？
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Correct answer
B. 3.0
Principle or equation
pH = -lg[H+]，对于弱酸，[H+] = c(酸) × 电离度。
Why this answer is correct
首先计算醋酸的物质的量：n = 0.30 g / 60 g/mol = 0.005 mol。溶液体积为0.500 L，所以浓度c = 0.005 mol / 0.500 L = 0.010 mol/L。电离度α = 1.0% = 0.010，因此[H+] = cα = 0.010 × 0.010 = 1.0×10^-4 mol/L。pH = -lg(1.0×10^-4) = 4.0。
Example
若0.10 mol/L醋酸电离度为1%，则[H+]=0.10×0.01=1.0×10^-3 mol/L，pH=3。
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