Reviewed CSCA Chemistry question · Standard

在25℃时，将0.40 g氢氧化钠固体溶于水，配成200 mL溶液。取该溶液10.0 mL，加水稀释至100 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）

12.0
13.0
11.0
10.0

Show the answer and explanation

Correct answer

A. 12.0

Principle or equation

pH = -lg c(H+)，c(H+) = Kw / c(OH-)。强碱溶液稀释时，OH-物质的量不变，浓度按体积比例稀释。

Why this answer is correct

原溶液c(NaOH)=0.40g/(40g/mol)/0.200L=0.050 mol/L。取10.0mL稀释至100mL，浓度变为0.050×(10.0/100)=0.0050 mol/L。c(OH-)=0.0050 mol/L，c(H+)=1.0×10^-14/0.0050=2.0×10^-12 mol/L，pH=-lg(2.0×10^-12)=11.7，最接近12.0。

Example

若将0.020 mol/L的NaOH溶液稀释10倍，c(OH-)=0.0020 mol/L，pH=11.3。

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.