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向 99.5mL 0.2mol/L的NaOH溶液中加入 100.5mL 0.1mol/L的H2SO4后,溶液的pH为多少?______.
人气:328 ℃ 时间:2020-04-22 22:23:07
解答
100.5mL 0.1mol/L的H2SO4 n(H+)=0.1005L×0.1mol/L×2=0.0201mol,99.5mL 0.2mol/L的NaOH溶液n(OH-)=0.0995L×0.2mol/L=0.0199mol,反应后溶液的体积为100.5mL+99.5mL=200mL=0.2L,则反应后c(H+)=
0.0201mol−0.0199mol
0.2L
=1×10-3,pH=-lgc(H+),pH=3,
故答案为:3.
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