题目
CHM-1020-LD01.2025SU PS II- Part I (Acids and Bases)
单项选择题
A vinegar sample had a pH of 3.88. This corresponds to [OH-] =
选项
A.1.3 × 10−4 M.
B.7.6 × 10−11 M.
C.7.6 × 103 M.
D.10.12 M.
E.8.7 × 10−4 M.
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标准答案
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思路分析
To relate pH to [OH-], recall that pH = -log10[H+], and at 25°C the product [H+][OH-] = 1.0 × 10^-14, so [OH-] = 1.0 × 10^-14 / [H+].
Option 1: 1.3 × 10^-4 M. This value is approximately [H+] for a pH around 3.89 (since [H+] ≈ 10^-pH ≈ 1.3 × 10^-4 M). ......Login to view full explanation登录即可查看完整答案
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