Calculate the pH of a solution that is 0.40 M HF and 1.00 M KF.?
回答 (1)
2018-02-21 10:41 am
Refer to: http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf
Ka for HF = 7.2 × 10⁻⁴ and pKa for HF = 3.14
(Different pKa values would lead to different answers.)
Consider the dissociation of HF :
HF(aq) ⇌ H⁺(aq) + F⁻(aq)
pH = pKa + log([F⁻]/[HF]) = 3.14 + log(1.00/0.40) = 3.54
Ka for HF = 7.2 × 10⁻⁴ and pKa for HF = 3.14
(Different pKa values would lead to different answers.)
Consider the dissociation of HF :
HF(aq) ⇌ H⁺(aq) + F⁻(aq)
pH = pKa + log([F⁻]/[HF]) = 3.14 + log(1.00/0.40) = 3.54
收錄日期: 2021-05-01 13:46:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180221020021AAbo4w5