Acetic Acid + H2O? 10 points to best answer.?
2018-03-08 11:26 pm
This experiment assumes that the pH = pKa (at the half-neutralization point) of the acid. Why? (hint: use the Henderson-Hasselbach equation.)
回答 (2)
2018-03-08 11:37 pm
✔ 最佳答案
CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l) …… KaAt the half-neutralization point, a half of CH₃COOH is converted to CH₃COO⁻ ions. Therefore, [CH₃COO⁻] = [CH₃COOH], and thus [CH₃COO⁻]/[CH₃COOH] = 1
Henderson-Hasselbach equation:
pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
pH = pKa + log(1)
Hence, pH = pKa
2018-03-08 11:32 pm
HH equation is pH = pKa + log [salt]/[acid]
At the equivalence point, the [salt] =[acid] and so the [salt]/[acid] is 1 and log of 1 is 0. Thus,
pH = pKa + 0 or pH = pKa
At the equivalence point, the [salt] =[acid] and so the [salt]/[acid] is 1 and log of 1 is 0. Thus,
pH = pKa + 0 or pH = pKa
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