A solution contains 10.0 mL of 0.100 M HAc and 10.0 mL of 0.100 M NaAc in a beaker. The pH of
this solution is measured to be 4.74. Compute the pKa of HAc.
Compute the pKa of HAc.?
2019-02-20 9:07 am
回答 (1)
2019-02-20 9:17 am
As HAc and Ac⁻ have equal volume and molarity, [HAc] = [Ac⁻] in the solution.
Consider the dissociation of HAc:
HAc(aq) ⇌ H⁺(aq) + Ac⁻(aq) …… pKa
According to Henderson-Hasselbalch equation:
pH = pKa + log([Ac⁻]/[HAc])
pKa = pH - log([Ac⁻]/[HAc]) = 4.74 - log(1) = 4.74
Consider the dissociation of HAc:
HAc(aq) ⇌ H⁺(aq) + Ac⁻(aq) …… pKa
According to Henderson-Hasselbalch equation:
pH = pKa + log([Ac⁻]/[HAc])
pKa = pH - log([Ac⁻]/[HAc]) = 4.74 - log(1) = 4.74
收錄日期: 2021-05-01 22:28:03
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