Determine the Ka for the acid HA given that the equilibrium concentrations are [HA]=2.31M, [A−]=0.27M, and [H3O+]=0.27M.?
回答 (1)
2021-03-12 12:31 pm
✔ 最佳答案
Balanced equation for the dissociation of HA:HA(aq) + H₂O(ℓ) ⇌ A⁻(aq) + H₃O⁺(aq) Ka = ?
At equilibrium:
Ka = [A⁻] [H₃O⁺] / [HA] = 0.27 × 0.27 / 2.31 = 0.032
收錄日期: 2021-04-23 23:07:48
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