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Exercise · Q29

Q.Starting from the equilibrium-constant expression for a weak acid HA\text{HA}, derive the Henderson-Hasselbalch equation, pH=pKa+log⁡[A−][HA]\text{pH} = \text{p}K_a + \log \dfrac{[\text{A}^-]}{[\text{HA}]}, and state the assumption it makes about the equilibrium concentrations of the acid and its conjugate base.

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Starting from the weak acid's own ionization equilibrium, HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-, the equilibrium constant is Ka=[H+][A−]/[HA]K_a = [\text{H}^+][\text{A}^-]/[\text{HA}]. Rearranging for [H+][\text{H}^+]: [H+]=Ka×[HA]/[A−][\text{H}^+] = K_a \times [\text{HA}]/[\text{A}^-]. Taking −log⁡10-\log_{10} of both sides: −log⁡[H+]=−log⁡Ka−log⁡([HA]/[A−])-\log[\text{H}^+] = -\log K_a - \log([\text{HA}]/[\text{A}^-]), i.e. pH=pKa+log⁡([A−]/[HA])\text{pH} = \text{p}K_a + \log([\text{A}^-]/[\text{HA}]) (flipping the sign of the log inverts the fraction). This derivation assumes that the equilibrium concentrations [HA][\text{HA}] and [A−][\text{A}^-] are well approximated by the concentrations of acid and salt as originally mixed — valid because the deliberately-added salt supplies a large concentration of the common ion A−\text{A}^-, which suppresses the weak acid's own (otherwise small) ionization almost to not …

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