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Chemistry · Ch 8 — Ionic Equilibrium

Hydrolysis of Salt of Strong Acid and Weak Base (Cationic Hydrolysis)

8.8.3

Hydrolysis of Salt of Strong Acid and Weak Base (Cationic Hydrolysis)

Consider HCl(aq)+NH4OH(aq)→NH4Cl(aq)+H2O(l)HCl(aq)+NH_4OH(aq) \rightarrow NH_4Cl(aq)+H_2O(l), which dissociates completely, NH4Cl(aq)→NH4+(aq)+Cl−(aq)NH_4Cl(aq) \rightarrow NH_4^+(aq)+Cl^-(aq). NH4+NH_4^+ is the conjugate acid of the weak base NH4OHNH_4OH, so it has a real tendency to donate a proton to water, NH4+(aq)+H2O(l)⇌NH4OH(aq)+H+(aq)NH_4^+(aq)+H_2O(l) \rightleftharpoons NH_4OH(aq)+H^+(aq), while Cl−Cl^- shows no such tendency toward H+H^+; the result is [H+]>[OH−][H^+]>[OH^-], so the solution is acidic (pH < 7) -- cationic hydrolysis. …

Misc 8.8.3-kh-derivationRelating Kh to Kb, and the pH expression

Worked out. By the same logic as the anionic case, Kh⋅Kb=KwK_h\cdot K_b=K_w, so Kh=KwKbK_h=\dfrac{K_w}{K_b}, and again Kh=h2CK_h=h^2C so [H+]=KhC=KwKbC[H^+]=\sqrt{K_h C}=\sqrt{\dfrac{K_w}{K_b}C}. Taking −log⁡10-\log_{10}: pH=−log⁡10[H+]=−12log⁡10Kw−12log⁡10C+12log⁡10Kb=7−12pKb−12log⁡10CpH=-\log_{10}[H^+]=-\tfrac12\log_{10}K_w-\tfrac12\log_{10}C+\tfrac12\log_{10}K_b=7-\tfrac12pK_b-\tfrac12\log_{10}C -- a weaker base (larger pKbpK_b) or a higher salt concentration both push the pH further be …

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