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While solving for weak acid and weak base salt hydrolysis, why do we take the degree of hydrolysis for both the anion and cation to be the same?

If we assume they are the same, when solving for $K_\mathrm{a}$ and $K_\mathrm{b}$ for the cation and anion using simultaneous equilibrium, I always get them to be the same, resulting in the solution always being neutral. I feel that both should have different degrees of hydrolysis.

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closed as unclear what you're asking by orthocresol Jan 7 '18 at 20:13

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  • $\begingroup$ I'm struggling to understand what you mean. Can you include a simple example calculation that gives an erroneously neutral solution? $\endgroup$ – Raeven0 Nov 12 '17 at 21:02
  • $\begingroup$ I currently can only find a definition of "degree of hydrolysis" $h$ in terms of protein hydrolysis, where it refers to the number of cleaved bonds. In a more general setting, for a generic reaction $$\ce{AB <=> A+ + B-},$$ one could define this as $$h = \frac{n(\ce{A+})}{n(\ce{AB})} = \frac{n(\ce{B-})}{n(\ce{AB})},$$ because it is obvious that the solution has to be neutral. I am really unsure what you're getting at, please add more context. $\endgroup$ – Martin - マーチン Nov 13 '17 at 10:14

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