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I understand that in a solution, the salt created by a neutralisation reaction can react again with water molecules, which can further affect the pH. Is it correct to say that the hydrolysis of salts is a secondary reaction after the neutralisation reaction?

Or, in the case where there is a strong base and weak acid (which are stoichiometrically balanced), it's just because the strong base fully dissociates and the weak acid only partially dissociates, so in the end there's just more $\ce{H+}$ than $\ce{OH-}$?

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    $\begingroup$ Review the guides Asking and How to ask. Not following the guidance may lead to lack of satisfying answers, objections, question down-voting or even question closure. $\endgroup$
    – Poutnik
    Apr 2 at 5:32
  • $\begingroup$ I have read it before but I don't understand how else I can improve my question because I did research as much as I could before asking, and it's a general case so should I have provided an example or something? I would appreciate some specific tips to improve my questions to make it easier for people to answer in the future. $\endgroup$
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2 Answers 2

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Reactions of neutralization

\begin{align} \ce{HA(aq) + OH-(aq) &-> A-(aq) + H2O} \tag{R1}\\ \ce{A-(aq) + H+(aq) &-> HA(aq)}\tag{R2}\\ \ce{H+(aq) + OH-(aq) &-> H2O} \tag{R3} \end{align}

dissociation

\begin{align} \ce{HA(aq) &-> A-(aq) + H+(aq)} \tag{R4}\\ \ce{H2O &-> H+(aq) + OH-(aq)} \tag{R5} \end{align}

and hydrolysis

$$\ce{A-(aq) + H2O -> HA(aq) + OH-(aq)} \tag{R6}$$

always occur simultaneously.

Reactions can be rewritten as the half number of equilibrium reactions instead:

\begin{align} \ce{A-(aq) + H2O &<=> HA(aq) + OH-(aq)} &K_\mathrm{b} \tag{R7} \\ \ce{ HA(aq) &<=> A-(aq) + H+(aq)} &K_\mathrm{a} \tag{R8}\\ \ce{H2O &<=> H+(aq) + OH-(aq)} &K_\mathrm{w} \tag{R9} \end{align}

bound by the equation of the equilibrium constant dependency

$$K_\mathrm{a} \cdot K_\mathrm{b} = K_\mathrm{w}$$

Which reaction direction dominates depends on concentration of involved species and respective dissociation constants. As for all acid-base reactions, equilibrium is established very fast by approaching it from either left either right side of the reaction. At the equilibrium, the rates of respective forward and backward reactions are equal and the net reaction rates are zero.

The same occurs in respective modifications for cases of strong acids and weak bases, or weak acids and weak bases.

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The various reactions happen concurrently [not simultaneously] depending on the relative rates, concentrations, activation energies and localized temperature differences. The main reaction in water is the self protonation: 2H2O = H3O+ + OH-. An acid transfers a proton to water faster than another water molecule does. A "strong" acid transfers a proton to water faster than H3O+ does. A weak acid transfers a proton to water at an intermediate rate. A base works in reverse. A weak base reacts faster than H2O and slower than OH-. This means that the addition of a weak base to water will remove H3O+ faster than pure water raising the pH. If this is done suddenly, say by brute force addition or a severe temperature jump, the kinetics can be measured. At more reasonable rates, as in a proper titration rate, the differences seem simultaneous but are concurrent. [It is possible [really easy] to titrate too fast and miss an endpoint.]

Finally at equilibrium the various concentrations satisfy the pertinent equilibrium constants.

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  • $\begingroup$ Reviewing the difference, simultaneously seems better than concurrently,, as the latter means, among other things, merely coincidence. See grammarhow.com/concurrent-vs-simultaneous $\endgroup$
    – Poutnik
    Apr 2 at 7:01
  • $\begingroup$ The reactions have different rates. The grammatical definition ignores molecular motions, the uncertainty principle and relativity. Thinking hard about it simultaneous events cannot be related or maybe not. [positron annihilation, electronic transitions, SN1reacttions etc are difficult] I tried to give a reasonable simple explanation of how these reactions happen. Yes here are many simultaneous reactions going on. $\endgroup$
    – jimchmst
    Apr 3 at 1:09

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