Salt hydrolysis

Dissociation is not hydrolysis. Any soluble ionic salt breaks apart completely into its cation and anion the instant it dissolves — that step is fast, total, and the same for every soluble salt. Hydrolysis is a separate, slower step that happens only to certain ions: it is the reaction of one of those ions with a water molecule, and whether it happens at all depends on the identity of the ion.

Pure water is already a delicate balance. Water undergoes its own tiny self-ionisation equilibrium, and it is this balance that a hydrolysing ion disturbs:

where: $K_w$ is the ionic product of water, and $[H^+]$, $[OH^-]$ are the equilibrium concentrations of hydrogen and hydroxide ions in mol/L. In pure water these are equal, giving pH 7. Any process that consumes or adds extra $H^+$ or $OH^-$ — such as an ion hydrolysing — shifts this balance and moves the pH away from 7.

The deciding question for any ion is one of parentage. Ask: is this ion the conjugate acid of a weak base, or the conjugate base of a weak acid? If yes, it hydrolyses, because it "wants" to pull back toward its weak, largely-unionised parent, and it does that by grabbing $H^+$ from water (releasing $OH^-$, basifying the solution) or donating $H^+$ to water (releasing $H^+$, acidifying it). If the ion instead descends from a strong acid or strong base, it has no such pull and simply sits in solution as a spectator, matching the four-case classification (strong–strong neutral, strong acid–weak base acidic, weak acid–strong base basic, weak–weak decided by relative $K_a$/$K_b$) already established for this topic.

One salt can carry two independent verdicts. A cation and an anion are judged separately — each is checked against its own parent. It is only in a weak-acid–weak-base salt that both verdicts are live at once, which is why that case alone needs the extra comparison of $K_a$ against $K_b$ to see which pull wins.

Salt hydrolysis is a competition between conjugate pairs to see which one is willing to give up its ion the least, and the outcome follows directly from a single idea: a weak acid or weak base only barely ionises in the first place, so its conjugate ion feels a strong pull to reverse that ionisation the moment it finds itself surrounded by water molecules, and it is exactly that reverse pull, applied to the water equilibrium itself, that shows up as a measurable shift in pH.

Step 1. Start from what "weak" means at the level of equilibrium. A weak acid like $\text{CH}_3\text{COOH}$ has a small $K_a$, meaning at equilibrium most of it sits as un-ionised molecules and only a small fraction has released $H^+$. Its conjugate base, $\text{CH}_3\text{COO}^-$, is therefore an ion that is thermodynamically "unhappy" being separated from its $H^+$ — it strongly favours recombining. Dropped into water, it grabs an $H^+$ from a water molecule to reform $\text{CH}_3\text{COOH}$, leaving behind an $OH^-$. That single act of grabbing is the hydrolysis reaction, and every $OH^-$ it produces is one that raises the solution's pH above 7.

Step 2. Run the identical logic on a weak base. $\text{NH}_4\text{OH}$ barely ionises into $\text{NH}_4^+$ and $OH^-$, so its conjugate acid $\text{NH}_4^+$ is equally "unhappy" and pulls a proton back off a nearby water molecule to reform $\text{NH}_4\text{OH}$, releasing $H^+$ into solution and lowering the pH below 7. Note the symmetry: a weak acid's conjugate base hydrolyses to make things basic, and a weak base's conjugate acid hydrolyses to make things acidic — the hydrolysing ion always pulls the pH in the opposite sense to its own weak parent's usual behaviour.

Step 3. Now explain why the conjugate of a strong acid or base never does this. $Cl^-$, the conjugate base of the strong acid $\text{HCl}$, has essentially zero pull to reclaim a proton, because $\text{HCl}$ itself ionises completely and never looks back; there is no equilibrium tension left in $Cl^-$ for water to disturb. The same reasoning makes $Na^+$ and $K^+$ inert toward hydrolysis, since $\text{NaOH}$ and $\text{KOH}$ are essentially fully ionised strong bases. This is why the strong-acid–strong-base case, and only that case, is reliably neutral.

Step 4. Finally, when both ions of a salt are "unhappy" conjugates — the weak–weak case — both pulls act simultaneously, one adding $H^+$ and the other adding $OH^-$, and the net pH is decided by whichever pull is stronger. The strength of each pull tracks the smallness of the parent's ionisation constant: a smaller $K_a$ means a stronger basifying pull from the conjugate base, and a smaller $K_b$ means a stronger acidifying pull from the conjugate acid. Comparing $K_a$ (of the parent acid) against $K_b$ (of the parent base) therefore tells you directly which pull dominates, without needing to calculate an actual pH.

Classify ammonium acetate and ammonium formate, and predict which is closer to neutral. $\text{CH}_3\text{COONH}_4$ is formed from $\text{CH}_3\text{COOH}$ ($K_a = 1.8\times10^{-5}$) and $\text{NH}_4\text{OH}$ ($K_b = 1.8\times10^{-5}$), as already established. $\text{HCOONH}_4$ (ammonium formate) is formed from formic acid, $\text{HCOOH}$ ($K_a = 1.8\times10^{-4}$), and the same weak base $\text{NH}_4\text{OH}$ ($K_b = 1.8\times10^{-5}$).

Step 1. Both salts dissociate completely: $\text{CH}_3\text{COONH}_4 \rightarrow \text{CH}_3\text{COO}^- + \text{NH}_4^+$ and $\text{HCOONH}_4 \rightarrow \text{HCOO}^- + \text{NH}_4^+$. In both, the cation is $\text{NH}_4^+$, the conjugate acid of a weak base, so it hydrolyses identically in each case, releasing $H^+$.

Step 2. In both salts the anion is also the conjugate base of a weak acid, so it too hydrolyses, releasing $OH^-$: $\text{CH}_3\text{COO}^- + H_2O \rightleftharpoons \text{CH}_3\text{COOH} + OH^-$ and $\text{HCOO}^- + H_2O \rightleftharpoons \text{HCOOH} + OH^-$. Both salts are therefore weak-acid–weak-base salts, case 4, and both need a $K_a$ versus $K_b$ comparison.

Step 3. For ammonium acetate, $K_a = K_b = 1.8\times10^{-5}$: the acidifying pull from $\text{NH}_4^+$ and the basifying pull from $\text{CH}_3\text{COO}^-$ are equal in strength, so they cancel and the solution is essentially neutral.

Step 4. For ammonium formate, $K_a = 1.8\times10^{-4}$ is ten times larger than $K_b = 1.8\times10^{-5}$. A larger $K_a$ means formic acid is less weak than acetic acid, so its conjugate base $\text{HCOO}^-$ has a comparatively weaker pull, while $\text{NH}_4^+$'s acidifying pull is unchanged. With $K_a > K_b$, the acidifying effect wins.

The verdict: ammonium acetate is closer to neutral pH, while ammonium formate is slightly acidic, even though both are weak-acid–weak-base salts with the same weak base — the difference comes entirely from how much weaker acetic acid is compared to formic acid.

Competitive stretch

A question may instead give you $K_a$ and $K_b$ as bare numbers without naming the acids, or ask you to rank three or four weak–weak salts from most acidic to most basic purely by comparing their $K_a/K_b$ ratios — the same Step 3/Step 4 comparison, just repeated pairwise, with no arithmetic beyond noting which constant is larger.

• Assuming "salt dissolved in water = neutral" for every salt — only the strong-acid–strong-base case is reliably neutral; any salt with a weak parent acid or base shifts the pH.
• Checking only one ion of a salt and stopping — a weak-acid–weak-base salt needs both ions checked, and the verdict needs the $K_a$-versus-$K_b$ comparison, not just "both hydrolyse."
• Hydrolysing the conjugate of a strong parent — $Cl^-$ from $\text{HCl}$, $Na^+$ from $\text{NaOH}$, and similar ions have essentially no pull to react with water and must be treated as spectators.
• Mixing up dissociation with hydrolysis — dissociation is the instantaneous, complete splitting of the salt into ions; hydrolysis is the separate, partial reaction of a susceptible ion with water afterward.

Ask the tutor: Want to try classifying a mixed set of five or six salts on the spot, including ones with Ka and Kb values given as raw numbers?