Visual walkthrough — Salt hydrolysis — pH of salt solutions (4 cases - SA - SB, SA - WB, WA - SB, WA - WB)

We will build: $\text{pH}=\frac{1}{2}(\text{p}{K}_w-\text{p}{K}_b-\log C)$ but we will not use a single one of those symbols until we have drawn it.

Step 1 — What is even in the beaker?

WHAT. We drop a salt like NH₄Cl (ammonium chloride) into pure water. Salt means it splits completely into two loose ions the instant it dissolves: a cation NH₄⁺ (positive) and an anion Cl⁻ (negative).

WHY. Before any chemistry, we must know the actors. A cation is just an atom-group missing electrons so it carries + charge; an anion carries − charge. Nothing else can happen until we know which pieces are floating around.

PICTURE. Two ion clouds in water. The Cl⁻ (mint) is calm — it came from a strong parent acid (HCl) and has no wish to react. The NH₄⁺ (coral) is restless — it came from a weak parent base (NH₃) and wants its parent back.

Step 2 — The one reaction that matters

WHAT. Only the restless ion reacts. NH₄⁺ hands a proton (H⁺) to a water molecule: $\underset{\text{restless cation}}{\underbrace{{\text{NH}}_4^{+}}}+\underset{\text{plenty of it}}{\underbrace{{\text{H}}_2\text{O}}}\rightleftharpoons \underset{\text{parent rebuilt}}{\underbrace{{\text{NH}}_3}}+\underset{\text{extra acid made}}{\underbrace{{\text{H}}_3{\text{O}}^{+}}}$

WHY. Each symbol is doing a job: NH₄⁺ is the proton donor, H₂O is the proton receiver, NH₃ is the weak parent that got rebuilt (proof the ion "remembered" it), and H₃O⁺ is the extra hydronium — this is the whole point, because extra H₃O⁺ means more acidic. The double arrow $\rightleftharpoons$ means it does not go all the way; it settles at a balance.

PICTURE. A proton (small yellow dot) jumping from NH₄⁺ onto H₂O, leaving behind NH₃ and creating H₃O⁺.

Step 3 — Measuring "how much it reacts": the equilibrium constant

WHAT. At balance, we write a ratio called the hydrolysis constant $K_h$: $K_h=\frac{[{\text{NH}}_3][{\text{H}}^{+}]}{[{\text{NH}}_4^{+}]}$

WHY this tool. Why a ratio of products over reactants? Because at equilibrium the forward and backward jumps happen at equal rate, and it turns out that particular fraction always settles to the same number for a given ion — no matter how much salt you started with. That fixed number is our anchor. The square brackets $[]$ mean "concentration of," measured in moles per litre (how crowded that species is). Water is left out because it is the ocean everything swims in — practically constant.

PICTURE. A balance scale: products (top pans NH₃ and H⁺) versus reactant (bottom pan NH₄⁺). $K_h$ is the reading on the scale.

Step 4 — We never measured $K_b$'s cousin, so trade it in

WHAT. We don't usually know $K_h$ directly, but we know the parent base's strength $K_b$. They are linked: $K_h=\frac{K_w}{K_b}$

WHY. Every conjugate acid–base pair obeys $K_a\cdot K_b=K_w$, where:

• $K_w=[{\text{H}}^{+}][{\text{OH}}^{-}]=10^{-14}$ is water's own tiny split-constant (a fixed fact at 25 °C),
• $K_b$ measures how strongly the base parent NH₃ grabs protons,
• and $K_h$ here is exactly the acid constant $K_a$ of NH₄⁺.

So $K_h=K_a=K_w/K_b$. We swap an unknown ($K_h$) for two known numbers ($K_w$, $K_b$). This is the single most useful trade in the whole derivation.

PICTURE. A see-saw: push $K_b$ up (stronger base parent) and $K_h$ drops down (less hydrolysis), because their product is pinned to the constant $K_w$.

Step 5 — Book-keeping with the ICE picture

WHAT. Start with concentration $C$ of NH₄⁺. Let a fraction $h$ of it react ($h$ = degree of hydrolysis, a number between 0 and 1). Then:

WHY. For every NH₄⁺ that reacts, we lose one NH₄⁺ and gain one NH₃ and one H⁺ — one-in, one-out, that's why all three "change" rows share the same $Ch$. Since NH₄⁺ is a weak reactor, $h\ll 1$, so $C(1-h)\approx C$. Each symbol: $C$ = crowding we poured in, $h$ = fraction that reacted, $Ch$ = actual amount of H⁺ produced.

PICTURE. A bar of height $C$ with a thin sliver $Ch$ shaved off the top and moved into the product columns.

Step 6 — Plug the boxes into the ratio

WHAT. Substitute the "end" row into $K_h$: $\frac{K_w}{K_b}=\frac{(Ch)(Ch)}{C}=C{h}^2$ Solve for $h$, then for $[{\text{H}}^{+}]=Ch$: $[{\text{H}}^{+}]=Ch=\sqrt{\frac{K_{w}C}{K_b}}$

WHY term-by-term. Numerator $[{\text{NH}}_3][{\text{H}}^{+}]=(Ch)(Ch)=C^2{h}^2$; denominator $[{\text{NH}}_4^{+}]\approx C$; the fraction cancels one $C$ to leave $C{h}^2$. Setting that equal to $K_w/K_b$ and rearranging isolates the thing we actually want — $[{\text{H}}^{+}]$. Notice more salt $C$ pushes $[{\text{H}}^{+}]$ up (inside the root), and a stronger base parent $K_b$ pushes it down (in the denominator). Both make physical sense.

PICTURE. The algebra as a flow: two $Ch$ boxes multiply, one $C$ divides out, square-root undoes the square.

Step 7 — Take the logarithm to reach pH

WHAT. pH is defined as $\text{pH}=-\log [{\text{H}}^{+}]$. Apply $-\log$ to both sides: $\text{pH}=-\log \sqrt{\frac{K_{w}C}{K_b}}=-\frac{1}{2}\log \left(\frac{K_{w}C}{K_b}\right)$ $\boxed{\text{pH}=\frac{1}{2}(\text{p}{K}_w-\text{p}{K}_b-\log C)}$

WHY the log tool. Concentrations like $10^{-6}$ are painful to compare, so pH squashes them onto a friendly scale where each step of 1 means a ten-fold change. Writing $\text{p}X=-\log X$: $-\log K_w=\text{p}{K}_w$, $+\log K_b=-\text{p}{K}_b$ (sign flips because $K_b$ is on the bottom), and $-\log C$ stays. The $\frac{1}{2}$ came straight from the square root in Step 6.

PICTURE. A number line: raw $[{\text{H}}^{+}]$ spanning many powers of ten collapses onto the neat 0–14 pH ruler; our answer lands below 7 (acidic), as promised.

Step 8 — Edge and mirror cases (never leave a gap)

WHAT. The exact same seven steps produce the other three boxes — only the actors change.

• Strong acid + strong base (NaCl): both ions inert, no Step 2 reaction at all. $K_h=0$, so $[{\text{H}}^{+}]=\sqrt{K_w}=10^{-7}$ → pH = 7.
• Weak acid + strong base (CH₃COONa): now the anion reacts, producing OH⁻ instead of H⁺. Mirror the derivation → $\text{pH}=\frac{1}{2}(\text{p}{K}_w+\text{p}{K}_a+\log C)$, above 7.
• Weak acid + weak base (CH₃COONH₄): both ions react and fight. The $C$ cancels out entirely → $\text{pH}=\frac{1}{2}(\text{p}{K}_w+\text{p}{K}_a-\text{p}{K}_b)$, direction set by which of $K_a$, $K_b$ wins.

WHY show them. So no reader meets a beaker we didn't draw. Degenerate case $h\to 0$ recovers neutral water; the two "weak on one side" cases are mirror images; the "weak on both" case is the tug-of-war.

PICTURE. Four beakers on one pH ruler, each marked with which ion is the restless one.

The one-picture summary

One frame carries the whole chain: restless ion → proton jump → $K_h$ ratio → trade for $K_w/K_b$ → ICE boxes → $\sqrt{K_{w}C/K_b}$ → take $-\log$ → pH below 7.

Reveal-check: