Salt hydrolysis — pH of salt solutions (4 cases - SA - SB, SA - WB, WA - SB, WA - WB): Question bank

Before the traps, let's build every symbol this page leans on — so no line surprises you.

Definition What "p" means and the key constants

The little "p" in front of anything means ==take the negative base-10 logarithm==. So $p X = - lo g_{10} X$ . Applied to our constants (all values below at the reference temperature 25 °C):

$K_{w} = [H^{+}] [OH^{-}] = 1 0^{- 14}$ is water's self-ionization constant. Then $p K_{w} = - lo g (1 0^{- 14}) = 14$ .
$K_{a}$ is the acid dissociation constant of the weak parent acid — big $K_{a}$ = strong acid. $p K_{a} = - lo g K_{a}$ ; a smaller $p K_{a}$ means a stronger acid.
$K_{b}$ is the base dissociation constant of the weak parent base. $p K_{b} = - lo g K_{b}$ ; smaller $p K_{b}$ = stronger base.
$pH = - lo g [H^{+}]$ and $pOH = - lo g [OH^{-}]$ , and they always add to $p K_{w} = 14$ at 25 °C.

Why logs at all? Concentrations span from $1$ down to $1 0^{- 14}$ — the log squashes that huge range into the friendly 0–14 pH scale.

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

Formula Where the pH formulas come from — the

$h$ , $K_{h}$ , $C$ chain Start a hydrolyzing salt at concentration $C$ . Let == $h$ == = the degree of hydrolysis = the fraction of ions that have reacted ( $0 \leq h \leq 1$ ). Build the mass balance (an "ICE" bookkeeping) for the anion case: $[CH_{3} COO^{-}] = C (1 - h), [CH_{3} COOH] = [OH^{-}] = C h$ WHAT: each reacted ion produces one CH₃COOH and one OH⁻, so both products equal $C h$ . WHY $C h$ not $h$ : $h$ is a bare fraction; multiply by the starting molarity $C$ to get an actual concentration. Substitute into $K_{h}$ : $K_{h} = \frac{( C h ) ( C h )}{C ( 1 - h )} = \frac{C h ^{2}}{1 - h} \approx h ≪ 1 C h^{2} \Rightarrow h = \frac{K _{h}}{C}$ WHY drop $1 - h$ : for a weak-parent ion almost nothing reacts, so $1 - h \approx 1$ . Then $[OH^{-}] = C h = C \frac{K _{h}}{C} = K_{h} C = \frac{K _{w} C}{K _{a}} .$ Convert to $[H^{+}] = K_{w} / [OH^{-}] = K_{w} K_{a} / C$ and take $- lo g$ : $pH = \frac{1}{2} (p K_{w} + p K_{a} + lo g C) (WA/SB, basic) .$ The SA/WB case is the mirror image ( $[H^{+}] = K_{w} C / K_{b}$ ), giving $pH = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)$ . The WA/WB case runs both reactions at once; the $C$ factors cancel, leaving $[H^{+}] = K_{w} K_{a} / K_{b}$ and $pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$ .

True or false — justify

An ion from a strong acid or strong base can still hydrolyze if the solution is concentrated enough.

False. Cl⁻, Na⁺, K⁺, NO₃⁻ have no tendency to grab H⁺ or OH⁻ at any concentration — their parents are 100% dissociated, so the reverse reaction has essentially zero driving force. Concentration changes how much a hydrolyzing ion reacts, but it cannot switch on hydrolysis in a non-hydrolyzing ion.

NaCl solution has pH exactly 7 at all temperatures.

False. pH = 7 only at 25 °C, where

K_{w} = 1 0^{- 14}

. NaCl never hydrolyzes, but

K_{w}

rises with temperature, so at (say) 50 °C neutral pH is below 7. NaCl stays neutral, not necessarily pH 7.

An acidic salt is a salt that contains an acidic hydrogen atom.

False. Here "acidic salt" means the solution turns acidic (pH < 7) because the cation hydrolyzes — e.g. NH₄Cl has no acidic H in the usual sense; its NH₄⁺ donates H⁺ to water. Don't confuse it with acid salts like NaHSO₄.

For a WA/SB salt, increasing concentration

C

raises the pH.

True. From

pH = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)

, larger

C

makes

lo g C

less negative, so pH rises (more basic). More anions means more OH⁻ produced overall.

For a WA/WB salt, doubling the concentration changes the pH.

False. Its formula

pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})

has no $C$ term — the H⁺ from the cation and OH⁻ from the anion scale together, so their ratio (and hence pH) is concentration-independent.

The degree of hydrolysis

h

is a probability and must lie between 0 and 1.

True.

h

is the fraction of the ion that has reacted with water, so

0 \leq h \leq 1

. The approximation

h ≪ 1

(used to drop the

1 - h

term) fails for very dilute or strongly-hydrolyzing salts, where

h

approaches 1.

A weak acid + weak base salt is always neutral.

False. It is only neutral when

K_{a} = K_{b}

. If

K_{a} > K_{b}

the cation's acid strength wins → acidic; if

K_{a} < K_{b}

the anion's base strength wins → basic. "Both weak" does not mean "cancel exactly."

Spot the error

A student writes

K_{h} = K_{a} \times K_{b}

for the hydrolysis of NH₄⁺. Find the mistake.

The correct relation is

K_{h} = K_{w} / K_{b}

(for a cation from a weak base). The product

K_{a} K_{b} = K_{w}

links a conjugate pair, not the two separate parents of a salt. NH₄⁺'s

K_{a}

K_{w} / K_{b}

, and that equals

K_{h}

Someone computes the pH of 0.1 M NH₄Cl as

pH = \frac{1}{2} (p K_{w} - p K_{b} + lo g C)

. What's wrong?

The sign on

lo g C

is flipped. The correct acidic-salt formula is

pH = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)

. With

lo g 0.1 = - 1

, the wrong version subtracts 1 where it should add 1, giving a pH off by a full unit.

A student says: "CH₃COO⁻ hydrolyzes by donating H⁺ to water, making the solution basic." Correct them.

Acetate is a base — it accepts H⁺ from water (grabbing it back to become CH₃COOH), which leaves behind excess OH⁻. Donating H⁺ would make it acidic and would require it to have an extra proton, which it doesn't.

In deriving the SA/WB pH, a student keeps the term

C (1 - h)

C (1 - h) \approx 0

. Where did it go wrong?

The approximation is

C (1 - h) \approx C

(since

h ≪ 1

, so

1 - h \approx 1

), not

\approx 0

. Setting it to zero would say no ammonium remains — absurd, since almost none has reacted.

A student concludes Na₂SO₄ is basic "because it came from a base." Diagnose.

Na₂SO₄ comes from the strong base NaOH and the strong acid H₂SO₄. Both parents are strong → neither ion hydrolyzes → neutral. "Came from a base" is meaningless unless you check whether that base is weak.

Someone claims NH₄CN is acidic because NH₄⁺ makes H⁺. What did they forget?

They ignored CN⁻, whose parent HCN is a very weak acid (tiny

K_{a}

), so CN⁻ is a strong conjugate base that grabs H⁺ hard. Since

K_{b} (CN^{-}) > K_{a} (NH_{4}^{+})

here, the solution is actually basic. In WA/WB you must compare both.

Why questions

Why does a salt of a strong acid and strong base give a neutral solution while its ions are floating around?

Because neither ion reacts with water — both parents were fully dissociated, so the ions have no "unfinished business." Water's own autoionization

H_{2} O ⇌ H^{+} + O H^{-}

is left undisturbed, keeping

[H^{+}] = [OH^{-}]

Why is the hydrolysis constant written as

K_{h} = K_{w} / K_{b}

and not just

K_{b}

itself?

K_{b}

describes the base forming from water; hydrolysis of its conjugate acid runs the reverse chemistry. Dividing

K_{w}

(water's own constant) by

K_{b}

gives the equilibrium constant for the ion-vs-water reaction, correctly small because the parent was weak.

Why does a weaker parent acid make its salt more basic?

A weaker acid clings to its proton more eagerly, so its conjugate base grabs H⁺ from water more strongly (larger

K_{h} = K_{w} / K_{a}

). More H⁺ removed from water → more leftover OH⁻ → higher pH. Weakness of the parent = strength of the conjugate.

Why does the WA/WB pH formula lose all dependence on concentration?

In the

K_{h}

substitution both products scale with

C

while the reactant also scales with

C

, so every

C

cancels — algebraically

[H^{+}] = K_{w} K_{a} / K_{b}

has no

C

left. Only the balance of

K_{a}

against

K_{b}

(relative strengths) survives.

Why do we take

[H^{+}] = C h

rather than

[H^{+}] = h

h

is a fraction (dimensionless); the actual amount of ion reacted is (fraction) × (starting concentration) =

C h

, which carries units of molarity. Each reacted ion releases one H⁺ (or OH⁻), so the produced concentration is

C h

Why can't we use

pH = - lo g C

(the strong-acid shortcut) for NH₄Cl?

That shortcut assumes complete dissociation into H⁺, but NH₄⁺ only partially hydrolyzes (

h ≪ 1

). Most of it stays as NH₄⁺, so far less H⁺ appears than

C

— the pH is much closer to 7 than

- lo g C

would predict.

Edge cases

What is the pH of a very dilute neutral salt, say

1 0^{- 8}

M NaCl?

Still essentially 7, not 8. The salt contributes nothing, and at such low "concentration" water's own

1 0^{- 7}

M H⁺ dominates completely. (This is the same trap as a

1 0^{- 8}

M strong acid — you can never make water basic by adding a neutral salt.)

As concentration of a WA/SB salt approaches zero, what happens to its pH?

It approaches 7 from above. Fewer hydrolyzing anions means less OH⁻ produced, so the tiny basic shift shrinks toward neutral. The formula's

lo g C

term drives pH down toward 7 as

C \to 0

(though water's floor prevents it going below 7).

Does the WA/WB "concentration-independent" formula still hold at extremely low concentration?

No. The derivation assumes the salt's own hydrolysis dominates over water's autoionization. Below roughly

1 0^{- 6}

–

1 0^{- 7}

M the H⁺ and OH⁻ from water itself (

1 0^{- 7}

M each) swamp the tiny salt contribution, so the pH drifts back toward 7 and the neat

\frac{1}{2} (p K_{w} + p K_{a} - p K_{b})

result breaks down.

For a WA/WB salt with

K_{a} = K_{b}

, is the solution exactly neutral?

Effectively yes at 25 °C:

pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b}) = \frac{1}{2} p K_{w} = 7

since the

K_{a}

and

K_{b}

terms cancel. Ammonium acetate (

K_{a} = K_{b} = 1.8 \times 1 0^{- 5}

) is the textbook example.

What happens to hydrolysis and pH as temperature rises?

K_{w}

increases with temperature, so hydrolysis constants (

K_{w} / K_{a}

K_{w} / K_{b}

) grow and hydrolysis intensifies. Neutral pH itself drops below 7, and hydrolyzing salts shift more strongly — so all pH values must be quoted with a temperature.

Can a salt of a strong acid and a strong base ever be non-neutral due to its own ions?

No. By definition both ions come from fully-dissociated strong parents and don't touch water. Any deviation from neutrality would come from temperature (

K_{w}

) or an impurity, never from the ions themselves.

Is the "

h ≪ 1

" approximation valid for extremely dilute or extremely weak-parent salts?

Not always. If

C

is tiny or

K_{h}

is large,

h

can approach 1, and dropping the

1 - h

term overestimates the reactant left. Then you must keep the full

K_{h} = C h^{2} / (1 - h)

and solve the quadratic instead of using

[H^{+}] = K_{h} C

Recall Quick self-test

What does the "p" in pKₐ mean? ::: Take the negative base-10 log: $p K_{a} = - lo g_{10} K_{a}$ . Write $K_{h}$ for acetate hydrolysis in terms of concentrations. ::: $K_{h} = \frac{[ CH _{3} COOH ] [ OH ^{-} ]}{[ CH _{3} COO ^{-} ]} = \frac{K _{w}}{K _{a}}$ . Sign of $lo g C$ in the acidic-salt (SA/WB) formula? ::: Minus: $pH = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)$ . Which single quantity is missing from the WA/WB pH formula? ::: Concentration $C$ — it is concentration-independent (until water's own ions dominate at very low $C$ ). Relation linking $h$ , $K_{h}$ , $C$ (weak-parent ion)? ::: $h = K_{h} / C$ , so $[H^{+}]$ or $[OH^{-}] = C h = K_{h} C$ .