Chemistry 2e · Acid-Base Equilibria

Hydrolysis of Salts

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

A is an ionic compound formed when an acid and a base neutralize each other. Dissolve table salt (NaCl) in water and you get a neutral solution — but many salts do not. Sodium acetate gives a basic solution; ammonium chloride gives an acidic one. The reason is : one or both of the salt's ions react with water, transferring protons and disturbing the [H⁺]/[OH^-] balance. Which ions hydrolyze — and how strongly — depends on whether they came from strong or weak acids and bases.

Why this matters

Salt hydrolysis explains everyday chemistry: why soapy water feels slippery (basic), why ammonium nitrate cold packs work, and why gardeners use lime to sweeten acidic soil. In biology, the pH of body fluids depends on dissolved salts. On exams you will predict whether a salt solution is acidic, basic, or neutral and calculate its pH — a skill that also sets up equivalence-point pH in Acid-Base Titrations.

The college version

Core Concepts

The four source combinations

The acidity of a salt solution is set by its ions' parentage:

Ion comes from	Example ions	Behavior in water
Strong acid (conjugate base of strong acid)	Cl^-, NO₃^-, Br^-, I^-, SO₄^2-	Spectator — no reaction with water
Weak acid (conjugate base of weak acid)	CH₃COO^-, CN^-, F^-, CO₃^2-	Acts as a weak base; makes solution basic
Strong base (conjugate acid of strong base)	Na⁺, K⁺, Ca²⁺, Ba²⁺	Spectator — no reaction with water
Weak base (conjugate acid of weak base)	NH₄⁺, CH₃NH₃⁺	Acts as a weak acid; makes solution acidic

Combining these: salt of strong acid + strong base → neutral (NaCl, KNO₃); weak acid + strong base → basic (NaCH₃COO, NaF); strong acid + weak base → acidic (NH₄Cl, NH₄NO₃); weak acid + weak base → depends on which constant is larger (compare K_a of the cation with K_b of the anion).

Anion hydrolysis: why acetate makes water basic

Acetate is the conjugate base of weak acetic acid, so it is a real base in water:

$CH 3 COO - (aq) + H 2 O (l) ⇌ CH 3 COOH (aq) + OH - (aq)$

The equilibrium constant for this reaction is the base ionization constant of the anion, obtained from the conjugate relationship:

$K b (CH 3 COO -) = K w K a (CH 3 COOH) = 1.0 \times 10 -14 1.8 \times 10 -5 = 5.6 \times 10 -10$

Producing OH^- raises the pH above 7. The weaker the parent acid, the stronger its anion acts as a base: fluoride (K_a(HF) = 6.3 × 10^-4, so K_b(F^-) = 1.6 × 10^-11) makes a less basic solution than acetate.

Cation hydrolysis: why ammonium salts are acidic

Ammonium is the conjugate acid of weak base ammonia, so it donates a proton to water:

$NH 4 + (aq) + H 2 O (l) ⇌ NH 3 (aq) + H 3 O + (aq)$

with

$K a (NH 4 +) = K w K b (NH 3) = 1.0 \times 10 -14 1.8 \times 10 -5 = 5.6 \times 10 -10$

Producing H₃O⁺ lowers the pH below 7. Note that this is the same K_a value you computed for ammonium in the previous topic — hydrolysis is just the conjugate-strength idea applied to a salt.

Salts of weak acid + weak base, and amphiprotic ions

When both ions hydrolyze, compare the numbers. For ammonium fluoride: K_a(NH₄⁺) = 5.6 × 10^-10 versus K_b(F^-) = K_w/K_a(HF) = 1.6 × 10^-11. The cation's acidity wins, so NH₄F is slightly acidic. Some ions can do either job: HCO₃^- and H₂PO₄^- are amphiprotic, able to accept or donate a proton. Their net effect is decided by comparing K_a and K_b; for bicarbonate, basicity wins (K_b = K_w/K_a1(H₂CO₃) = 2.3 × 10^-8 beats K_a = 5.6 × 10^-11).

How It Works / Step-by-Step Process

Split the salt into its cation and anion.
Identify the parent acid of the anion and the parent base of the cation.
Classify each parent as strong or weak; label each ion spectator, acidic, or basic.
Predict the solution: spectator + spectator = neutral; basic anion present = basic; acidic cation present = acidic; both active = compare constants.
To find pH, write the hydrolysis reaction, compute the ion's K_a or K_b from the conjugate relationship, set up an ICE table, and solve (checking the 5% rule).

Common Confusions

Do Not Confuse	With	Difference
"Salt of a strong acid"	"Acidic salt solution"	The phrase describes the anion's parent. NaCl is a salt of strong acid HCl yet is neutral.
Na⁺ hydrolyzing	K⁺/Na⁺ being active ions	Group 1/2 cations are spectators because their parent bases are strong; they do not react with water.
Salt dissolving	Salt hydrolyzing	Dissolving is just dispersal into ions; hydrolysis is a subsequent proton-transfer reaction with water.
Neutral solution pH = 7	"Neutral" meaning no acid/base character	For salts, neutral means spectator ions only; pH stays 7 only at 25 °C (pure water pH shifts with temperature).
K_a of HCO₃^-	K_a1 of H₂CO₃	Bicarbonate's acidity uses K_a2 (5.6×10⁻¹¹); its basicity uses K_w/K_a1.

Eli explains

The same idea, in plain words

Explain it like I’m 10

A salt is like a team of two players: a "cation" and an "anion." Some players just stand around watching (the ones from strong acids and strong bases), while others quietly react with water — one kind makes the water more sour, the other more soapy. Check where each player came from: if both parents are strong, the water stays neutral; otherwise the weak parent's child changes the pH.

Worked example

Example 1: pH of sodium acetate

Calculate the pH of a 0.10 M sodium acetate solution.

Acetate hydrolyzes as a base; sodium is a spectator. First compute K_b:

$K b (CH 3 COO -) = K w K a (CH 3 COOH) = 1.0 \times 10 -14 1.8 \times 10 -5 = 5.6 \times 10 -10$

Set up the ICE table for CH₃COO^- + H₂O ⇌ CH₃COOH + OH^-. Let x = [OH^-]:

$K b = [CH 3 COOH][OH -] [CH 3 COO -] = x 2 0.10 - x \approx x 2 0.10$

$x = \sqrt K b \times 0.10 = \sqrt (5.6 \times 10 -10)(0.10) = \sqrt 5.6 \times 10 -11 = 7.5 \times 10 -6 M$

$pOH = - log (7.5 \times 10 -6) = 5.13, pH = 14.00 - 5.13 = 8.87$

The solution is basic, as predicted. Check: x/C = 7.5 × 10^-5, far below 5%, so the approximation is valid.

Example 2: pH of ammonium chloride

Calculate the pH of a 0.050 M NH₄Cl solution.

Ammonium hydrolyzes as an acid; chloride is a spectator. First compute K_a:

$K a (NH 4 +) = K w K b (NH 3) = 1.0 \times 10 -14 1.8 \times 10 -5 = 5.6 \times 10 -10$

For NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, let x = [H₃O⁺]:

$K a = x 2 0.050 - x \approx x 2 0.050$

$x = \sqrt (5.6 \times 10 -10)(0.050) = \sqrt 2.8 \times 10 -11 = 5.3 \times 10 -6 M$

$pH = - log (5.3 \times 10 -6) = 5.28$

The solution is acidic, as predicted for a salt of a strong acid and weak base.

Example 3: predicting without calculating

Classify these 0.10 M solutions as acidic, basic, or neutral: KNO₃, NaF, NH₄NO₃, NaHCO₃.

KNO₃: K⁺ (spectator) + NO₃⁻ (spectator) → neutral.
NaF: Na⁺ spectator + F⁻ (conjugate base of weak HF, K_b = 1.6 × 10^-11) → basic.
NH₄NO₃: NH₄⁺ (K_a = 5.6 × 10^-10) + NO₃⁻ spectator → acidic.
NaHCO₃: Na⁺ spectator; bicarbonate is amphiprotic. Compare K_b = K_w/K_a1 = 2.3 × 10^-8 with K_a = K_a2 = 5.6 × 10^-11. Since K_b > K_a, the solution is basic (why baking soda solutions are slightly alkaline).

Key takeaways

Neutral salts come from strong acid + strong base: NaCl, KBr, KNO₃, Na₂SO₄.
Basic salts come from weak acid + strong base: the anion is a real base (acetate, fluoride, carbonate, cyanide).
Acidic salts come from strong acid + weak base: the cation is a real acid (ammonium, alkylammonium).
Weak acid + weak base salts: compare K_a of cation vs K_b of anion — the bigger constant wins.
Hydrolysis constants come from conjugate pairs: K_b(A^-) = K_w/K_a(HA), K_a(BH⁺) = K_w/K_b(B).
Group 1/2 metal cations are spectators; OH^- and O^2- themselves make solutions strongly basic.
A "salt of a strong acid" is not automatically acidic — the phrase means its anion is neutral.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

Predict whether 0.10 M KCN is acidic, basic, or neutral. (For HCN, K_a = 4.9 × 10^-10.)
Show answer
Basic. K⁺ is a spectator; CN⁻ is the conjugate base of weak HCN with K_b = K_w/4.9 × 10^-10 = 2.0 × 10^-5, so it hydrolyzes to produce OH^-.
Why is Na₂SO₄ neutral even though sulfate carries a 2− charge?
Show answer
Both ions are conjugates of strong parents: Na⁺ of NaOH, SO₄²⁻ of HSO₄⁻/H₂SO₄ (a strong acid). Neither reacts with water, so pH stays ~7.
A salt's cation is the conjugate acid of a weak base and its anion is the conjugate base of a weak acid. What determines whether the solution is acidic or basic?
Show answer
Compare K_a of the cation with K_b of the anion; the larger constant controls the pH. If they are equal, the solution is close to neutral.
Which solution has the higher pH: 0.10 M NaF or 0.10 M NaCH₃COO? (For HF, K_a = 6.3 × 10^-4; for acetic acid, K_a = 1.8 × 10^-5.)
Show answer
Sodium acetate, because acetate has the larger K_b (5.6 × 10^-10 vs. 1.6 × 10^-11 for fluoride) — the weaker the parent acid, the stronger the conjugate-base anion.
Write the hydrolysis reaction for CO₃^2- in water and identify the conjugate acid formed.
Show answer
CO₃^2- + H₂O ⇌ HCO₃^- + OH^-; the conjugate acid is bicarbonate.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

salt: Ionic compound from an acid–base neutralization.
hydrolysis: Reaction of an ion with water that produces H₃O⁺ or OH^-.
spectator ion: Ion that does not react with water (conjugate of a strong acid or base).
amphiprotic ion: Ion that can act as either an acid or a base (HCO₃^-, H₂PO₄^-).
salt hydrolysis constant: K_b of a conjugate-base anion or K_a of a conjugate-acid cation.

Sources & references

openstax.org — Chemistry 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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In 30 seconds

Why this matters

The college version

Core Concepts

The four source combinations

Anion hydrolysis: why acetate makes water basic

Cation hydrolysis: why ammonium salts are acidic

Salts of weak acid + weak base, and amphiprotic ions

How It Works / Step-by-Step Process

Common Confusions

The same idea, in plain words

Worked example

Example 1: pH of sodium acetate

Example 2: pH of ammonium chloride

Example 3: predicting without calculating

Key takeaways

Check yourself

Keep learning

Key vocabulary

Related lessons

Sources & references