Ionization of acids and bases: pH, Ka and Kb

How far do acids and bases ionise?

Unless stated otherwise, calculations below use dilute ideal aqueous solutions at 298 K, with activities approximated by concentrations relative to 1 mol L⁻¹. Concentration expressions are approximations; neutral pH is pK_w/2 at the temperature in question.

In plain words: this is the arithmetic of acidity. Strong acids such as HCl give up their proton completely, so [H⁺] equals their concentration (unless the acid is extremely dilute). Weak acids give up only a fraction, set by K_a, so you need a small equilibrium calculation. Water is always there in the background, keeping [H⁺][OH⁻] fixed, and the pH scale turns these tiny concentrations into everyday numbers.

Strong and weak acids and bases

Perchloric (HClO₄), hydrochloric, hydrobromic, hydriodic, nitric and sulphuric acid’s first ionisation are strong: almost complete in water. The second ionisation of sulphuric acid is weaker and need not be complete. LiOH, NaOH, KOH, CsOH and Ba(OH)₂ are strong bases (NCERT pages 192–193). In Brönsted–Lowry terms, a strong acid is a good proton donor, and in HA + H₂O ⇌ H₃O⁺ + A⁻ the equilibrium moves towards the weaker acid and weaker base.

So strong acids have very weak conjugate bases: ClO₄⁻, Cl⁻, Br⁻, I⁻, NO₃⁻ and HSO₄⁻ are much weaker bases than water. Weak acids such as nitrous acid, hydrofluoric acid and acetic acid have stronger conjugate bases, and NH₂⁻, O²⁻ and H⁻, the conjugate bases of extremely weak acids, are very good proton acceptors (page 193).

Indicators. Some water-soluble organic compounds, such as phenolphthalein and bromothymol blue, are weak acids whose acid form, HIn, and conjugate base, In⁻, have different colours: HIn(aq) + H₂O(l) ⇌ H₃O⁺(aq) + In⁻(aq). Shifting this equilibrium with acid or base changes the colour, which is why they are used in acid–base titrations and to estimate [H⁺] (page 193).

The ionic product of water

Water acts as an acid and a base at once (NCERT Section 6.11.1, page 193): H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq). Since [H₂O] is constant, it is taken into a new constant, the ionic product of water:

[H⁺] in pure water at 298 K has been measured as 1.0 × 10⁻⁷ M, and [OH⁻] is equal to it. K_w depends on temperature, like any equilibrium constant. Pure water is 1000/18.0 = 55.56 M (NCERT writes 55.55), so there are about 2 hydronium–hydroxide ion pairs per 10⁹ water molecules. Each pair contains two water-derived ions, so this is not a count of only two participating water molecules. A solution is acidic if [H₃O⁺] > [OH⁻], neutral if they are equal and basic if [H₃O⁺] < [OH⁻].

The pH scale

NCERT defines the pH of a solution as the negative logarithm to base 10 of the activity of the hydrogen ion (Section 6.11.2, pages 193–194). In dilute solutions (below about 0.01 M) the activity is approximately the molarity divided by 1 mol L⁻¹ (the accuracy depends on ionic strength), so pH = −log [H⁺] (page 194). For example, 10⁻² M HCl has pH 2, and NaOH with [OH⁻] = 10⁻⁴ M has [H₃O⁺] = 10⁻¹⁰ M and pH 10. Taking −log of K_w:

At 298 K a neutral solution has pH 7, an acidic one pH < 7 and a basic one pH > 7. One pH unit means a factor of 10 in [H⁺]; two units, a factor of 100. pH paper with four strips gives the pH to about 0.5; some pH meters display increments of 0.001 (page 194), but display resolution is not guaranteed accuracy; calibration, temperature and electrode behaviour matter.

NCERT slip (page 195): NCERT’s Table 6.5 of common pH values lists lime water as 10.5. Beyond the NCERT text: lime water is a saturated solution of calcium hydroxide, of which about 1.73 g L⁻¹ dissolves at 20 °C (Wikipedia, “Calcium hydroxide”), and calcium hydroxide saturated at 25 °C has pH 12.4 (Hazardous Substances Data Bank, via PubChem), so 10.5 is far too low. From the solubility, [OH⁻] is about 0.047 M and the ideal pH about 12.7; the measured 12.4 is lower because at this strength activity is below concentration (reasoned). In an exam that asks for NCERT’s table value, give 10.5. If an exercise supplies a concentration or solubility product, calculate from those data with its stated approximation; a measured pH also depends on temperature and on CO₂ absorbed from the air (reasoned).

Ionisation constants of weak acids

For a monoprotic weak acid HX with no added common ion and negligible water ionisation, of initial concentration c with degree of ionisation α (NCERT Section 6.11.3, pages 195–196): HX(aq) + H₂O(l) ⇌ H₃O⁺(aq) + X⁻(aq), with c(1 − α), cα and cα at equilibrium.

The larger K_a, the stronger the acid; pK_a = −log K_a. If α is small, K_a ≈ cα², so α ≈ √(K_a/c): a weak acid ionises more when it is diluted. Dilution raises α but still lowers [H⁺] = cα: 0.1 M acetic acid has 1.3 × 10⁻³ M H⁺, 0.001 M only 1.2 × 10⁻⁴ M.

K_a values at 298 K (NCERT Table 6.6, page 195): HF 3.5 × 10⁻⁴, HNO₂ 4.5 × 10⁻⁴, HCOOH 1.8 × 10⁻⁴, benzoic acid 6.5 × 10⁻⁵, acetic acid 1.74 × 10⁻⁵, niacin 1.5 × 10⁻⁵, HClO 3.0 × 10⁻⁸, HCN 4.9 × 10⁻¹⁰, phenol 1.3 × 10⁻¹⁰.

NCERT’s two books differ on phenol. Table 6.6 gives its K_a as 1.3 × 10⁻¹⁰ (pK_a 9.89), while NCERT Class 12 Chemistry, Unit 7, Table 7.3 (page 207) gives pK_a 10.0, that is K_a = 1.0 × 10⁻¹⁰, the value the Class 11 book’s own Exercise 6.44 (page 213) uses. In a problem, use the value the question gives.

NCERT’s method for the pH of a weak electrolyte (page 196): identify the Brönsted acids and bases present; write all the proton-transfer reactions; take the one with the larger K as the primary reaction; tabulate initial, change (in terms of α) and equilibrium concentrations; solve for α; find the concentrations; and finally pH = −log [H₃O⁺]. The percent dissociation is ([HA] dissociated/[HA] initial) × 100 (NCERT Eq. 6.32).

Ionisation of weak bases

For a weak base MOH ⇌ M⁺ + OH⁻ (NCERT Section 6.11.4, pages 197–198), K_b = [M⁺][OH⁻]/[MOH] = cα²/(1 − α), and pK_b = −log K_b. Amines, derivatives of ammonia, are weak bases; ammonia gives OH⁻ by NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq).

K_b values at 298 K (NCERT Table 6.7, page 197): dimethylamine 5.4 × 10⁻⁴, triethylamine 6.45 × 10⁻⁵, ammonia 1.77 × 10⁻⁵, quinine 1.10 × 10⁻⁶, pyridine 1.77 × 10⁻⁹, aniline 4.27 × 10⁻¹⁰, urea 1.3 × 10⁻¹⁴.

NCERT’s two books differ on triethylamine. Table 6.7 gives its K_b as 6.45 × 10⁻⁵ (pK_b 4.19), while NCERT Class 12 Chemistry, Unit 9, Table 9.3 (page 267) gives pK_b 3.25, that is K_b = 5.6 × 10⁻⁴. The other amines in Table 6.7 agree with Table 9.3 (dimethylamine pK_b 3.27, ammonia 4.75), so only triethylamine differs. In a problem, use the value the question gives.

Relation between K_a and K_b

For NH₄⁺ (acid, K_a = 5.6 × 10⁻¹⁰) and NH₃ (base, K_b = 1.8 × 10⁻⁵), adding the two ionisation equations gives 2H₂O ⇌ H₃O⁺ + OH⁻, and the constants multiply (NCERT Section 6.11.5, pages 198–199):

This is a case of a general rule: the constant for a net reaction obtained by adding reactions is the product of their constants, K_net = K₁ × K₂ × … (NCERT Eq. 6.35). Knowing one of K_a and K_b gives the other, and a strong acid has a weak conjugate base.

Polybasic acids and polyacidic bases

Oxalic, sulphuric and phosphoric acids have more than one ionisable proton (NCERT Section 6.11.6, pages 199–200). A dibasic acid H₂X ionises in two steps, H₂X ⇌ H⁺ + HX⁻ (K_a1) and HX⁻ ⇌ H⁺ + X²⁻ (K_a2).

(NCERT Table 6.8, page 200, 298 K, with the ascorbic first constant corrected.) The printed ascorbic value 7.4 × 10⁻⁴ is inconsistent with pK_a1 ≈ 4.17, reported in a primary structural study by Xu and colleagues (Journal of Biological Inorganic Chemistry, 2020, 25:875–885), which gives K_a1 ≈ 6.8 × 10⁻⁵. The carbonic first constant is the conventional apparent constant counting dissolved CO₂ with H₂CO₃, not the intrinsic constant of molecular H₂CO₃ alone. For the listed acids, successive constants decrease, because a positive proton is harder to pull away from a negative ion: from HCO₃⁻ than from H₂CO₃, from HPO₄²⁻ than from H₂PO₄⁻. When successive constants are well separated and the solution is not extremely dilute, almost all the H₃O⁺ comes from the first step. (NCERT gives the reason as “H₂A being a strong acid”; H₂A need not be strong, only much stronger than HA⁻, K_a1 ≫ K_a2, as weak carbonic acid shows.)

Bases can likewise accept protons in steps: CO₃²⁻ takes one to give HCO₃⁻ (K_b1 = K_w/K_a2 = 1.8 × 10⁻⁴), then a second to give H₂CO₃ (K_b2 = K_w/K_a1 = 2.3 × 10⁻⁸).

Factors affecting acid strength

The extent of dissociation of HA depends on the strength and the polarity of the H–A bond (NCERT Section 6.11.7, page 200). For related hydrides, a weaker H–A bond (lower bond enthalpy) and a more polar one both make HA a stronger acid.

• Down a group of related hydrides, weakening bonds are a major factor: as A grows, the H–A bond weakens and acid strength increases, HF << HCl << HBr << HI; H₂S is a stronger acid than H₂O.
• Across a period of related hydrides, bond polarity and conjugate-base stability are major factors: as the electronegativity of A increases, so does acid strength, CH₄ < NH₃ < H₂O < HF.

Solvation and conjugate-base stability also contribute to acid strength. HCl, HBr and HI are all essentially fully ionised in water, so their intrinsic strength order is not distinguished by simply comparing ionisation percentages in water.

Common ion effect

Adding acetate ions to acetic acid, CH₃COOH ⇌ H⁺ + CH₃COO⁻, lowers [H⁺]; adding H⁺ pushes the equilibrium back towards un-ionised acid (NCERT Section 6.11.8, pages 200–201).

It can be defined as a shift in equilibrium on adding a substance that provides more of an ionic species already present in the dissociation equilibrium.

NCERT Chemistry Part I (Class 11), Section 6.11.8, page 201

It is Le Chatelier’s principle at work. With 0.05 M acetate in 0.05 M acetic acid (K_a = 1.8 × 10⁻⁵), [H⁺] = K_a × 0.05/0.05 = 1.8 × 10⁻⁵ M and pH = 4.74, far less acidic than 0.05 M acetic acid alone (work that out in Q4).

Hydrolysis of salts

Salt ions either stay simply hydrated or react with water to re-form the weak acid or base, which is hydrolysis (NCERT Section 6.11.9, pages 201–202). Ions of strong acids and bases (Na⁺, K⁺, Ca²⁺, Ba²⁺, Cl⁻, Br⁻, NO₃⁻, ClO₄⁻) hydrolyse negligibly in this introductory model, so their salts are approximately neutral at 298 K. The following rules cover simple salts with the stated conjugate partners; amphiprotic ions need their competing acid/base equilibria:

• weak acid, strong base (CH₃COONa): CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, so pH > 7;
• strong acid, weak base (NH₄Cl): NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, so pH < 7;
• weak acid, weak base (CH₃COONH₄): both ions hydrolyse; neglecting water ionisation and activity effects, the usual hydrolysis approximation is independent of concentration and

Solved examples

Common mistakes to avoid

Quick revision summary

• K_w = [H⁺][OH⁻] = 10⁻¹⁴ at 298 K; pH + pOH = 14.
• Weak acid: K_a = cα²/(1 − α) ≈ cα²; weak base: K_b likewise; pK = −log K.
• Conjugate pair: K_a × K_b = K_w, pK_a + pK_b = 14.
• For the listed polybasic acids the constants decrease; first-step dominance requires well-separated constants and suitable concentration.
• Acid strength: down a group, bond strength decides (HF < HI); across a period, electronegativity decides (CH₄ < HF).
• Common ion holds back ionisation; salts of weak acids are basic, of weak bases acidic; weak–weak salts: pH = 7 + ½(pK_a − pK_b).

Questions students ask

Does the ionic product of water change with temperature?

Yes, the ionic product of water, K_w = [H⁺][OH⁻], increases with temperature, because the ionisation of water, 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq), absorbs heat: it is the reverse of neutralisation. K_w is about 1.1 × 10^⁻¹⁵ at 0 °C, 1.0 × 10^⁻¹⁴ at 25 °C and 5 × 10^⁻¹³ at 100 °C, so the pH of pure, neutral water falls from about 7.5 to 7.0 to 6.1. Written with concentrations, K_w has the unit mol^² L^⁻², though it is usually quoted as a pure number.

What is the difference between the ionisation constant and the ionic product of water?

In NCERT’s concentration bookkeeping, writing H₂O ⇌ H⁺ + OH⁻ gives K_c = [H⁺][OH⁻]/[H₂O] ≈ 10^⁻¹⁴/55.5 = 1.8 × 10^⁻¹⁶ mol L^⁻¹ at 298 K. Absorbing the nearly constant water concentration gives the concentration product K_w ≈ 10^⁻¹⁴ mol^² L^⁻². This is a concentration-standard convention; thermodynamic water autoionisation uses activities with solvent-water activity approximately 1, and its dimensionless constant is not obtained by treating 55.5 M as a pure-water activity.

What are the pH formulas for the hydrolysis of salts?

For a simple salt sufficiently concentrated that water autoionisation is negligible, with small hydrolysis and ideal behaviour, the usual approximation is as follows. For a weak-acid/strong-base salt such as sodium acetate, the anion hydrolyses with hydrolysis constant K_h = K_w/K_a and degree of hydrolysis h ≈ √(K_h/c), which gives pH = 7 + ½pK_a + ½log c at 298 K, where c is the salt concentration in mol L^⁻¹. For 0.010 M sodium benzoate (K_a = 6.5 × 10^⁻⁵, pK_a = 4.19), pH = 7 + 2.09 − 1.00 ≈ 8.09. For a salt of a strong acid and a weak base, such as NH₄Cl, K_h = K_w/K_b and pH = 7 − ½pK_b − ½log c.

Can pH be negative or more than 14?

Yes, pH can be below 0 or above 14, because the scale has no fixed ends: pH = −log [H⁺] is negative whenever [H⁺] is more than 1 mol L^⁻¹. On this simple definition 2.0 M HCl has pH = −log 2.0 = −0.30, and at 298 K 2.0 M NaOH has pOH = −0.30 and pH = 14.30. In such concentrated solutions the activity of H⁺ differs a lot from its molarity, so these values are only approximate; the familiar 0 to 14 range simply covers solutions up to 1 M.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find the pH of 0.010 M NaOH at 298 K.

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   NaOH is a strong base, so [OH⁻] = 0.010 M, pOH = 2 and pH = 14 − 2 = 12.

2. Q2. Find [H⁺] and the pH of 0.10 M HCN (K_a = 4.9 × 10⁻¹⁰, NCERT Table 6.6).

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   [H⁺] ≈ √(4.9 × 10⁻¹⁰ × 0.10) = √(4.9 × 10⁻¹¹) = 7.0 × 10⁻⁶ M, so pH ≈ 5.15. α = 7.0 × 10⁻⁵, tiny, so the shortcut is safe.

3. Q3. K_b of pyridine is 1.77 × 10⁻⁹. Find K_a and pK_a of the pyridinium ion, its conjugate acid.

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   K_a = K_w/K_b = 10⁻¹⁴/1.77 × 10⁻⁹ = 5.65 × 10⁻⁶; pK_a = 5.25 (and pK_a + pK_b = 5.25 + 8.75 = 14).