Acid-Base Chemistry

In 1923, the Danish chemist J.N. Bronsted and the English chemist T.M. Lowry independently expanded the Arrhenius theory, which was limited to water as the solvent. They pointed out that acid–base reactions are proton-transfer reactions, and acids and bases can be defined in terms of the proton (H⁺) donors and acceptors.

According to the Bronsted–Lowry concept, an acid is the species donating a proton in a proton-transfer reaction and a base is the species accepting the proton.

When HCl dissolves in water, an H⁺ ion is transferred from HCl to water, where it attaches to a lone pair on the oxygen atom and forms H₃O⁺. HCl (the acid) has donated H⁺ and H₂O (the base) has accepted it:

HCl_(g) + H₂O_(l) ⇌ H₃O⁺_(aq) + Cl⁻_(aq)

A species formed after a Bronsted base accepts a proton from the acid is called the conjugate acid. The hydronium ion (H₃O⁺) is the conjugate acid of water. A species formed when an acid donates a proton to a base is called the conjugate base. HCl–Cl⁻ and H₂O–H₃O⁺ are conjugate acid–base pairs — the two members of a pair differ by exactly one proton.

When ammonia dissolves in water, an H⁺ released from H₂O attaches to the nitrogen lone pair; NH₄⁺ and OH⁻ are formed:

NH_3(g) + H₂O_(l) ⇌ NH₄⁺_(aq) + OH⁻_(aq)

Here H₂O (the acid) has donated the proton and NH₃ (the base) has accepted it. NH₃–NH₄⁺ and H₂O–OH⁻ are conjugate acid–base pairs. So water is amphoteric: it acts as a base in one case and as an acid in the other. An amphoteric substance is a species that can act as either an acid or a base (it can lose or gain a proton), depending on the other reactant.

📌 Note (book's own point): the Bronsted–Lowry theory can be applied to acids in solvents other than water, or even in solventless reactions. Gaseous ammonia and HCl give solid NH₄Cl: HCl_(g) + NH_3(g) → NH₄Cl_(s). HCl acts as an acid because it donates a proton to NH₃, and NH₃ acts as a base even though no hydroxide ion OH⁻ is present — something the Arrhenius theory could not explain.

Table 9.1 — Conjugate acid–base pairs of common species: HNO₃ + H₂O ⇌ H₃O⁺ + NO₃⁻; H₂SO₄ + H₂O ⇌ H₃O⁺ + HSO₄⁻; H₂O + CO₃²⁻ ⇌ HCO₃⁻ + OH⁻.

Quick Check 9.1 (solved in brief): (a) Conjugate pairs — (i) HCN/CN⁻ and H₂O/H₃O⁺; (ii) CH₃COOH/CH₃COO⁻ and H₂O/H₃O⁺; (iii) H₂S/HS⁻ and H₂O/H₃O⁺; (iv) HCl/Cl⁻ and HCO₃⁻/H₂CO₃. (b) Lewis acids/bases — (i) SO₃ is the Lewis acid (accepts a pair from O²⁻, the Lewis base): SO₃ + O²⁻ → SO₄²⁻; (ii) HCl + H₂O → H₃O⁺ + Cl⁻: H₂O is the Lewis base, H⁺ (from HCl) the Lewis acid; (iii) BF₃ + NH₃ → F₃B←NH₃: BF₃ is the Lewis acid, NH₃ the Lewis base.

In 1923, G.N. Lewis proposed a generalized definition of acid–base behaviour in which acids and bases are identified by their ability to accept or donate a pair of electrons and form a coordinate covalent bond.

A Lewis acid is any species (molecule or ion) that can accept a pair of electrons, and a Lewis base is any species (molecule or ion) that can donate a pair of electrons. A Lewis acid–base reaction occurs when a base donates a pair of electrons to an acid, forming an acid–base adduct.

Example 1 — silver ion and ammonia: each of two ammonia molecules, Lewis bases, donates a pair of electrons to a positively charged silver ion, the Lewis acid. The sum of charges on the left-hand side is +1, so the acid–base adduct on the right-hand side must also carry a charge of +1:

2NH₃ + Ag⁺ → [H₃N→Ag←NH₃]⁺

Example 2 — boron trifluoride: the boron atom in BF₃ has only six electrons in its valence shell. Since the boron atom has an incomplete octet, it can behave as an electron-pair acceptor. As a result, BF₃ is a very good Lewis acid and reacts with many Lewis bases; a fluoride ion is the Lewis base here, donating one of its lone pairs:

F⁻ + BF₃ → [BF₄]⁻

The negative charge on the adduct is −1, the sum of charges on the left-hand side (−1 + 0), so charge is conserved. Notice that no proton is transferred in either example: the Lewis concept covers acid–base reactions that the Bronsted–Lowry concept (proton transfer) cannot describe — that is why BF₃ and AlCl₃ are Lewis acids but not Bronsted–Lowry acids (see MCQ VI).

Pure water is a poor conductor of electricity but its conductance is measurable. Water undergoes self-ionization reversibly:

H₂O_(l) + H₂O_(l) ⇌ H₃O⁺_(aq) + OH⁻_(aq)

Net reaction: H₂O_(l) ⇌ H⁺_(aq) + OH⁻_(aq)

The equilibrium constant for this reaction can be written as K_c = [H⁺][OH⁻]/[H₂O] = 1.8 × 10⁻¹⁶ mol dm⁻³. The concentration of H₂O in pure water is 1000 g dm⁻³ divided by 18 g mol⁻¹, giving 55.5 mol dm⁻³. Since water is present in very large excess and very few of its molecules undergo ionization, its concentration remains effectively constant. That constant concentration is taken to the left-hand side and multiplied with K_c to give another constant called K_w:

K_w = K_c[H₂O] = [H⁺][OH⁻] = 1.8 × 10⁻¹⁶ × 55.5 = 1.01 × 10⁻¹⁴ at 25 °C

K_w is called the ionic product of water or the dissociation constant of water. Its value increases almost 75 times when temperature is increased from 0 °C to 100 °C (Table 9.2: 0 °C → 0.11 × 10⁻¹⁴; 10 °C → 0.30 × 10⁻¹⁴; 25 °C → 1.0 × 10⁻¹⁴; 40 °C → 3.00 × 10⁻¹⁴; 100 °C → 7.5 × 10⁻¹⁴).

For neutral water, [H⁺] = [OH⁻], so [H⁺][H⁺] = 10⁻¹⁴ and [H⁺]² = 10⁻¹⁴ (at 25 °C), giving [H⁺] = 10⁻⁷ mol dm⁻³ and [OH⁻] = 10⁻⁷ mol dm⁻³.

When some quantity of acid or base is added to water, K_w remains the same, but [H⁺] and [OH⁻] are no longer equal: adding a small amount of acid gives [H⁺] > [OH⁻]; adding a few drops of a base gives [OH⁻] > [H⁺]. During both additions, K_w remains 10⁻¹⁴ at 25 °C.

In all the aqueous solutions, the concentrations of H⁺ and OH⁻ ions are too low to be conveniently expressed and used in calculations. In 1909, Sorensen, a Danish biochemist, introduced the term pH and pOH. So, the scales of pH and pOH were developed:

pH = −log[H⁺]    and    pOH = −log[OH⁻]

For neutral water, pH = −log[10⁻⁷] = 7 and pOH = −log[10⁻⁷] = 7. Also, pH + pOH = 14. The value of pH normally varies between 0 → 14 at 25 °C.

📌 Interesting Information (book's own note): solutions of negative pH and having values more than 14 are also known — the 0–14 range is the normal range, not an absolute limit.

Figure 9.1 — pH values of some common substances: HCl (1.0 M, pH 0.0), stomach acid (pH 1.0–3.0), lemon juice (pH 2.2–2.4), vinegar (pH 2.4–3.4), carbonated water (pH 3.9), beer (pH 4.0–4.5), milk (pH 6.4), blood (pH 7.4), seawater (pH 7.0–8.5), baking soda (0.1 M, pH 8.4), milk of magnesia (pH 10.5), household ammonia (pH 11.9), NaOH (1.0 M, pH 14.0).

Sample Problem 9.1: In a solution, the pH is 9.2. Determine the ionic product of water, K_w, at 25 °C. Solution: pOH = 14 − pH = 14 − 9.2 = 4.8. Taking antilog: [H⁺] = 10^−9.2 ≈ 6.3 × 10⁻¹⁰ M; [OH⁻] = 10^−4.8 ≈ 1.6 × 10⁻⁵ M. K_w = [H⁺][OH⁻] = (6.3 × 10⁻¹⁰)(1.6 × 10⁻⁵) = 1.01 × 10⁻¹⁴. (The book prints the antilog steps with the minus signs dropped in typesetting — "10^pH" and "10^9.2" — the values used, 6.3 × 10⁻¹⁰ and 1.6 × 10⁻⁵, show the intended 10^−pH and 10^−pOH; corrected here.)

Sample Problem 9.2: K_w = 1.0 × 10⁻¹⁴ at 25 °C; [H⁺] = 1.0 × 10⁻⁷ M. Find [OH⁻], pH and pOH. Solution: [OH⁻] = K_w/[H⁺] = 1.0 × 10⁻¹⁴/1.0 × 10⁻⁷ = 1.0 × 10⁻⁷ M; pH = −log(1.0 × 10⁻⁷) = 7; pOH = −log(1.0 × 10⁻⁷) = 7. The solution is neutral, as expected.

Quick Check 9.2 (solved in brief): (a) Mixing equal volumes of pH 4.0 and pH 10.0 solutions: the pH 4.0 solution has [H⁺] = 10⁻⁴ M and the pH 10.0 solution has [OH⁻] = 10⁻⁴ M; equal volumes neutralize each other, so the mixture is neutral and K_w = 1.0 × 10⁻¹⁴ at 25 °C (K_w does not change on mixing). (b) K_w = 1.0 × 10⁻¹⁴, [OH⁻] = 2.5 × 10⁻⁵ M: [H⁺] = 1.0 × 10⁻¹⁴/2.5 × 10⁻⁵ = 4.0 × 10⁻¹⁰ M; pH = −log(4.0 × 10⁻¹⁰) ≈ 9.4. (c) 0.30 M HNO₃ (strong acid, fully dissociated): [H⁺] = 0.30 M; pH = −log(0.30) ≈ 0.52; [OH⁻] = 1.0 × 10⁻¹⁴/0.30 ≈ 3.3 × 10⁻¹⁴ M; pOH = 14 − 0.52 = 13.48.

Acids and bases, when dissolved in water, may or may not be completely dissociated. Many acids are weak electrolytes and ionize to an extent which is much less than 100%. The value of K_c, called the dissociation constant of an acid, is the quantitative measure of the strength of the acid.

For an acid HA dissolved in water: HA + H₂O ⇌ H₃O⁺ + A⁻. The equilibrium expression, with the constant [H₂O] absorbed, gives K_a = [H₃O⁺][A⁻]/[HA]. This equation is used to calculate K_a for any acidic solution if we know the pH or [H⁺] and the initial concentration of the acid, and to calculate the equilibrium concentration of H₃O⁺ and A⁻ if we know the initial concentration and K_a.

Strength rule (book): when K_a < 10⁻³, the acid is weak; K_a = 1 to 10⁻³, moderately strong; K_a > 1, strong.

Table 9.3 — K_a values of some acids: HCl very large (10⁶, very strong); HNO₃ very large (10³, very strong); H₂SO₄ very large (10³, very strong, first step); HSO₄⁻ 1.3 × 10⁻² (strong); HF 6.7 × 10⁻⁵ (weak); CH₃COOH 1.85 × 10⁻⁵ (weak); H₂CO₃ 4.4 × 10⁻⁷ (weak); H₂S 1.0 × 10⁻⁷ (weak); NH₄⁺ 5.7 × 10⁻¹⁰ (weak); HCO₃⁻ 4.7 × 10⁻¹¹ (weak); H₂O 1.8 × 10⁻¹⁶ (very weak). Larger K_a means a stronger acid — and, as Short Q(d) asks, a weaker conjugate base.

9.5.1 Calculating H₃O⁺ ions from K_a: weak acids do not completely dissociate, so [H₃O⁺] is not equal to the initial concentration. Let the initial concentration of HA be C mol dm⁻³. At equilibrium: [H₃O⁺] = x, [A⁻] = x, [HA] = C − x. So K_a = x²/(C − x). For weak acids, x is usually very small compared to C, so C − x ≈ C, and K_a ≈ x²/C, giving x ≈ √(K_a × C) — the concentration of H₃O⁺.

Sample Problem 9.3: Calculate [H₃O⁺] in a 0.1 M solution of acetic acid, K_a = 1.8 × 10⁻⁵. Solution: [H₃O⁺] ≈ √(1.8 × 10⁻⁵ × 0.1) = √(0.18 × 10⁻⁵) ≈ 1.34 × 10⁻³ mol dm⁻³. (The book prints "1.34 ⊗ 10⁻³" with a misprinted symbol; the value is 1.34 × 10⁻³.)

Quick Check 9.3 (solved in brief): (a) 0.10 M formic acid HCOOH, pH 2.38: [H⁺] = 10^−2.38 ≈ 4.17 × 10⁻³ M; K_a = [H⁺]²/(C − [H⁺]) ≈ (4.17 × 10⁻³)²/0.0958 ≈ 1.8 × 10⁻⁴. (b) 0.1 M HNO₂, K_a = 4 × 10⁻⁴: [H₃O⁺] ≈ √(4 × 10⁻⁴ × 0.1) = √(4 × 10⁻⁵) ≈ 6.3 × 10⁻³ mol dm⁻³. (c) Vinegar: 0.837 M CH₃COOH, [H₃O⁺] = 3.86 × 10⁻³ mol dm⁻³: K_a = (3.86 × 10⁻³)²/(0.837 − 0.00386) ≈ 1.79 × 10⁻⁵ (≈ the book's 1.85 × 10⁻⁵).

The suppression of ionization of a weak electrolyte by adding a common ion to it is called the common ion effect. We are familiar with purification of sodium chloride by passing hydrogen chloride gas through brine (saturated solution of NaCl).

Sodium chloride is fully ionized in the solution: NaCl_(s) ⇌ Na⁺_(aq) + Cl⁻_(aq), with K_c = [Na⁺][Cl⁻]/[NaCl]. HCl also ionizes in solution: HCl_(g) ⇌ H⁺_(aq) + Cl⁻_(aq). On passing HCl gas, the concentration of Cl⁻ ion is increased, therefore NaCl crystallizes out of the solution to maintain the constant value of the equilibrium constant. The addition of a common ion to the solution of a less soluble electrolyte suppresses its ionization and the concentration of unionized species increases, which may come out as a precipitate: Na⁺_(aq) + Cl⁻_(aq) ⇌ NaCl_(s).

More examples of common ion effect:

(i) The solubility of a less soluble salt KClO₃ in water is suppressed by the addition of a more soluble salt KCl by common ion effect. K⁺ is a common ion. The ionization of KClO₃ is suppressed and it settles down as precipitate: KClO_3(s) ⇌ K⁺_(aq) + ClO₃⁻_(aq); KCl_(s) ⇌ K⁺_(aq) + Cl⁻_(aq).

(ii) Similarly, the dissociation of a weak acid H₂S in water can be suppressed by the addition of stronger acid HCl. H⁺ is a common ion. H₂S becomes less dissociated in acidic solution. In this way, low concentration of S²⁻ ion is produced. This low concentration of S²⁻ ions helps to do the precipitation of radicals of II group basic radicals during salt analysis: H₂S_(g) ⇌ 2H⁺_(aq) + S²⁻_(aq); HCl_(aq) ⇌ H⁺_(aq) + Cl⁻_(aq).

(iii) An addition of NH₄Cl in NH₃ solution suppresses the concentration of OH⁻ due to the presence of a large excess of NH₄⁺ from NH₄Cl. Actually, NH₄Cl is a strong electrolyte. The combination of these two substances is used as a group reagent in III group basic radicals: NH₄Cl_(aq) ⇌ NH₄⁺_(aq) + Cl⁻_(aq); NH₃ + H₂O_(aq) ⇌ NH₄⁺_(aq) + OH⁻_(aq).

Common ion effect finds many applications in the qualitative analysis and the preparation of buffers (the next section).

The solutions which resist the change in their pH when a small amount of an acid or a base is added to them, are called buffer solutions. They have a specific constant value of pH and their pH values do not change on dilution and even keeping for a long time.

Buffer solutions are mostly prepared by mixing two substances:

(i) By mixing a weak acid and its salt with a strong base. Such solutions give acidic buffers with pH less than 7. Mixture of acetic acid and sodium acetate is one of the best examples of acidic buffers.

(ii) By mixing a weak base and its salt with a strong acid. Such solutions will give basic buffers with pH more than 7. Mixture of NH₄OH and NH₄Cl is one of the best examples of basic buffers.

Take the example of an acidic buffer consisting of CH₃COOH and CH₃COONa. CH₃COOH being a weak electrolyte undergoes very little dissociation. When CH₃COONa is added to CH₃COOH solution, then the dissociation of CH₃COOH is suppressed, due to common ion effect of CH₃COO⁻: CH₃COOH_(l) + H₂O_(l) ⇌ CH₃COO⁻_(aq) + H₃O⁺_(aq); CH₃COONa_(aq) ⇌ CH₃COO⁻_(aq) + Na⁺_(aq).

If one goes on adding CH₃COONa in CH₃COOH solution, then the added concentrations of CH₃COO⁻ decrease the dissociation of CH₃COOH and the pH of solution increases. Table 9.4 — Effect of addition of acetate ions on the pH of acetic acid solution: 0.10 M CH₃COOH alone: 1.3% dissociated, pH 2.89; + 0.05 M CH₃COO⁻: 0.036%, pH 4.44; + 0.10 M: 0.018%, pH 4.74; + 0.15 M: 0.012%, pH 4.92. Greater the concentration of acetic acid as compared to CH₃COONa, lesser is the pH of solution.

Actually, a buffer mentioned above is a large reservoir of CH₃COOH and CH₃COO⁻ components. When an acid or H₃O⁺ ions are added to this buffer, they will react with CH₃COO⁻ to give back acetic acid and hence the pH of the solution will almost remain unchanged. The reason is that CH₃COOH being a weak acid will prefer to remain undissociated. Similarly, the buffer solution consisting of NH₄Cl and NH₄OH, can resist the change of pH and pOH, when acid or base is added from outside. When a base or OH⁻ ions are added in it, they will react with H₃O⁺ to give back H₂O and the pH of the solution again will remain almost unchanged.

9.7.1 Calculating the pH of a buffer: Consider a weak acid HA and its salt NaA with a strong base say NaOH. The pH of a buffer solution can be calculated by using Henderson's equation:

pH = pK_a − log([Acid]/[Salt])  =  pH = pK_a + log([Salt]/[Acid])

This relationship is called Henderson's equation. It shows that two factors evidently govern the pH of a buffer solution: first is the pK_a of the acid used and the second is the ratio of the concentrations of the salt and the acid. The best buffer is prepared by taking equal concentration of salt and acid. So, pH is controlled by pK_a of the acid. For example, for acetic acid – sodium acetate buffer, if [CH₃COOH] = [CH₃COONa], then pH = pK_a + log(1) = pK_a + 0 = pK_a = 4.74. It means that the pH of this buffer is just equal to the pK_a of the acid.

📌 Interesting Information (book's own note): the body's blood buffering system, involving bicarbonates, helps maintain a stable pH. The compositions of injections are buffered to ensure that no change in blood's pH occur.

Quick Check 9.4 (solved in brief): (a) Common ion effect on solubility: adding a common ion lowers the solubility of a slightly soluble salt (Section 9.6 and 9.8.1b). (b) Saturated AgCl + NaCl solution: (i) the concentration of Ag⁺ decreases — the common Cl⁻ ion shifts AgCl ⇌ Ag⁺ + Cl⁻ left and AgCl precipitates; (ii) yes, this is exactly the common ion effect. (c) A buffer maintains pH stability because it is a large reservoir of a weak acid and its conjugate base (or weak base and its conjugate acid): added H₃O⁺ is removed by the base component and added OH⁻ by the acid component, so pH almost does not change. (d) Buffer of 0.50 M HF and 0.45 M F⁻ salt (K_a of HF = 6.8 × 10⁻⁴): pK_a = −log(6.8 × 10⁻⁴) ≈ 3.17; initial pH = 3.17 + log(0.45/0.50) ≈ 3.12. After adding 0.40 g NaOH to 1.0 dm³ (0.40/40 = 0.010 mol): HF falls to 0.49 M and F⁻ rises to 0.46 M, so pH = 3.17 + log(0.46/0.49) ≈ 3.14 — almost unchanged, as a buffer should. (e) Buffer of 0.11 M CH₃COONa and 0.09 M CH₃COOH (K_a = 1.85 × 10⁻⁵): pK_a ≈ 4.73; pH = 4.73 + log(0.11/0.09) ≈ 4.82.

When a soluble ionic compound is dissolved in water, like NaCl, it dissociates completely into ions. But for slightly soluble salts the dissociation is not complete at equilibrium stage. For example, when PbSO₄ is shaken with water the solution contains Pb²⁺, SO₄²⁻ and undissociated PbSO₄. It means an equilibrium exists between solid solute, PbSO₄, and the dissolved ions, Pb²⁺ and SO₄²⁻.

Lead sulphate is a well-known sparingly soluble compound and it dissociates to a very small extent like PbCl₂: PbSO_4(s) ⇌ Pb²⁺_(aq) + SO₄²⁻_(aq), with K_c = [Pb²⁺][SO₄²⁻]/[PbSO₄].

Being a sparingly soluble salt, the concentration of lead sulphate (PbSO₄) almost remains constant. Bring [PbSO₄] on L.H.S. with K_c: K_c[PbSO₄] = [Pb²⁺][SO₄²⁻]; if K_c[PbSO₄] = K_sp, then K_sp = [Pb²⁺][SO₄²⁻] = 1.6 × 10⁻⁸ at 25 °C.

K_sp is called the solubility product of PbSO₄. It is the product of molar solubilities of two ions at equilibrium stage. Similarly, for another sparingly soluble salt, PbCl₂, K_sp = [Pb²⁺][Cl⁻]².

K_sp is usually a very small quantity at room temperature. The value of K_sp is temperature dependent. So, the solubility product is the product of the concentrations of ions raised to an exponent equal to the co-efficient of the balanced equation. The value of K_sp is a measure of the dissociation of sparingly soluble salt.

Table 9.5 — K_sp values of some ionic compounds: AgBr 5.0 × 10⁻¹³; AgCl 1.8 × 10⁻¹⁰; Al(OH)₃ 3 × 10⁻³⁴; BaCO₃ 1.1 × 10⁻¹⁰; CaCO₃ 3.3 × 10⁻⁹; CaF₂ 3.2 × 10⁻¹¹; CuS 8 × 10⁻³⁴; Fe₂S₃ 1.4 × 10⁻⁸⁵; MgCO₃ 3.5 × 10⁻⁸; MnS 3 × 10⁻¹¹; PbCrO₄ 2.3 × 10⁻¹³; PbSO₄ 1.6 × 10⁻⁸ (each as the product of its ions raised to their coefficients, e.g. [Al³⁺][OH⁻]³ for Al(OH)₃).

(a) Determination of solubility from K_sp: for this purpose, we need the formula of the compound and K_sp value. Then the unknown molar solubility S is calculated and the concentration of the ions is determined. Table 9.6 — Relation between solubility and solubility products of some salts: PbSO₄ (2 ions): K_sp 1.69 × 10⁻⁴ (as printed; note this table value differs from Table 9.5's 1.6 × 10⁻⁸ — the table's own solubility column, 1.3 × 10⁻⁴ g dm⁻³, is in grams, not moles, so the two tables are not directly comparable; see caveat in the exercise section), solubility 1.3 × 10⁻⁴ g dm⁻³; Ca(OH)₂ (3 ions): K_sp 6.5 × 10⁻⁶, solubility 1.175 × 10⁻² g dm⁻³; CaF₂ (3 ions): K_sp 3.2 × 10⁻¹¹, solubility 2.0 × 10⁻⁴ g dm⁻³; Ag₂CrO₄ (3 ions): K_sp 2.6 × 10⁻¹², solubility 8.7 × 10⁻⁵ g dm⁻³.

Sample Problem 9.4: Ca(OH)₂ is a sparingly soluble compound. Its solubility product is 6.5 × 10⁻⁶. Calculate the solubility of Ca(OH)₂. Solution: let the solubility be S mol dm⁻³. Ca(OH)_2(aq) ⇌ Ca²⁺_(aq) + 2OH⁻_(aq). The concentration of OH⁻ is double that of Ca²⁺, so K_sp = [Ca²⁺][OH⁻]² = S × (2S)² = 4S³. Hence 4S³ = 6.5 × 10⁻⁶; S³ = 6.5 × 10⁻⁶/4 = 1.625 × 10⁻⁶; S = (1.625)^1/3 × 10⁻² = 1.18 × 10⁻² mol dm⁻³. Hence, at equilibrium, 1.18 × 10⁻² mol dm⁻³ of Ca²⁺ and 2 × 1.18 × 10⁻² = 2.36 × 10⁻² mol dm⁻³ of OH⁻ are present. (The book's ICE table is misprinted — it shows "S + S" for the equilibrium row — but the working and answer use the correct S and 2S; corrected here.)

(b) Common Ion Effect: the presence of a common ion decreases the solubility of a slightly soluble ionic compound. To explain it, consider a saturated solution of PbCrO₄, a sparingly soluble ionic salt: PbCrO_4(aq) ⇌ Pb²⁺_(aq) + CrO₄²⁻_(aq). Now add Na₂CrO₄, a soluble salt. CrO₄²⁻ is the common ion. It combines with Pb²⁺ to form more insoluble PbCrO₄. So equilibrium is shifted to the left to keep K_sp constant, and solubility falls.

(c) Predicting Precipitation: the solubility product can also help in predicting whether the precipitation of a salt will occur or not. For example, the solubility product of CaSO₄ is 2 × 10⁻⁵. If we add 10⁻² mol dm⁻³ solution of Ca²⁺ to 10⁻² mol dm⁻³ solution of SO₄²⁻ ions at 25 °C (equal volumes, so each concentration halves on mixing): [Ca²⁺] = [SO₄²⁻] = 10⁻²/2 = 5.0 × 10⁻³ mol dm⁻³. Ionic product = (5.0 × 10⁻³)(5.0 × 10⁻³) = 2.5 × 10⁻⁵ mol² dm⁻⁶ > K_sp of CaSO₄. As the ionic product of concentrations is greater than K_sp, therefore CaSO₄ will precipitate out.

Rule: ionic product > K_sp → supersaturated → precipitation YES; ionic product = K_sp → saturated → NO; ionic product < K_sp → unsaturated → NO.

Quick Check 9.5 (solved in brief): (a) AgCl, K_sp = 1.77 × 10⁻¹⁰ at 25 °C, in 0.10 M NaCl: the common Cl⁻ ion gives [Ag⁺] = K_sp/[Cl⁻] = 1.77 × 10⁻¹⁰/0.10 = 1.77 × 10⁻⁹ mol dm⁻³ — far below its pure-water solubility √(1.77 × 10⁻¹⁰) ≈ 1.33 × 10⁻⁵ mol dm⁻³. (Note: the question's wording "a solution already contains M of 0.10 M AgCl" is garbled in the book; it is read here as 0.10 M NaCl added, matching the Quick Check 9.4 pattern and the chapter's common-ion treatment.) (b) CaSO₄ with 10⁻³ mol dm⁻³ of each of Ca²⁺ and SO₄²⁻ mixed: if these are the concentrations after mixing, ionic product = 10⁻⁶ < K_sp (2 × 10⁻⁵), so NO precipitation; if they are the pre-mixing concentrations of equal volumes, each halves to 5 × 10⁻⁴ and the ionic product 2.5 × 10⁻⁷ is still below K_sp — either way, no precipitation. (c) Potassium carbonate solution is basic: (i) CO₃²⁻ is the conjugate base of the weak acid HCO₃⁻, so it hydrolyses (see Section 9.9.3) and produces OH⁻, making the solution alkaline; (ii) CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻.

When a salt dissolves in water, it dissociates into its constituent ions. These ions can interact with water, affecting the solution's pH depending on the nature of the acid and base from which the salt is derived. This reaction of an ion with water is called hydrolysis (the book uses the term in the SLO and in 9.9; the ion "reacts with water" to give H₃O⁺ or OH⁻).

9.9.1 Salts of strong acids and strong bases: for salts derived from strong acids and strong bases (e.g. sodium chloride, NaCl), the conjugate base of a strong acid (e.g. Cl⁻ from HCl) is very weak and does not significantly react with water. The conjugate acid of a strong base (e.g. Na⁺ from NaOH) is also very weak and does not significantly react with water: NaCl_(s) ⇌ Na⁺_(aq) + Cl⁻_(aq). Na⁺ is the conjugate acid of NaOH (a strong base) and does not affect the pH. Cl⁻ is the conjugate base of HCl (a strong acid) and does not affect the pH. The solution remains neutral.

9.9.2 Salts of strong acids and weak bases: for salts derived from strong acids and weak bases (e.g. ammonium chloride, NH₄Cl), the conjugate base of the strong acid (e.g. Cl⁻) does not react with water. The conjugate acid of the weak base (e.g. NH₄⁺ from NH₃) reacts with water to produce H₃O⁺ ions, making the solution acidic: NH₄Cl_(s) ⇌ NH₄⁺_(aq) + Cl⁻_(aq); NH₄⁺_(aq) + H₂O_(l) ⇌ NH_3(g) + H₃O⁺_(aq). The solution is acidic due to the NH₄⁺ ion.

9.9.3 Salts of weak acids and strong bases: the example is sodium acetate. CH₃COONa_(s) + H₂O_(l) ⇌ CH₃COO⁻_(aq) + Na⁺_(aq). CH₃COO⁻ is the conjugate base of CH₃COOH (a weak acid) and reacts with water: CH₃COO⁻_(aq) + H₂O_(l) ⇌ CH₃COOH_(aq) + OH⁻_(aq). Na⁺ is the conjugate acid of NaOH (a strong base) and does not affect the pH. The solution is basic due to the CH₃COO⁻ ion. (This is also why potassium carbonate solution is alkaline — Quick Check 9.5c: CO₃²⁻ hydrolyses the same way.)

9.9.4 Salts of weak acids and weak bases: for salts derived from weak acids and weak bases (e.g. ammonium acetate, CH₃COONH₄), both the conjugate acid (NH₄⁺) and the conjugate base (CH₃COO⁻) affect the pH: NH₄CH₃COO_(aq) ⇌ NH₄⁺_(aq) + CH₃COO⁻_(aq); NH₄⁺ hydrolyses to produce H₃O⁺: NH₄⁺_(aq) + H₂O_(l) ⇌ NH_3(aq) + H₃O⁺_(aq); CH₃COO⁻ hydrolyses to produce OH⁻: CH₃COO⁻_(aq) + H₂O_(aq) ⇌ CH₃COOH_(aq) + OH⁻_(aq). The resultant pH of the solution depends on the relative strengths of the conjugate acid and conjugate base. The solution may be acidic, basic, or nearly neutral, depending on which reaction is more dominant.

📌 Interesting Information (book's own note, p. 200): the shell of the nautilus is composed mainly of calcium carbonate. The nautilus adjusts conditions so shell material is formed when the concentration of calcium ions and carbonate ions in seawater are high enough to precipitate calcium carbonate — a real-life Ksp/precipitation story.

An indicator is a substance that changes colour to mark a titration's endpoint. Acid–base indicators exhibit one colour in acid and another in base. Most indicators used in acid–base titration are weak organic acids or bases. In solution, a weak-acid indicator (HIn) can be represented by the following equilibrium: HIn ⇌ H⁺ + In⁻. In⁻ is the symbol of the anion part of the indicator. Because the reaction is reversible, both HIn and In⁻ are present. The colours displayed result from the fact that HIn and In⁻ have different colours.

In acidic solutions, any In⁻ ions that are present act as Bronsted bases and accept protons from the acid. The indicator is then present in largely unionized form, HIn. In basic solutions, the OH⁻ ions from the base combine with the H⁺ ions produced by the indicator. The indicator molecules further ionize to equalize the loss of H⁺ ions. The indicator is thus present largely in the form of its anion, In⁻. The solution now shows the base-indicating colour, which for litmus is blue. Figure 9.2 shows unionized and ionized acid–base indicators and their ionized forms existing in an equilibrium.

Different indicators change colour at different pH values. The colour depends on the relative amounts of HIn and In⁻ at a given pH. For example, methyl red changes from red to yellow between pH 4.4 and 6.2. At pH 4.4, the indicator exists mostly as HIn molecules, which appear red in the solution. Figure 9.3 — Range and colour changes of some common acid–base indicators: methyl orange (red → yellow) and phenolphthalein (colourless → pink/red) across the pH scale.

9.10.1 Selecting a suitable indicator: the two general criteria for an indicator to be used in a titration are: (i) the pH at the end of the titration should be close to the indicator's neutral point; (ii) the indicator should indicate a sharp colour change near the equivalence point of the titration. Each pH indicator changes colour over a defined range of pH, known as its range. An indicator changes colour over a range of about 2 pH units.

9.10.2 Titration curve and equivalence point: a pH curve is a graph of the pH of the solution versus the volume of titrant added. The equivalence point is the point at which the amount of titrant added is just enough to neutralize the analyte solution completely. Titration curves show how the pH of an acidic or basic solution changes as a basic or acidic solution is added to it. We can use the titration curve to choose an indicator that will show when the titration is complete and we reach the equivalence point. The end point of the titration occurs when the indicator changes colour. We choose an indicator with an end point close to the equivalence point.

1. Strong acid–strong base titration curve (Figure 9.4, HCl with NaOH): initially, the solution contains only the strong acid. Since the acid is strong, it completely dissociates, leading to a high concentration of H⁺ ions and the pH is very low. As NaOH is added, OH⁻ ions from NaOH begin to neutralize the H⁺ ions from HCl and pH of the solution rises. The equivalence point is reached when the amount of OH⁻ added is stoichiometrically equal to the amount of H⁺ originally present in the acid. The pH at the equivalence point in a strong acid–strong base titration is 7.0. The titration curve at this stage shows a steep rise in pH, changing quickly from acidic to neutral. The pH value at the endpoint changes about from 4.0 to 10.0. These titrations have a pH of 7.0 at equivalence, so, such phenolphthalein as indicator can be used as they show different colours in this range.

2. Strong acid–weak base titration curve (Figure 9.5, HCl with aqueous NH₃): initially, the solution contains only the weak base and it only partially dissociates, with a lower concentration of OH⁻ ions compared to a strong base. The initial pH will be greater than 7 but lower than the pH of a strong base. As HCl is added, the concentration of OH⁻ decreases, causing a further decrease in pH. Before reaching the equivalence point, the solution is in a buffer-like region where the pH changes more gradually. The presence of the weak base and its conjugate acid (from the salt formed) creates a buffering effect, which helps to moderate the pH change as acid is added. The pH at the equivalence point will be less than 7 (pH = 5.27) because the conjugate acid slightly dissociates, releasing H⁺ ions making the solution acidic (NH₄⁺ → NH₃ + H⁺). After the equivalence point, the pH of the solution decreases rapidly. This is because the strong acid dissociates completely in water, providing a high concentration of hydrogen ions (H⁺), which significantly lowers the pH. Methyl orange has its colour change in this range, therefore, it can be used as an indicator for strong acid–weak base titrations.

Table 9.7 — pH ranges of common indicators: methyl orange: acid colour orange, base colour yellow, pH range 3.2–4.5, type of titration strong acid–strong base / strong acid–weak base; phenolphthalein: acid colour colourless, base colour red, pH range 8.2–10.0, type of titration strong acid–strong base.

Quick Check 9.6 (solved in brief): (a) End point vs equivalence point: the equivalence point is the stoichiometric point — titrant just enough to neutralize the analyte completely (Section 9.10.2); the end point is the experimental point where the indicator changes colour. A good titration chooses an indicator whose end point is close to the equivalence point, but the two are not the same thing. (b) An indicator changes colour because its two forms, HIn and In⁻, have different colours, and the acid/base shifts the HIn ⇌ H⁺ + In⁻ equilibrium: in acid it is mostly HIn (acid colour), in base mostly In⁻ (base colour). (c) Weak acid–strong base titration: the equivalence point is basic (pH > 7, by the mirror image of the strong acid–weak base case — the conjugate base of the weak acid makes the solution basic), so an indicator changing in the basic range, phenolphthalein (8.2–10.0), is suitable. ⚠️ The book gives no dedicated strong-acid/weak-base-style worked curve for weak acid–strong base titrations; this answer is derived from the chapter's own hydrolysis rule (Section 9.9.3: the conjugate base of a weak acid gives OH⁻) and the indicator-range rule, and is flagged as derived. (d) Weak acid–weak base titration: ⚠️ the book gives no dedicated treatment of this case — there is no sharp pH jump at equivalence and no indicator range in Table 9.7 covers it, so no indicator from the chapter's table can be recommended with confidence; this is stated honestly rather than inventing one. (In general practice such titrations have no sharp endpoint, which is why the book's table omits them.)