The four classes of inorganic compounds
Oxides are compounds of an element with oxygen, like CaO or CO₂. Metal oxides are usually basic; non-metal oxides are usually acidic; some, like Al₂O₃ and ZnO, are amphoteric (they react with both acids and bases).
Acids give H⁺ ions in water (HCl, H₂SO₄, HNO₃). Bases give OH⁻ ions or accept H⁺ (NaOH, Ca(OH)₂, NH₃). A base that dissolves in water is an alkali. Salts are made of a metal (or NH₄⁺) ion and an acid's leftover ion (NaCl, CuSO₄, NH₄NO₃).
Acid–base ideas you should know
- Arrhenius: acid gives H⁺ in water; base gives OH⁻.
- Brønsted–Lowry: acid donates a proton; base accepts a proton. NH₃ is a base because it takes H⁺.
- Lewis: acid accepts an electron pair; base donates one.
Naming acids, bases and salts
- Binary acids: hydro- + root + -ic: HCl hydrochloric acid, HBr hydrobromic acid.
- Oxoacids: more oxygen = -ic, less oxygen = -ous: H₂SO₄ sulfuric, H₂SO₃ sulfurous; HNO₃ nitric, HNO₂ nitrous.
- Bases: metal + hydroxide: NaOH sodium hydroxide, Fe(OH)₃ iron(III) hydroxide (Roman numeral for the charge).
- Salts: metal first, then the acid ending changes: -ic → -ate, -ous → -ite, hydro-…-ic → -ide. So H₂SO₄ gives sulfates, H₂SO₃ sulfites, HCl chlorides.
- Acid salts still have H: NaHCO₃ sodium hydrogencarbonate. Basic salts still have OH: Mg(OH)Cl.
The genetic link between the classes
One element can be followed through the classes, like a family line:
Metal: Ca → CaO → Ca(OH)₂ → CaCO₃ (2Ca + O₂ → 2CaO; CaO + H₂O → Ca(OH)₂; Ca(OH)₂ + CO₂ → CaCO₃ + H₂O)
Non-metal: S → SO₂ → H₂SO₃ → Na₂SO₃
The two lines meet in a salt. Salts form in many ways: acid + base, acid + metal, acid + basic oxide, base + acidic oxide, and acid + carbonate.
Introduction to titration
A titration finds the unknown concentration of a solution by reacting it with a solution of known concentration (the standard solution or titrant).
- Pipette an exact volume (say 25.0 mL) of the unknown acid into a conical flask. Add 2–3 drops of indicator.
- Fill the burette with the standard base. Read the start volume at the bottom of the meniscus, at eye level.
- Add base while swirling. Near the end add it drop by drop.
- Stop at the first permanent colour change: the end point. Record the final reading. Repeat until two results agree within 0.10 mL (concordant).
Equivalence point = moles of acid and base match exactly by the equation. End point = when the indicator changes. A good indicator makes them almost the same.
Titration curves and choosing an indicator
Plot pH against volume of base added. You get an S-shaped titration curve.
- Strong acid + strong base (HCl + NaOH): pH starts near 1, jumps steeply from about 3 to 11, equivalence at pH 7. Phenolphthalein or methyl orange both work.
- Weak acid + strong base (CH₃COOH + NaOH): starts near pH 3, has a flat buffer region, equivalence above 7 (about 8.7). Use phenolphthalein (changes 8.2–10). At half-equivalence, pH = pKa.
- Strong acid + weak base (HCl + NH₃): equivalence below 7 (about 5). Use methyl orange (changes 3.1–4.4).
- Weak + weak: no sharp jump, so no indicator works well; use a pH meter.
Rule: choose an indicator whose colour-change range sits inside the steep part of the curve.
Titration calculations
Moles = concentration (mol/L) × volume (L). Use the balanced equation's ratio.
For HCl + NaOH (1 : 1): CaVa = CbVb.
For H₂SO₄ + 2NaOH: moles NaOH = 2 × moles H₂SO₄, so 2CaVa = CbVb.
Before equivalence (strong acid): [H⁺] = (leftover moles of acid) ÷ (total volume). After: [OH⁻] = (extra moles of base) ÷ (total volume), then pH = 14 − pOH (at 25 °C).
Key formulas and definitions
- n = C × V (V in litres)
- 1 : 1 reaction: C₁V₁ = C₂V₂
- a·CₐVₐ = b·C_bV_b (a, b = number of H⁺ / OH⁻ each gives)
- pH = −log[H⁺]; pH + pOH = 14 (25 °C)
- Weak acid at half-equivalence: pH = pKa
Worked examples
1. Name: (a) HNO₂ (b) K₂SO₄ (c) Fe(OH)₂ (d) NaHCO₃.
(a) nitrous acid (less O → -ous). (b) potassium sulfate (sulfuric → sulfate). (c) iron(II) hydroxide. (d) sodium hydrogencarbonate, an acid salt.
2. Write the genetic chain for magnesium ending in magnesium chloride.
2Mg + O₂ → 2MgO; MgO + H₂O → Mg(OH)₂; Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O.
3. 25.0 mL of HCl needs 20.0 mL of 0.100 M NaOH. Find [HCl].
n(NaOH) = 0.100 × 0.0200 = 0.00200 mol = n(HCl). C = 0.00200 ÷ 0.0250 = 0.0800 M.
4. 20.0 mL of H₂SO₄ is neutralised by 32.0 mL of 0.250 M NaOH. Find [H₂SO₄].
n(NaOH) = 0.250 × 0.0320 = 0.00800 mol. H₂SO₄ : NaOH = 1 : 2, so n(H₂SO₄) = 0.00400 mol. C = 0.00400 ÷ 0.0200 = 0.200 M.
5. 25.0 mL of 0.100 M HCl. Find the pH after adding 20.0 mL of 0.100 M NaOH.
Acid 0.00250 mol, base 0.00200 mol. Leftover H⁺ = 0.000500 mol in 45.0 mL. [H⁺] = 0.000500 ÷ 0.0450 = 0.0111 M. pH = 1.95.
6. Same flask, after 30.0 mL of NaOH. Find the pH.
Base 0.00300 mol, extra OH⁻ = 0.000500 mol in 55.0 mL. [OH⁻] = 0.00909 M, pOH = 2.04, pH = 11.96. The pH jumped from about 2 to 12 over just 10 mL – the steep part of the curve.
Common mistakes
- Forgetting to convert mL to L before working out moles.
- Using C₁V₁ = C₂V₂ for H₂SO₄ + NaOH; the 1 : 2 ratio must be used.
- Thinking the equivalence point is always at pH 7. Only strong acid + strong base gives 7.
- Reading the burette from above or below the line of the meniscus (parallax error), or not rinsing the burette with the titrant first.