HSC Chemistry Formula Sheet

HSC tip: Always convert mass to grams before substituting. For compounds, add up all atomic masses: M(H₂O) = 2(1.008) + 16.00 = 18.016 g mol⁻¹.

HSC tip: Specify what is being counted. 1 mol of H₂ contains 6.022 × 10²³ molecules, but 2 × 6.022 × 10²³ atoms.

Common mistake: NSW HSC uses STP = 0°C and 100 kPa (not 1 atm). Always use 22.7 L mol⁻¹ unless the question specifies SLC. Check your data sheet.

Unit trap: Volume must be in litres (L), not mL. 250 mL = 0.250 L. Write the conversion explicitly in your working.

Why it works: Dilution doesn't change the number of moles, only the volume increases. Since n = cV, n stays constant: c₁V₁ = c₂V₂.

Example: % of O in H₂SO₄ (M = 98): % = (4 × 16.00 / 98.09) × 100 = 65.3%

HSC note: Both values must be in the same unit (both grams or both moles). Always calculate theoretical yield first using the limiting reagent.

Common use: Used in gravimetric analysis and back-calculations. If a sample is 96.2% pure, then m_pure = 0.962 × m_sample.

4-step method: (1) Write balanced equation. (2) Find moles of known. (3) Apply mole ratio. (4) Convert moles of unknown to required unit.

Sign convention: If temperature rises, the reaction is exothermic (heat released to solution, Q is positive here, but ΔH_rxn is negative). Always state whether energy is released or absorbed.

The negative sign: If Q is positive (solution warmed, heat released by reaction), ΔH must be negative (exothermic). The sign flip ensures the system's perspective is correct. Always convert Q to kJ before dividing.

The value that is always zero: ΔH°_f of an element in its standard state is 0 by definition, so O₂(g), C(graphite) and N₂(g) contribute nothing. Products minus reactants, in that order — reversing it is the most common sign error in this question type.

Unlike ΔH°_f, S° is never zero for an element — absolute entropy is only zero for a perfect crystal at 0 K. Sanity-check the sign against the equation: more moles of gas on the right means ΔS° should come out positive.

The unit trap that costs the mark: ΔH° is in kJ mol⁻¹ but ΔS° is in J K⁻¹ mol⁻¹, so ΔS° must be converted to kJ K⁻¹ mol⁻¹ first. Temperature must be in kelvin, never °C. When ΔH° and ΔS° share a sign, spontaneity depends on T — that is the whole basis of "at what temperature does this become spontaneous", solved by setting ΔG° = 0.

What to leave out: Pure solids and pure liquids do not appear in the expression, so only include gases and aqueous species. A large K means products are favoured at equilibrium; it says nothing about how fast equilibrium is reached.

How to use it: Compare Q with K. If Q < K the reaction moves forward (makes more product); if Q > K it moves in reverse; if Q = K the system is already at equilibrium. This is the calculation behind a "which way will it shift" question.

Worked link to solubility: The undissolved solid is left out of the expression. For AgCl, K_sp = [Ag⁺][Cl⁻] = s², so s = √K_sp. For a 1:2 salt such as CaF₂, K_sp = (s)(2s)² = 4s³ — the coefficients become powers, so you cannot compare K_sp values across different salt ratios.

The step students skip: Mixing two solutions dilutes both, so recalculate each concentration for the combined volume before you substitute. Then Q_sp > K_sp means a precipitate forms, Q_sp = K_sp means the solution is exactly saturated, and Q_sp < K_sp means no precipitate.

Strong versus weak: For a strong monoprotic acid, [H⁺] equals the concentration of the acid, because it ionises completely. A weak acid does not — you need K_a and an ICE table. One pH unit is a tenfold change in [H⁺], so pH 3 is 100 times more acidic than pH 5.

Route for a base question: Find [OH⁻] from the concentration of the base, take pOH, then use pH + pOH = 14.00 at 25 °C. Watch dibasic bases such as Ca(OH)₂, which give two moles of OH⁻ per mole of base.

The temperature trap: "pH + pOH = 14" and "neutral means pH 7" are both only true at 25 °C. Self-ionisation is endothermic, so heating water raises K_w and lowers the neutral pH — the water is still neutral, because [H⁺] still equals [OH⁻].

The standard weak-acid method: Build an ICE table with x = [H⁺], so K_a = x²/(c − x). For a weak acid you may approximate c − x ≈ c, then check the approximation holds (ionisation under about 5%). Do not use this on a strong acid.

Why it is used for ranking: K_a values span many powers of ten, so pK_a puts them on a readable scale. One pK_a unit is a tenfold difference in K_a. The direction reverses, so the strongest acid in a list has the smallest pK_a — the most common slip in this question type.

It only works for a pair: K_a and K_b must belong to the same conjugate acid-base pair, such as CH₃COOH and CH₃COO⁻. This is how you get K_b for a base the question only gives you K_a for, and it is why a stronger acid always has a weaker conjugate base.

Marks are lost here, not in the algebra: Use the average of concordant titres only (within about 0.10 mL), convert every volume from mL to L, and check the ratio — H₂SO₄ with NaOH is 1 : 2, not 1 : 1. For a back titration, find the moles that reacted with the excess and subtract.

Not the same as yield: Percentage yield measures how much of the possible product you actually got; atom economy measures how much of the reactant mass ends up in the product by design. A reaction can have 95% yield and poor atom economy, which is the standard "explain why a high-yield route can still be wasteful" question.

Read the direction: A lower E-factor is better, which is the opposite of atom economy and percentage yield. Because it counts solvent, a route with excellent atom economy can still have a high E-factor — that contrast is the point of the metric.

Why the calibration curve exists: you rarely know ε, so you plot absorbance against known standards and read the unknown off the straight line. This underpins both colourimetry (finding an equilibrium concentration for a K_eq calculation) and AAS. The line stays straight only at low concentration — a very concentrated sample must be diluted first.

Where the 1 mg L⁻¹ shortcut comes from: 1 L of dilute aqueous solution has a mass of about 1 kg = 10⁶ mg, so 1 mg of solute in 1 L is 1 part per million. It holds for water-quality work, not for concentrated or non-aqueous solutions. Use ppb × 10⁹ for trace heavy metals.

Exam advice: These values are provided on the NESA HSC Chemistry reference sheet. You don't need to memorise them, but you do need to know when and how to use them. The official sheet also carries a periodic table, solubility rules and a table of standard potentials, which are not reproduced here.