SACE Stage 2 Chemistry exam strategy: The 2026 guide

Seawater has $[\text{H}^+] = 7.9 \times 10^{-9}$ mol L$^{-1}$. Find its pH and explain how rising atmospheric carbon dioxide changes it.

$\text{pH} = -\log(7.9 \times 10^{-9}) = 8.10$.

$\text{CO}_2(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_2\text{CO}_3(aq) \rightleftharpoons \text{H}^+(aq) + \text{HCO}_3^-(aq)$

More dissolved $\text{CO}_2$ shifts these equilibria to the right, increasing $[\text{H}^+]$ and lowering the pH. The extra $\text{H}^+$ reacts with carbonate ions ($\text{H}^+ + \text{CO}_3^{2-} \rightarrow \text{HCO}_3^-$), making less carbonate available for shells and favouring the dissolution of calcium carbonate.

1.00 mol of $\text{N}_2\text{O}_4$ is placed in a sealed 2.00 L flask: $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$. At equilibrium, 0.40 mol of $\text{NO}_2$ is present. Calculate Kc.

Convert moles to concentrations before substituting, and use the mole ratio (1 : 2) for the change row.

Burning 0.50 g of ethanol ($M = 46.07$ g mol$^{-1}$) raises the temperature of 200 g of water by 15.0 °C. Estimate the molar enthalpy of combustion.

$q = mc\Delta T = 200 \times 4.18 \times 15.0 = 12\,540$ J $= 12.5$ kJ. $n(\text{ethanol}) = \dfrac{0.50}{46.07} = 0.0109$ mol.

$\Delta H \approx -\dfrac{12.5}{0.0109} \approx -1.16 \times 10^3$ kJ mol$^{-1}$.

The accepted value is about $-1367$ kJ mol$^{-1}$; the experimental value is smaller in magnitude because heat is lost to the surroundings and combustion may be incomplete, a systematic error worth naming in an evaluation question. Per gram, the accepted value is $1367 \div 46.07 \approx 29.7$ kJ g$^{-1}$.

A water sample contains 25 mg of nitrate ions per litre. Express this in ppm and in mol L$^{-1}$ ($M(\text{NO}_3^-) = 62.0$ g mol$^{-1}$).

For dilute aqueous solutions, 1 mg L$^{-1}$ is 1 ppm, so the concentration is 25 ppm. In moles: $\dfrac{0.025}{62.0} = 4.0 \times 10^{-4}$ mol L$^{-1}$.