Define a buffer solution. Which of the following pairs of chemicals would act as buffer solution: H3PO4 / NaH2PO4; H2SO4 / Na2SO4; NH3 / NH4Cl.

The balanced chemical equation is:

$$2\text{SO}_{2(g)} + \text{O}_{2(g)} + \text{H}_{2}\text{O}_{(l)} \longrightarrow 2\text{H}_{2}\text{SO}_{4(l)}$$

From the stoichiometry, 1 mole of O$_{2}$ reacts to produce 2 moles of H$_{2}$SO$_{4}$.
Given 1.5 moles of O$_{2}$:

$$\text{Moles of H}_{2}\text{SO}_{4} = 1.5 \text{ mol O}_{2} \times \frac{2 \text{ mol H}_{2}\text{SO}_{4}}{1 \text{ mol O}_{2}} = 3.0 \text{ mol H}_{2}\text{SO}_{4}$$

Given atomic masses: H = 1.0, O = 16.0, S = 32.0

$$\text{Molar mass of H}_{2}\text{SO}_{4} = (2 \times 1.0) + (1 \times 32.0) + (4 \times 16.0) = 2.0 + 32.0 + 64.0 = 98.0 \text{ g/mol}$$

$$\text{Mass of H}_{2}\text{SO}_{4} = \text{Moles} \times \text{Molar mass} = 3.0 \text{ mol} \times 98.0 \text{ g/mol} = 294.0 \text{ g}$$

The mass of H$_{2}$SO$_{4}$ that can be prepared is $\boxed{\text{294.0 g}}$.

3(c)(i)

• (α) less than 7: NH$_{3(aq)}$ with H$_{2}$SO$_{4(aq)}$
• (β) equal to 7: KOH$_{(aq)}$ with HCl$_{(aq)}$
• (γ) greater than 7: NaOH$_{(aq)}$ with H$_{2}$CO$_{3(aq)}$

3(c)(ii)

• For (α) less than 7 (NH$_{3}$ with H$_{2}$SO$_{4}$): This is a titration of a weak base with a strong acid. At the equivalence point, the resulting salt (ammonium sulfate) contains the ammonium ion (NH$_{4}^{+}$), which is the conjugate acid of a weak base. The NH$_{4}^{+}$ ion hydrolyzes in water to produce H$^{+}$ ions, making the solution acidic (pH < 7).
• For (β) equal to 7 (KOH with HCl): This is a titration of a strong base with a strong acid. At the equivalence point, the resulting salt (potassium chloride) is formed from a strong acid and a strong base. Neither the K$^{+}$ nor the Cl$^{-}$ ions hydrolyze significantly in water, so the solution remains neutral (pH = 7).
• For (γ) greater than 7 (NaOH with H$_{2}$CO$_{3}$): This is a titration of a strong base with a weak acid. At the equivalence point, the resulting salt (sodium carbonate or bicarbonate) contains the carbonate or bicarbonate ion, which is the conjugate base of a weak acid. These ions hydrolyze in water to produce OH$^{-}$ ions, making the solution basic (pH > 7).

3(d)(i)
The factors that determine whether a given ionic solid would dissolve in water include:

• Lattice energy: The energy required to break the ionic bonds in the solid. Lower lattice energy generally favors dissolution.
• Hydration energy: The energy released when ions are surrounded by water molecules. Higher hydration energy generally favors dissolution.
• Temperature: The solubility of most ionic solids increases with increasing temperature.

3(d)(ii)

• (α) Na$_{2}$CO$_{3(aq)}$ + ZnCl$_{2(aq)}$ $\longrightarrow$ 2NaCl$_{(aq)}$ + ZnCO$_{3(s)}$: This is a precipitation reaction (or double displacement reaction) because two aqueous solutions react to form an insoluble solid (precipitate).
• (β) Zn$_{(s)}$ + CuSO$_{4(aq)}$ $\longrightarrow$ ZnSO$_{4(aq)}$ + Cu$_{(s)}$: This is a single displacement reaction (or redox reaction) because a more reactive metal (zinc) displaces a less reactive metal (copper) from its compound.
• (γ) CH$_{4(g)}$ + 2O$_{2(g)}$ $\longrightarrow$ CO$_{2(g)}$ + 2H$_{2}$O$_{(l)}$: This is a combustion reaction because a hydrocarbon (methane) reacts rapidly with oxygen to produce carbon dioxide and water, releasing energy.