HSC Exam Practice

Sample response. A buffer solution is a solution that resists significant changes in pH when small amounts of strong acid or strong base are added, or when the solution is diluted with a small amount of water. It must contain: (1) a weak acid (HA) and (2) its conjugate base (A⁻), both present simultaneously in comparable concentrations.

Marking notes. 1 mark for “resists significant pH change on addition of small amounts of acid or base” (accept: “minimises”; reject: “prevents” or “neutralises” as sole descriptor). 1 mark for identifying a weak acid as one component. 1 mark for identifying its conjugate base as the other component (both must be present simultaneously for this mark).

Sample response. (i) When HCl is added (H⁺ added): H⁺ + CH₃COO⁻ → CH₃COOH. The conjugate base consumes the added H⁺, preventing a large pH decrease. (ii) When NaOH is added (OH⁻ added): OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O. The weak acid consumes the added OH⁻, preventing a large pH increase.

Marking notes. 1 mark per correct equation (must show correct reactants and products; arrows; ionic form). Deduct if equation shows H⁺ reacting with HA (common reversal error).

Sample response. Buffer capacity is maximised when [A⁻] = [HA] because at this ratio, both components are present in equal and maximum relative amounts. Any added acid is consumed by A⁻ and any added base is consumed by HA, so neither component is limiting. The buffer can absorb the maximum load of added acid or base before one component is exhausted. By Henderson-Hasselbalch: when [A⁻] = [HA], log([A⁻]/[HA]) = log(1) = 0, so pH = pK_a. Buffer capacity is therefore maximised at pH = pK_a.

Marking notes. 1 mark for stating that [A⁻] = [HA] means neither component is limiting / both available in maximum equal amounts. 1 mark for explaining this allows maximum absorption of either added acid (consuming A⁻) or base (consuming HA). 1 mark for stating pH = pK_a when [A⁻] = [HA], referencing Henderson-Hasselbalch.

Sample response. Buffer capacity is the total moles (or concentration) of strong acid or base a buffer can absorb before significant pH change occurs; it depends on the absolute concentrations of HA and A⁻ present. A more concentrated buffer has greater capacity even if its pH range is the same. Effective buffer range is the pH range over which a buffer resists pH change effectively; by Henderson-Hasselbalch (pH = pK_a + log([A⁻]/[HA])), this range extends approximately pK_a ± 1 pH units, corresponding to [A⁻]/[HA] ratios of 10:1 to 1:10. Outside this range, one component is present in such low proportion that it cannot adequately resist pH changes. Buffer capacity is a quantity; effective buffer range is a pH interval.

Marking notes. 1 mark for defining buffer capacity as a quantity (moles absorbed before failure; depends on concentration). 1 mark for defining effective buffer range as a pH interval (pK_a ± 1). 1 mark for distinguishing them: capacity = how much the buffer can absorb; range = over what pH values the buffer works.

Sample response. Henderson-Hasselbalch: pH = pK_a + log([HCO₃⁻]/[H₂CO₃]). At pH 7.40: 7.40 = 6.10 + log([HCO₃⁻]/[H₂CO₃]), so log([HCO₃⁻]/[H₂CO₃]) = 1.30, giving a ratio of 10^1.30 = 20:1. Although pH 7.40 is well outside the closed-solution effective range of pK_a ± 1 (= 5.10–7.10), the blood system is “open”: the lungs continuously expel CO₂ produced from H₂CO₃ decomposition (H₂CO₃ → CO₂ + H₂O), which regenerates buffer capacity by removing the acid component. The respiratory rate regulates [H₂CO₃] dynamically: faster breathing lowers [H₂CO₃], increasing the ratio and raising pH; slower breathing raises [H₂CO₃], decreasing the ratio and lowering pH. This open-system coupling allows effective buffering at pH 7.40 despite the large gap from pK_a.

Marking notes. 1 mark for correctly calculating [HCO₃⁻]/[H₂CO₃] = 20:1 using Henderson-Hasselbalch. 1 mark for identifying that pH 7.40 is outside the closed-solution effective range (pK_a ± 1 = 5.10–7.10). 1 mark for explaining the open-system mechanism (lungs expel CO₂, removing H₂CO₃, regenerating capacity). 1 mark for linking respiratory rate to dynamic pH regulation (Le Chatelier via CO₂ expulsion).

Sample response. In the partial neutralisation method, a calculated volume of strong base (e.g. NaOH) is added to a weak acid (HA) solution. The neutralisation reaction HA + OH⁻ → A⁻ + H₂O goes to completion (strong base fully consumed). The moles of conjugate base formed equal the moles of OH⁻ added: n(A⁻)formed = n(OH⁻)added. The moles of weak acid remaining equal the initial moles minus the moles of base added: n(HA)remaining = n(HA)initial − n(OH⁻)added. The Henderson-Hasselbalch equation is then applied using the mole ratio n(A⁻)/n(HA), since both species occupy the same solution volume. The amount of base added must be less than the initial moles of HA to ensure both components are present after the reaction.

Marking notes. 1 mark for the correct reaction equation (HA + OH⁻ → A⁻ + H₂O, goes to completion). 1 mark for both expressions: n(A⁻) = n(OH⁻) and n(HA)remaining = n(HA)initial − n(OH⁻). 1 mark for stating the condition that n(OH⁻) must be less than n(HA) to form a buffer (not the equivalence point).

Part (a) sample response. pK_a = −log(1.77 × 10⁻⁴) = 4 − log(1.77) = 4 − 0.248 = 3.75. pH = 3.75 + log(0.080/0.120) = 3.75 + log(0.667) = 3.75 − 0.176 = 3.57.

Part (a) marking notes (3 marks). 1 mark: pK_a = 3.75 (accept 3.7–3.8). 1 mark: correct ratio 0.080/0.120 = 0.667 inserted into Henderson-Hasselbalch. 1 mark: pH = 3.57 (accept 3.56–3.58 for rounding).

Part (b) sample response. n(HCOOH) = 0.120 × 0.500 = 0.0600 mol; n(HCOO⁻) = 0.080 × 0.500 = 0.0400 mol. n(NaOH) = 0.500 × 0.0100 = 5.00 × 10⁻³ mol. Reaction: OH⁻ + HCOOH → HCOO⁻ + H₂O. n(HCOOH) after = 0.0600 − 5.00 × 10⁻³ = 0.0550 mol. n(HCOO⁻) after = 0.0400 + 5.00 × 10⁻³ = 0.0450 mol. pH = 3.75 + log(0.0450/0.0550) = 3.75 + log(0.818) = 3.75 − 0.087 = 3.66. pH increases from 3.57 to 3.66 (rises by 0.09 units). Adding NaOH converts HA to A⁻, increasing [A⁻]/[HA], so log([A⁻]/[HA]) increases and pH rises.

Part (b) marking notes (3 marks). 1 mark: correct n(NaOH) = 0.00500 mol and correct post-reaction moles (n(HCOOH) = 0.0550, n(HCOO⁻) = 0.0450). 1 mark: pH = 3.66 (accept 3.65–3.67). 1 mark: pH increases (not decreases) with correct Henderson-Hasselbalch reasoning ([A⁻]/[HA] ratio increases).

Part (c) sample response. The buffer fails when all the weak acid (HCOOH) is consumed by OH⁻. Maximum n(NaOH) = n(HCOOH) initial = 0.0600 mol. Beyond this point, no HA remains to resist further base addition and pH rises dramatically. (The buffer can also fail if all HCOO⁻ is consumed by acid: maximum n(HCl) = n(HCOO⁻) = 0.0400 mol.)

Part (c) marking notes (1 mark). 1 mark for 0.0600 mol NaOH with correct justification (all HA consumed). Partial credit if student identifies 0.0400 mol for acid capacity instead.

Part (a) sample response. pH = pK_a + log([HCO₃⁻]/[H₂CO₃]) = 6.10 + log(14.2/1.13) = 6.10 + log(12.57) = 6.10 + 1.099 = 7.21. This is close to the recorded value of 7.25 (difference of 0.04 units; acceptable within rounding). The calculation confirms the Henderson-Hasselbalch equation models blood acid-base status accurately.

Part (a) marking notes (2 marks). 1 mark for correct calculation: 6.10 + log(14.2/1.13) = 7.20–7.21 (accept any answer in range 7.19–7.22; the table value of 7.25 is the “measured” value — students verify against it). 1 mark for stating the calculated pH is close to the table pH of 7.25 and noting this confirms the Henderson-Hasselbalch model.

Part (b) sample response. Patient Y’s [H₂CO₃] = 0.66 mmol/L, approximately half Patient W’s normal value of 1.20 mmol/L. H₂CO₃ forms from dissolved CO₂: CO₂(aq) + H₂O ⇌ H₂CO₃. If [H₂CO₃] decreases, by Le Chatelier’s principle the equilibrium shifts left, meaning [CO₂(aq)] has also decreased. This indicates that Patient Y is expelling CO₂ faster than it is being produced — consistent with hyperventilation (rapid, deep breathing, e.g. from anxiety, high altitude, or pain). Reducing [H₂CO₃] increases the [HCO₃⁻]/[H₂CO₃] ratio, raising blood pH above 7.45 (alkalosis).

Part (b) marking notes (3 marks). 1 mark for identifying [CO₂] has decreased (from Le Chatelier applied to CO₂ + H₂O ⇌ H₂CO₃). 1 mark for identifying hyperventilation (excessive CO₂ expulsion) as the cause. 1 mark for explaining how this raises the Henderson-Hasselbalch ratio and pH (increasing [HCO₃⁻]/[H₂CO₃] raises log term, raises pH above 7.45).

Sample response. The ocean carbonate buffer system involves the equilibrium CO₂(aq) + H₂O ⇌ H₂CO₃ ⇌ HCO₃⁻ + H⁺ ⇌ CO₃²⁻ + 2H⁺. The dominant buffering pair at ocean pH (~8.1) is H₂CO₃/HCO₃⁻, described by Henderson-Hasselbalch: pH = pK_a1 + log([HCO₃⁻]/[H₂CO₃]). At pre-industrial pH 8.2 and pK_a1 = 6.35, the ratio [HCO₃⁻]/[H₂CO₃] = 10^{(8.20−6.35)} = 708:1. Increasing atmospheric CO₂ from ~280 ppm (pre-industrial) to over 420 ppm (current) dissolves additional CO₂ in ocean water, shifting the equilibrium CO₂ + H₂O ⇌ H₂CO₃ to the right (Le Chatelier). This increases [H₂CO₃] and hence [H⁺], decreasing the [HCO₃⁻]/[H₂CO₃] ratio and lowering ocean pH. Since the industrial revolution, average ocean pH has fallen from ~8.2 to ~8.1 — a 26% increase in [H⁺]. The buffer has slowed but has not prevented this pH change, demonstrating that buffer capacity is finite. At the GBR, AIMS monitoring data show ocean pH has fallen from 8.18 (1980s) to 8.03 (2010s) as atmospheric CO₂ rose from 340 to 405 ppm. This pH decrease threatens reef-building corals: CaCO₃(s) + H⁺ ⇌ Ca²⁺(aq) + HCO₃⁻(aq). Increased [H⁺] shifts this equilibrium right, accelerating CaCO₃ dissolution and making calcification (CaCO₃ deposition) energetically more costly for corals. At pH below 7.9, aragonite (the CaCO₃ polymorph corals use) becomes undersaturated and net dissolution exceeds deposition, preventing reef growth and causing structural collapse. The carbonate buffer system therefore provides limited but significant protection: it has dampened the pH decrease by absorbing much of the added H⁺ via HCO₃⁻ + H⁺ → H₂CO₃. However, the rate of CO₂ addition exceeds the rate at which the ocean can replenish carbonate capacity (which comes from slow dissolution of seafloor CaCO₃, over centuries). The evidence — continued GBR pH decline despite buffering — demonstrates that the buffer capacity is being overwhelmed. Without rapid reduction in CO₂ emissions, the GBR carbonate buffer will fail to protect coral ecosystems from further acidification.

Marking notes. 1 mark — correct carbonate equilibrium and identification of H₂CO₃/HCO₃⁻ as the buffering pair. 1 mark — applies Henderson-Hasselbalch: increasing [H₂CO₃] lowers the ratio and hence pH (Le Chatelier). 1 mark — cites specific GBR/AIMS data or named Australian context (GBR, pH 8.2 → 8.1, AIMS); accept any correct Australian reference. 1 mark — explains CaCO₃ dissolution equilibrium and link to increased [H⁺] threatening corals (shell dissolution / calcification energy cost). 1 mark — defines or applies buffer capacity correctly: the total H⁺ the buffer can absorb before pH change becomes significant; ocean capacity is finite. 1 mark — evaluative claim: the buffer has slowed acidification but cannot prevent it at current CO₂ trajectories (evidence: observed continued pH decline). 1 mark — logical, coherent extended response linking chemistry to ecology with correct use of at least two: equilibrium equations, Henderson-Hasselbalch, Le Chatelier, buffer capacity, named Australian example.