IB Chemistry SL Paper 2 Practice: Chemical Equilibrium and Acid-Base Calculations

2026-11-10

IB Chemistry SL · Paper 2

IB Chemistry SL Paper 2 Practice: Chemical Equilibrium and Acid-Base Calculations: 6 questions, 36 parts, 100 marks, 75 minutes.

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Question 1 (18 marks): The Haber Process and Kc

Nitrogen and hydrogen react at 673 K and 200 atm over an iron catalyst: N2(g) + 3H2(g) in equilibrium with 2NH3(g), forward reaction exothermic by 92 kJ/mol.

#	Question	Answer and marking notes
1	(a) Write the Kc expression for this reaction.	Kc = [NH3]^2 / ([N2][H2]^3)
2	(b) In a 2.00 dm3 vessel at equilibrium there are 0.600 mol N2, 1.800 mol H2 and 0.400 mol NH3. Calculate Kc.	[N2]=0.300, [H2]=0.900, [NH3]=0.200 mol/dm3. Kc = (0.200)^2 / (0.300 x 0.900^3) = 0.0400 / 0.2187 = 0.183 mol-2 dm6 (0.18 accepted)
3	(c) State the units of Kc in (b).	(mol dm-3)^2 over (mol dm-3)^4 gives mol-2 dm6; the units depend on the stoichiometry.
4	(d) A further 0.200 mol of N2 is added at constant volume. Calculate Q and state the direction of shift.	New [N2] = 0.800/2.00 = 0.400. Q = 0.0400/(0.400 x 0.729) = 0.137. Q is less than Kc (0.183), so the equilibrium shifts to the right and NH3 increases.
5	(e) The volume is reduced from 2.00 dm3 to 1.00 dm3 at constant temperature. Explain the effect on the yield of NH3 and state whether Kc changes.	Concentrations double; there are 4 mol of gas on the left and 2 mol on the right, so the system shifts towards the side with fewer gas molecules (the right). The yield of NH3 rises. Kc is unchanged because temperature is constant.
6	(f) The temperature is raised from 673 K to 773 K. Explain the effect on Kc and on the equilibrium yield of NH3.	The forward reaction is exothermic, so raising the temperature shifts the equilibrium in the endothermic (reverse) direction. The yield of NH3 falls and Kc decreases.
7	(g) More iron catalyst is added. Explain the effect on the position of equilibrium and on Kc.	A catalyst lowers the activation energy of both directions equally. It only shortens the time taken to reach equilibrium; it does not change the position of equilibrium or Kc.

Question 2 (16 marks): Equilibrium Calculation for N2O4 and NO2

At 350 K, dinitrogen tetroxide dissociates: N2O4(g) in equilibrium with 2NO2(g). 0.200 mol of N2O4 is sealed in a 1.00 dm3 vessel and at equilibrium [NO2] = 0.0600 mol/dm3.

#	Question	Answer and marking notes
8	(a) Write the Kc expression.	Kc = [NO2]^2 / [N2O4]
9	(b) Use an ICE table to find the equilibrium concentration of N2O4.	Let the change be x: NO2 goes from 0 to 2x, so 2x = 0.0600 and x = 0.0300; [N2O4] = 0.200 – 0.0300 = 0.170 mol/dm3.
10	(c) Calculate Kc at 350 K.	Kc = (0.0600)^2 / 0.170 = 0.00360 / 0.170 = 0.0212 mol/dm3.
11	(d) Calculate the percentage dissociation of N2O4.	Dissociation = 0.0300 / 0.200 x 100% = 15.0%.
12	(e) The volume is halved at constant temperature. Predict the direction of shift and the change in colour, and explain.	Halving the volume doubles the concentrations, so Q = (0.120)^2/0.340 = 0.0424, which exceeds Kc; the equilibrium shifts left, NO2 falls and N2O4 rises, so the brown colour fades after an initial darkening.
13	(f) If the forward reaction were endothermic, how would Kc and the degree of dissociation change on heating?	An endothermic reaction shifts forward on heating, so Kc increases and the degree of dissociation rises.

Question 3 (18 marks): pH of Strong and Weak Acids

At 298 K, Kw = 1.0 x 10^-14. For ethanoic acid, CH3COOH, Ka = 1.8 x 10^-5.

#	Question	Answer and marking notes
14	(a) Calculate the pH of 0.0200 mol/dm3 hydrochloric acid (a strong acid).	[H+] = 0.0200 mol/dm3; pH = -log(0.0200) = 1.70.
15	(b) Calculate [H+] and the pH of 0.0200 mol/dm3 ethanoic acid.	Using Ka = [H+]^2/c with [H+] much less than c: [H+] = sqrt(1.8e-5 x 0.0200) = 6.0 x 10^-4 mol/dm3; pH = 3.22.
16	(c) Explain why the ethanoic acid has a higher pH than the hydrochloric acid of the same concentration.	Ethanoic acid is a weak acid and is only partly ionised, so [H+] is well below the acid concentration; hydrochloric acid is fully ionised, so [H+] equals its concentration and its pH is lower.
17	(d) Calculate the degree of ionisation of 0.0200 mol/dm3 ethanoic acid.	Degree of ionisation = 6.0e-4 / 0.0200 x 100% = 3.0%.
18	(e) The acid is diluted ten times to 0.00200 mol/dm3. Calculate the new pH and explain why the pH rises by less than one unit.	[H+] = sqrt(1.8e-5 x 0.00200) = 1.90 x 10^-4, so pH = 3.72, a rise of only 0.50. Dilution increases the degree of ionisation, which partly replaces the lost [H+].
19	(f) A solution at 298 K has pH 11.50. Calculate its [OH-] and [H+].	pOH = 14.00 – 11.50 = 2.50; [OH-] = 10^-2.50 = 3.16 x 10^-3 mol/dm3; [H+] = Kw/[OH-] = 3.16 x 10^-12 mol/dm3.

Question 4 (16 marks): Titration Curve and Indicator Choice

25.0 cm3 of 0.100 mol/dm3 ethanoic acid (Ka = 1.8 x 10^-5, Kw = 1.0 x 10^-14 at 298 K) is titrated with 0.100 mol/dm3 sodium hydroxide.

#	Question	Answer and marking notes
20	(a) Calculate the volume of NaOH needed to reach the equivalence point.	n(CH3COOH) = 0.100 x 0.0250 = 2.50 x 10^-3 mol; the reaction is 1:1, so V(NaOH) = 2.50e-3 / 0.100 = 0.0250 dm3 = 25.0 cm3.
21	(b) Calculate the pH at the equivalence point.	Total volume 50.0 cm3, so [CH3COO-] = 2.50e-3 / 0.0500 = 0.0500 mol/dm3. Kb = Kw/Ka = 5.56 x 10^-10; [OH-] = sqrt(5.56e-10 x 0.0500) = 5.27 x 10^-6; pOH = 5.28 and pH = 8.72.
22	(c) Which indicator should be used? State the colour change.	Phenolphthalein (range pH 8.3-10.0), colourless to pink; the equivalence pH of 8.72 lies inside this range.
23	(d) Describe the shape of the titration curve (initial pH, buffer region, near the equivalence point).	Initial pH about 2.9 for the weak acid; a buffer region from 0 to about 20 cm3 where pH rises slowly; a sharp rise near 25.0 cm3 from about pH 7 to pH 11; equivalence pH 8.72; then it approaches about pH 12.7 for the excess NaOH.
24	(e) Explain why methyl orange (pH 3.1-4.4) cannot be used for this titration.	The equivalence pH of 8.72 is far above the methyl orange range; methyl orange would already have changed colour near pH 4, well before the equivalence point, so the titre would be far too small.

Question 5 (18 marks): Determining Ka from Titration and pH

A student titrates 25.0 cm3 of a monoprotic weak acid HA of unknown concentration with 0.1000 mol/dm3 NaOH; the mean titre is 24.60 cm3. The pH of the original acid solution is 2.90 at 298 K.

#	Question	Answer and marking notes
25	(a) Calculate the concentration of HA from the titration data.	n(NaOH) = 0.1000 x 0.02460 = 2.460 x 10^-3 mol; the reaction is 1:1, so c(HA) = 2.460e-3 / 0.0250 = 0.0984 mol/dm3.
26	(b) Calculate [H+] in the original solution from pH 2.90.	[H+] = 10^-2.90 = 1.26 x 10^-3 mol/dm3.
27	(c) Calculate Ka for HA.	Ka = [H+]^2 / c = (1.26e-3)^2 / 0.0984 = 1.61 x 10^-5.
28	(d) Calculate pKa for HA.	pKa = -log(1.61e-5) = 4.79.
29	(e) Calculate the percentage ionisation of HA at this concentration.	Ionisation = 1.26e-3 / 0.0984 x 100% = 1.28%.
30	(f) The calculation in (c) assumed that [H+] is much smaller than c. Use the figures to judge whether this is valid, and state how Ka would be found without the assumption.	1.26e-3 is about 1.3% of 0.0984, well below 5%, so the assumption is valid. Without it, Ka = x^2/(c – x) must be solved, or the denominator corrected to 0.0971, giving Ka of about 1.63 x 10^-5, some 1% higher.

Question 6 (14 marks): Applying and Interpreting Equilibrium Constants

The following questions concern the meaning of the equilibrium constant and its use in predicting the direction of change.

#	Question	Answer and marking notes
31	(a) Explain why Kc changes only with temperature and not with concentration, pressure or catalyst.	Kc is a ratio of equilibrium concentrations, and the position of equilibrium depends only on temperature for a given reaction; changing concentration or pressure moves the system to a new position that still satisfies the same Kc, and a catalyst does not shift the position.
32	(b) What is the effect of adding a catalyst on the position of equilibrium?	None; both rates increase equally, so only the time to reach equilibrium is shortened.
33	(c) Kc for Question 1 is 0.183. Write Kc for the reverse reaction 2NH3(g) in equilibrium with N2(g) + 3H2(g).	Kc(reverse) = 1 / 0.183 = 5.47 (units mol2 dm-6).
34	(d) At one instant [NH3] = 0.400, [N2] = 0.100 and [H2] = 0.300 mol/dm3. Calculate Q and state the direction of shift.	Q = (0.400)^2 / (0.100 x 0.300^3) = 0.160 / 0.00270 = 59.3. Q greatly exceeds Kc (0.183), so the reaction shifts left and NH3 falls.
35	(e) In the industrial process, ammonia is continuously condensed and removed. Explain the effect on the yield.	Removing product makes Q smaller than Kc, so the equilibrium keeps shifting right and the overall yield of ammonia rises (Le Chatelier).
36	(f) Industry uses both high pressure and a moderately low temperature. Explain this compromise.	High pressure favours the right-hand side, which has fewer gas molecules, and raises the yield; a low temperature favours the exothermic forward reaction and raises Kc, but too low a temperature makes the rate far too slow and the catalyst ineffective, so about 673 K is used.

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