1. Chemical equilibrium & acids and bases Applying Le Chatelier's Principle to acid & base equilibrium

2. Answering equilibrium questions Check that you have answered the question eg colour change, yield etc Reason: Explain how favouring the reaction identified above relieves the stress (eg reverse reaction is endothermic therefore absorbs heat and relieves the stress of high temperature.) Response: State which reaction is favoured (forward or reverse). If a graph or data is given explain how you know that this reaction was favoured (eg the % yield of product increased or decreased). Stress: Identify stress (eg increase temperature, increase pressure etc)

3. Apply principles to acids and hence predict the effect on pH Calculation of pH is not required. A qualitative understanding of pH decreasing with increasing [H 3 O + ] is sufficient. Application of equilibrium principles to acids & bases Relate the strength of an acid or base to the degree of its ionisation and its K a or K b value..

4. HSO 4 - (aq) + H 2 O (l)  SO 4 2- (aq) + H 3 O + (aq) When the temperature is decreased the pH of the solution decreases. Use this information to deduce whether the forward reaction is exothermic or endothermic? Explain fully. Stress: Decrease temperature Response: Favour forward reaction as shown by the decrease in pH which indicates an increase in [H 3 O + ] Reason: The forward reaction must be exothermic produces heat which relieves stress of low temperature.

5. HSO 4 - (aq) + H 2 O ( ℓ )  SO 4 2- (aq) + H 3 O + (aq) ΔH<0 (K a = 1,1 x 10 -2 ) Is HSO 4 - a strong or weak acid? Explain. Answer: HSO 4 - is a weak acid since K a is low (<1) indicating that it does not ionise completely. How is K a affected by a decrease in temperature? Answer: Increase. A decrease in temperature favours the forward (exothermic) reaction, leading to an increase in [products] and a decrease in [reactants].

6. HSO 4 - (aq) + H 2 O ( ℓ )  SO 4 2- (aq) + H 3 O + (aq) How would the pH be affected if a few crystals of Na 2 SO 4 ions were dissolved in the solution? Stress: Add SO 4 2- Response: Favour reverse reaction Reason: The reverse reaction uses SO 4 2- ions  relieves stress (common ion effect) Answer to question? pH will increase as [H 3 O + ] decreases in the reverse reaction.

7. Calculate the concentration of OH - ions in the aqueous equilibrium mixture if the concentration of H 3 O + ions at 25 o C is 2,5 x 10 -3 mol.dm -3. K w = [H 3 O + ].[OH - ] = 1 x 10 -14 [OH - ] = 1 x 10 -14 2,5 x 10 -3 = 4 x 10 -12 mol.dm -3 HSO 4 - (aq) + H 2 O ( ℓ )  SO 4 2- (aq) + H 3 O + (aq)

8. H 2 X (aq) + 2OH - (aq  X 2- (aq) +2H 2 O ( ℓ ) ΔH>0 No. of moles time t1t1 X 2- H2XH2X OH - X 2- OH - H2XH2X With reference to the changes shown in the graph use Le Chatelier’s Principle to explain what stress was introduced at t 1 and state how it affected the pH.

9. Stress: X 2- added as shown by sudden increase in no. of moles of X 2- at t 1. Response: Reverse reaction favoured as shown by the increase in the no. of moles of OH - and H 2 X after t 1. Reason: The reverse reaction uses up X 2- and relieves the stress. Effect on pH? pH increases since [OH - ] increases in the reverse reaction (which would lead to a decrease in [H 3 O + ]).

10. Hydrolysis of a salt – a reaction with water where water itself is decomposed. Eg. Is an aqueous solution of CH 3 COONa acidic, basic or neutral? Explain your answer with reference to the interaction between the ions of the salt and the ions present in water.

11. Quick inspection method (suitable for answering multiple choice and for identifying acid and base required to make the salt.) CH 3 COONa Na + STRONG base (NaOH) CH 3 COO - weak acid (CH 3 COOH) Therefore, CH 3 COONa is a weak base salt (and will upset the equilibrium between the ions in water leading to an excess of OH - ions.) H+H+ OH -

12. Full explanation (cognitive level 4) Consider an aqueous solution of CH 3 COONa CH 3 COONa (s)  CH 3 COO - (aq) + Na + (aq) H 2 O  H + + OH - Consider the pairs of ions formed; 1.CH 3 COO - and H + would form the weak acid CH 3 COOH which is not fully ionised. 2.Na + and OH - would form the strong base NaOH which remains fully dissociated. H + ions are removed from the solution, upsetting the eqm in water (favouring the forward reaction for the ionisation of water) and leading to an excess of OH - ions pH>7

13. Acid-base calculations (titrations and other) Use of mole ratios rather than ‘formula’ Eg (1) 20 cm 3 of KOH (0,3 mol.dm -3 ) just neutralised 12 cm 3 of H 2 SO 4. Calculate the concentration of the acid. Moles of KOH n = C.V = 0,3 x 0,02 = 0,006 mol KOH Mole ratio (acid : base)H 2 SO 4 : KOH 1 : 2 0,003 : 0,006 Conc. of H 2 SO 4 C = n = 0,003 V 0,012 = 0,25 mol.dm -3 H 2 SO 4 + 2KOH  K 2 SO 4 + 2H 2 O

14. Acid-base calculations (titrations and other) Eg (2) 0,28 g of KOH is dissolved in distilled water. 10 cm 3 of sulphuric acid exactly neutralises the KOH. The products formed are potassium sulphate and water. Write a balanced equation for the reaction and calculate the concentration of the acid. H 2 SO 4 +2KOH  K 2 SO 4 + 2H 2 O 0,28 g KOH n = m/M = 0,28/ 56 = 0,05 mol mol ratioH 2 SO 4 : KOH 1 : 2 thus, 0,025 : 0,05 Conc. of H 2 SO 4 ? C = n/V = 0,025 / 0,01 = 0,25 mol.dm -3

15. X 2(g) + 2Y (s)  2XY (g) ΔH>0 What change was made at 40 s? Choose from ONE of the following; increase pressure, increase temperature, remove XY, decrease pressure, decrease temperature). Top line at start= forward reaction

16. Stress: Increase temperature It couldn’t have been an increase in pressure as this would have favoured the reverse reaction which leads to a decrease in the number of moles of gas (IGNORE SOLID). It couldn’t have been remove XY as this would have decreased the rate of both reactions (less molecules  less effective collisions per second.) Response: Favour forward reaction. Both reactions increased in rate but the forward reaction (broken line) increased the most. Reason: The forward reaction is endothermic absorbs heat which relieves stress of high temperature

17. X 2(g) + 2Y (s)  2XY (g) ΔH>0 What change was made at 60 s? Choose from ONE of the following; increase pressure, increase temperature, decrease pressure, decrease temperature).

18. Stress: Decrease pressure Response: Favour forward reaction. Both reactions decreased in rate but the forward reaction (broken line) decreased the least. Reason: The forward reaction produces more moles of gas (IGNORE SOLID)  more collisions with sides of container which increases pressure and relieves the stress.

19. X 2(g) + 2Y (s)  2XY (g) ΔH>0 Explain the effect on the equilibrium of adding more of the solid Y to the reaction vessel. A solid has NO EFFECT on the equilibrium since it is a pure substance  does not have a concentration.

20. H 2 (g) + I 2 (g) ⇌ 2HI (g) ΔH < 0 What change was made at 35 s?

21. Stress: Increase temperature Response: Favour reverse reaction as shown by the decrease in [HI] and increase in [H 2 ] and [I 2 ] Reason: The reverse reaction is endothermic  absorbs heat which relieves stress of high temperature

22. H 2 (g) + I 2 (g) ⇌ 2HI (g) ΔH < 0 What change was made at 55 s? H 2 was added, favouring the forward reaction to use it up

23. Practical demonstration of equilibrium using cobalt chloride solution To prepare solution Dissolve cobalt chloride crystals in ethanol and slowly add water until the solution resembles the same purple colour as methylated spirits.

24. CoCℓ 4 2 (eth.) +6H 2 O (eth.)  Co(H 2 O) 6 2+ eth.) +4 Cℓ - (eth.) ΔH<0 BluePink Mixture is heated. State and explain what is observed Stress: Increase in temperature Response: Favour reverse reaction Reason: The reverse reaction is endothermic and absorbs the extra heat energy therefore relieving the stress. Observation: Solution turns blue since the reverse reaction produces CoCl 4 2- Solvent = ethanol (eth.)

25. NaCℓ is dissolved in mixture. Stress: Increase [Cℓ – ] (common ion effect) Response: Favour reverse reaction Reason: The reverse reaction uses Cℓ – ions therefore relieving the stress. Observation: Solution turns blue since the reverse reaction produces CoCl 4 2- CoCℓ 4 2 (eth.) +6H 2 O (eth.)  Co(H 2 O) 6 2+ eth.) +4 Cℓ - (eth.) ΔH<0 BluePink

26. Water is added to the reaction mixture. Stress: Add water Response: Favour forward reaction Reason: The forward reaction uses water (to make the hydrated cobalt ion) therefore relieving the stress. Observation: Solution turns pink since the forward reaction produces Co(H 2 O) 6 2- CoCℓ 4 2 (eth.) +6H 2 O (eth.)  Co(H 2 O) 6 2+ eth.) +4 Cℓ - (eth.) ΔH<0 BluePink