1. ACIDS AND BASES Questions may involve any of the following: description of acids and bases in terms of proton transfer calculations involving K w and pH (K w = [H 3 O + ][OH  ] = 10 -14 given) comparison of properties of aqueous solutions of strong and weak acids and bases (properties restricted to conductivity, rate of reaction and pH)  recognition of acid-base properties of ionic compounds that contain only one ion that changes the pH of water.

2. Acids were originally classified in terms of the following behaviour in aqueous solution: acids turn blue litmus red acids have a pH less than 7 acids neutralise bases acids react with metal carbonates which fizzes and produces CO 2 gas. acids react with active metals to produce hydrogen gas (but not with Cu, Ag or Au). acids taste sour.

3. Bases were originally classified in terms of the following behaviour in aqueous solution: turn red litmus blue, have a pH greater than 7, neutralise acids, taste bitter, feel soapy

4. Bronsted-Lowry acids and bases. Bronsted acids are classified as those substances that donate a proton (H + ) to a base, a proton acceptor. Proton donation to a water molecule forms H 3 O + (hydronium) ions. HA(aq) + H 2 O(l) H 3 O + (aq) + A  (aq) acid base proton donor proton acceptor Proton (H+) donation

5. Bronsted-Lowry acids and bases. Similarly proton donation from water to a base produces OH - ions. H 2 O(l) + B(aq) BH + (aq) + OH  (aq) acid base proton donor proton acceptor Proton (H+) donation Solution is now basic ie pH > 7 because of the formation of OH- ions

6. Conjugate acids and bases. Conjugate acid-base pairs. If 2 species differ by just 1 proton they are classed as a conjugate acid-base pair. H 2 O(l) + B(aq) BH + (aq) + OH  (aq) acid 1 base 2 acid 2 base 1 The conjugate acid/base pairs in the above reaction are: Acid Base H 2 O / OH BH + / B

7. Remember water can act as an acid by donating an H+ or a base by accepting an H+

8. Conjugate acid-base pairs If two species differ by just one proton they are classed as a conjugate acid-base pair. Examples of acid-base pairs are H 2 SO 4 /HSO 4 , and NH 4 + /NH 3. Write the reaction of H 2 SO 4 with water H 2 SO 4 + H 2 O HSO 4 - + H 3 O + Acid Base H 2 SO 4 / HSO 4 - H 3 O + / H 2 O The acid is always the species with the additional proton. It can also be said that HSO 4 - is the conjugate base of H 2 SO 4. The conjugate acid/base pairs in the above reaction are:

9. Some substances can behave as both acids and bases. They are classed as amphiprotic eg. H 2 O or HCO 3 . HCO 3 - + OH  H 2 O + CO 3 2  acid 1 base 2 acid 2 base 1 HCO 3  + H 2 O OH  + H 2 CO 3 base 1 acid 2 base 2 acid 1 Is HCO 3 – acting as an acid or a base in the above reaction ? Here HCO 3 – is acting as base because it is accepting a proton in the above reaction

10. Exercises: 1. Complete the acid/base pairs for each of the following: (a) acids - H 2 CO 3 HSO 4  H 2 O HCl bases - HCO 3 - SO 4 2- OH - Cl - (b) bases - HSO 4  SO 4 2  H 2 O CH 3 COO  acids - H 2 SO 4 HSO 4 - H 3 O + CH 3 COOH Which of the species listed above can be classified as amphiprotic?

11. Turn to page154 in your lab book – Acids and bases complete the questions 1. Water is very poor conductor of electricity. Explain why. Pure water ionises to a very slight extent to produce the following concentrations [H 3 O + ] = 1 x 10 -7 molL -1 and [OH-] = 1 x 10 -7 molL -1 such concentrations have a very small charge carrying capacity

12. 2. Write a chemical equation to show what happens in a neutralisation reaction. H 3 O + (aq) + OH- (aq) 2H 2 O

13. 3. CompoundSpecies present Is solution acidic, basic or neutral ? Formula of contributing ion Ethanoic acid Sodium carbonate Ammonium chloride Sodium ethanoate Sodium nitrate Potassium chloride Sodium hydrogen sulfate H 3 O +, CH 3 COO - acidic basic neutral acidic H3O+H3O+ CO 3 2- NH 4 + CH 3 COO - _ _ HSO 4 - Na+, CO 3 2-, (OH-) * NH 4 +, Cl - Na +, CH 3 COO - Na+, NO 3 - K+, Cl - Na+, HSO 4 -

14. 4. Identify by symbol the conjugate base for the following (a) HF (b) H 2 O (c) HSO 3 – (d) H 2 CO 3 F - OH – SO 3 2- HCO 3 -

15. 5. Identify by symbol the conjugate acid for the following (a) I - (b) NO 3 - (c) HSO 4 – (d) H 2 O HI HNO 3 H 2 SO 4 H3O +H3O +

16. 6. Hydrogen Chloride and ammonia. (a) HCl + NH 3 NH 4 + + Cl  acid 1 base 2 acid 2 base 1 (b) Water and the hydrogen sulfate ion H 2 O + HSO 4  H 3 O + + SO 4 2- base 1 acid 2 acid 1 base 2

17. Discussion Questions for Strong and weak acid Expt 1.The balanced equations for the reactions involved are: 2CH 3 COOH + CaCO 3 Ca(CH 3 COO) 2 + H 2 O + CO 2 2HCl + CaCO 3 CaCl 2 + H 2 O + CO 2 2. How do the rates of reaction compare for acids of the same concentration? The rates involving the HCl solutions were much greater than those involving CH 3 COOH solutions

18. Dissociation constant of water, K w Pure water is a very weak conductor of electricity showing it contains a very low concentration of ions. These ions are produced by the dissociation of water molecules as follows: H 2 O + H 2 O H 3 O + + OH 

19. Dissociation constant of water, K w H 2 O + H 2 O H 3 O + + OH  This equilibrium lies well to the left so that most of the water is in the molecular form. Since each dissociation produces both a hydronium ([H 3 O + ]) and a hydroxide (OH - ) ion it follows that [H 3 O + ] = [OH  ]. At 25 o C, the concentration of both species (ie H 3 O+ and OH-) = 1 x 10 -7 mol L -1.

20. K w is the dissociation constant for pure water and it is given by the equation K w = [H 3 O + ] x [OH  ] = (1 x 10 -7 ) x (1 x 10 -7 ) = 1 x 10 -14 In this course we will not assign K w units

21. If an acid is added to pure water then the [H 3 O + ] increases and at the same time [OH - ] decreases so that the product of the two concentrations remains unchanged. In other words given an H 3 O+ or OH- concentration we can use the Kw = [H 3 O+] [OH-] expression to find an unknown concentration Remember Kw = 1x 10 -14 This concept is very important – and useful

22. For example if [H 3 O + ] = 2.5 x 10 -3 mol L -1 find the [OH-] [OH  ] == 4.0 x 10 -12 mol L -1 Rearrange K w = [H 3 O + ] x [OH  ] to find the OH- concentration

23. Complete the following table using the previous Kw formula. [H 3 O + ][OH  ]acid or base? 2.4 x 10 -3 0.0175 0.0036 5.4 x 10 -10 4.17 x10 -12 5.71 x 10 -13 2.78 x 10 -12 1.85 x 10 -5 acid base acid

24. pH Scale Because the H+ (or H 3 O+) concentration is a logrithmic scale and not a linear scale we use the pH scale to express the acidity of a solution So the pH of an aqueous solution is given by the expression pH = -log 10 [H 3 O + ] pOH Scale This is the same for an OH- ion concentration too. Likewise the pOH of an aqueous solution is given by the expression pOH = -log 10 [OH - ]

25. pH Scale The pH of an aqueous solution is given by the expression pH = -log 10 [H 3 O + ] In pure water, [H 3 O + ] = 1 x 10 -7 mol L -1, so pH = -log 10 (1 x 10 -7 ) = 7.0 Any solution with a pH = 7.0 is a neutral solution Since [H 3 O + ] = [OH  ], ie. neither [H 3 O + ] or [OH  ] is in excess. The pH of an acidic solution is 7.

26. Mathematically pH and pOH are related to each other Remember in the dissociation of pure H 2 O [H 3 O + ] x [OH  ] = K w (1 x 10 -7 ) x (1 x 10 -7 ) = 1 x 10 -14 -log 10 (1 x 10 -7 ) + -log 10 [H 3 O + ]+ -log 10 [OH - ] = -log 10 [1 x 10 -14 ] -log 10 (1 x 10 -7 ) = 14 7 + 7 = 14 pH + pOH = 14 This is handy to know!

27. pH Calculations (remember these!) pH = -log [H 3 O + ] [H 3 O + ] = 10 -pH pOH = -log [OH - ] [OH - ] = 10 -pOH AND

28. Example calculation Finding pH Find the pH of a O.1molL -1 HCl solution For any strong acid in water the H 3 O + concentration = the concentration of the acid So pH = -log [H 3 O + ] pH = -log [0.1] pH = 1 We can also find the pOH using the formula pH + pOH = 14 ie 1 + 13 = 14 therefore the pOH of the solution is 13

29. Example Finding H 3 O+ concentration from pH Calculate the H 3 O + concentration from a pH of 3.2 pH = -log [H 3 O + ] rearranged: [H 3 O + ] = 10 -pH [H 3 O + ] = 10 -3.2 [H 3 O + ] = 6.31 x 10 -4 molL-1

30. Example Finding OH- concentration from pOH Calculate the OH- concentration from a pOH of 8.4 pOH = -log [OH-] rearranged: [OH-] = 10 -pOH [OH-] = 10 -8.4 [OH-] = 3.98 x 10 -9 molL-1

31. [H 3 O + ] [OH  ] pOHpH 2.4 x 10 -3 0.0175 0.0036 9.3 Exercise: Complete the following table. 2.62 4.17x10 -12 1.76 5.75x10 -13 2.78x10 -12 11.6 5.01x10 -10 2.00x10 -5 4.7 12.24 2.4 11.38

32. Complete the pH problems on page 166 lab manual Calculate the pH of the following [H 3 O + ]: a) 0.1 molL -1 b) 0.001molL -1 c) 1 x 10 -7 molL -1 d)3 x 10 -7 molL -1 e) 5 x 10 -4 molL -1 pH = -log [H 3 O + ] = - log 0.1 = 1 pH = -log [H 3 O + ] = - log 1x10 -3 = 3 pH = -log [H 3 O + ] = - log 1x10 -7 = 7 pH = -log [H 3 O + ] = - log 3x10 -7 = 6.5 pH = -log[H 3 O + ] = -log 5 x10 -4 = 3.3

33. 2. Using the relationship: K w =[H 3 O + ][OH-] = 10 -14, calculate the pH of solutions with the following [OH-] a) 0.1 molL -1 b) 0.001molL -1 c) 1 x 10 -7 molL -1 d)3 x 10 -7 molL -1 e) 5 x 10 -4 molL -1

34. Strong Acids These are Acids that dissociate (ie react fully) in H 2 O to form a lot of H 3 O+ ions in solution to form solutions with very low pH ie 1- 2. Examples HCl, HNO 3 and H 2 SO 4 HCl reacting with water is written as HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl  (aq) How could you tell by just looking at the equation that this is a strong acid?

35. Weak Acids These are Acids that don’t dissociate (react) fully in H 2 O and produce a low number of H 3 O+ ions in solution to give higher pH values than strong acids eg pH 3 - 5. Examples CH 3 COOH (ethanoic acid) HCOOH (methanoic acid) and NH 4 + (ammonium ion from a NH 4 Cl salt) CH 3 COOH reacting with water is written as CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO - (aq) How could you tell by just looking at the equation that this is a weak acid?

36. Demo of Weak and strong acids Why the difference in conductivity of the two acids? Why the difference in conductivity of the two acids? Turn to page 158 in Lab book: The Relative Strength of Acids Complete the experiment recording your observations in the table on page 159 and then complete the discussion questions on page 159

37. Discussion Questions pg 159 3. Explain the different rates of reaction obtained when solutions of HCl and CH 3 COOH of the same concentration, reacted with CaCO 3 Rates of these reactions depend on [[H 3 O + ] ions in solution. The stronger acid HCl has completely dissociated and contains a higher initial [H 3 O + ] than the weak acid CH 3 COOH

38. Discussion Questions pg 159 This next question involves a very important concept you need to understand It may pay to remember it

39. 4. 10 mL of 1 molL -1 HCl and 10 mL of 1 molL -1 CH 3 COOH are mixed separately with NaOH solution. Which acid will require the greater amount of NaOH for complete reaction? Explain.

40. 4. Both acids will need the amount of NaOH. In the case of CH 3 COOH adding OH - ions will move the equilibrium to the RHS CH 3 COOH + H 2 O H 3 O + + CH 3 COO - This is because H 3 O + ions are removed by the addition of OH- ions and eventually, CH 3 COOH will give up all its potential H+ ions. SAME Removed by the addition of OH- ions

41. further explanation 10 mL of 1 molL -1 HCl and 10 mL of 1 molL -1 CH 3 COOH are mixed separately with NaOH solution. Both acids will produce the same amount of H 3 O + ions in solution with the NaOH The (strong acid) HCl produces the H 3 O + ions almost immediately and reacts quickly with the OH- ions. The (weak acid) CH 3 COOH will produce the same number of H 3 O + ions at a slower rate taking more time to react with the OH- ions.

42. 5. How does a strong acid differ from a weak acid chemically? Strong acids react faster than weak acids because initially there is a greater [H 3 O + ] available for reaction.

43. Example calculation Find the pOH of a 0.012molL-1 NaOH solution For any strong base in water the OH- concentration = the concentration of the base So pOH = -log [OH-] pOH = -log [0.012] pOH = 1.92 We can also find the pH using the formula pH + pOH = 14 ie 12.08 + 1.92 = 14 therefore the pOH of the solution is 13