1. Topic 18- Acids and bases 18.1 Calculations involving acids and bases 18.2 Buffer solutions 18.3 Salt hydrolysis 18.4 Acid-base titrations 18.5 Indicators

2. 18.1 Calculations involving acids and bases pH = -log[H + ] pOH = -log[OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH pH + pOH = 14 [H + ]=[H 3 O + ]

4. Calculate the pH in following solutions: 1M HClpH = -log[1] = 0 0.001 M HNO 3 pH = -log [0.001] = 3 0.5 M H 2 SO 4  2H + pH = -log [2*0.5]= 0 0.15 M NaOHpOH =-log[0.15]= 0.82 pH = 14-0.82 = 13.18

5. Calculate the [H + ] in following solutions pH = 5,5[H + ] = 10 -5.5 = 3.2*10 -6 pH =- 1[H + ] =10 -(-1) = 10 1 = 10 M

6. Autoprotolysis of water H 2 O + H 2 O H 3 O + + OH - K c = [H 3 O + ]. [OH - ] [H 2 O]. [H 2 O] The concentration of water is not changing- it is constant K w = K. [H 2 O]. [H 2 O] = [H 3 O + ]. [OH - ] K w = the dissociation constant of water

7. K w = dissociation constant of water H 2 O + H 2 O H 3 O + + OH - In 25 o C pure water: [H 3 O + ] = 10 -7 mol/dm 3 [OH - ] = 10 -7 mol /dm 3 K w = [H + ]*[OH - ] = 10 -7* 10 -7 = 10 -14 mol 2 /dm 6 -lg K w = -lg ([H + ]*[OH - ]) = -lg [H + ]+ -lg[OH - ]=-lg 10 -14 pK w = pH + pOH = 14

8. K w = 1,0. 10 -14 vid 25 C The effect of temperature on the dissociation constant Temperature ( ◦ C) KwKw pK w 00,11. 10 -14 14,96 150,45. 10 -14 14,35 251,0. 10 -14 14,00 505,5. 10 -14 13,26 10051. 10 -14 12,29 H 2 O + H 2 O H 3 O + + OH -  H> 0

9. HA + H 2 O H 3 O + + A - K= [H 3 O + ]. [A - ] [HA]. [H 2 O] K a =K. [H 2 O] = [H 3 O + ]. [A - ] [HA] Weak acids K a = acid dissociation constant K a => acid strength => higher value => stronger acid. pK a = -log K a => lower value => stronger acid

10. KaKa pK a Hydrochloric acidHCl10 4 -4 Sulphuric acidH 2 SO 4 10 3 -3 Nitric acidHNO 3 1 Oxalic acidH2C2O4H2C2O4 0.061.25 Phosphoric acidH 3 PO 4 0.0072.15 Salicylic acidC7H6O3C7H6O3 0.0012.97 Citric acidC6H8O7C6H8O7 7. 10 -4 3.13 Ascorbic acid C 6 H 4 O 2 (OH) 4 1. 10 -4 4.17 Acetic acid HAcCH 3 COOH1.8. 10 -5 4.75 Carbonic acidH 2 CO 3 4.2. 10 -7 6.37 Ammonium ionNH 4 + 5.5. 10 -10 9.26 Some acid dissociation constants

11. BH + H 2 O BH 2 + + OH - K= [BH 2 + ]. [OH - ] [BH]. [H 2 O] K b =K. [H 2 O] = [BH 2 + ]. [OH - ] [BH] Weak bases K b = base dissociation constant K b => base strength => higher value => stronger base pK b = -logK b => lower value => stronger base

12. KbKb pK b Litium hydroxideLiOH2.3-0.36 Sodium hydroxideNaOH0.630.2 AmmoniaNH 3 1.8. 10 -5 4.74 Hydrogen carbonate ion HCO 3 - 2.4. 10 -8 7.62 Acetate ionCH 3 COO - 5.5. 10 -10 9.25 Nitrate ionNO 3 - 10 -15 15 Chloride ionCl - 10 -18 18 Bases

13. K w connects K a and K b for an acid/ base pair K a * K b = K w = 10 -14 pK a + pK b = pK w = 14 (at 25 ºC)

14. Calculate pKa of ethanoic acid, HAc We know that c= 0.01M We measure pH HAc+H 2 O H 3 O + +Ac - C start 0.1 0 0 C eq 0.1- 10 -pH 10 -pH 10 -pH K a = [H 3 O + ]*[Ac - ] / [HAc] = 10 -pH * 10 -pH /0.1- 10 -pH= pKa= -log K a =

15. Calculate pH in 0,1 M ethanoic acid, HAc HAc H 3 O + +Ac - C start 0.1 0 0pKa = 4.75 C eq 0.1-X X X(see CDB) K a = [H 3 O + ]*[Ac - ] / [HAc] = 10 -4.75 = X 2 /0.1-X ~ X 2 /0.1 if x is small X 2 = 0.1* 10 -4.75 X = (0.1* 10 -4.75 ) ½ pH = -log [X] = -log[(0.1* 10 -4.75 ) ½ ] = 2.88

16. 18.2 Buffer solutions In pure water pH= 7. Addition of small amounts of acid or base gives big changes in pH That can be great problem, especially in biological systems. But there are ways to make a solution that can be quite pH stable. A buffer resist changes in pH when a strong acid or base is added

17. A Buffer: a mixture of a weak acid and its conjugate base The equilibrium: HA (aq) H + (aq) + A - (aq) If strong acid is added => reaction goes to the left (Le Chatelier’s principle) => little Change in pH. “The strong acid is transformed to a weak acid. If a strong base is added => OH - reacts with H + => reaction goes to the right => Restore the [H + ] => little Change in pH.

18. How to prepare a buffer: Mix a weak acid and its conjugate base e.g. CH 3 COOH and CH 3 COO - Na +. Mix a weak base and its conjugate acid e.g. NH 3 and NH 4 Cl. Add strong base to an excess of weak acid. Add strong acid to an excess of weak base.

19. The Hydrogen ion concentration and pH can be calculated with the acid dissociation constant: [H + ] = K a * [HA] /[A - ] pH = pKa -log([HA] /[A - ]) You have to be able to derive it yourself!

20. Exercises 1. Which combination will form a buffer solution? A. 100 cm 3 of 0.10 mol dm –3 hydrochloric acid with 50 cm 3 of 0.10 mol dm –3 sodium hydroxide. B. 100 cm 3 of 0.10 mol dm –3 ethanoic acid with 50 cm 3 of 0.10 mol dm –3 sodium hydroxide. C. 50 cm 3 of 0.10 mol dm –3 hydrochloric acid with 100 cm 3 of 0.10 mol dm –3 sodium hydroxide. D. 50 cm 3 of 0.10 mol dm –3 ethanoic acid with 100 cm 3 of 0.10 mol dm –3 sodium hydroxide. 2. Buffer solutions resist small changes in pH. A phosphate buffer can be made by dissolving NaH 2 PO 4 and Na 2 HPO 4 in water, in which NaH 2 PO 4 produces the acidic ion and Na 2 HPO 4 produces the conjugate base ion. (i)Deduce the acid and conjugate base ions that make up the phosphate buffer and state the ionic equation that represents the phosphate buffer. (ii) Describe how the phosphate buffer minimizes the effect of the addition of a strong base, OH – (aq), to the buffer. Illustrate your answer with an ionic equation. (iii) Describe how the phosphate buffer minimizes the effect of the addition of a strong acid, H + (aq), to the buffer. Illustrate your answer with an ionic equation.

21. Answers 1. Which combination will form a buffer solution? B. 100 cm 3 of 0.10 mol dm –3 ethanoic acid with 50 cm 3 of 0.10 mol dm –3 sodium hydroxide. You have both ethanoic acid and sodium ethanoate 2. (i) Acid: H 2 PO 4 – ; (Conjugate) base: HPO 4 2– ; H 2 PO 4 – (aq) H + (aq) + HPO 4 2– (aq); (ii)strong base/OH – replaced by weak base (H 2 PO 4 2–, and effect minimized) / strong base reacts with acid of buffer / equilibrium in (i) shifts in forward direction; OH – (aq) + H 2 PO 4 – (aq) → H 2 O(l) + HPO 4 2– (aq); (iii) strong acid/H + replaced by weak acid (H 2 PO 4 –, and effect minimized) / strong acid reacts with base of buffer / equilibrium in (i) shifts in reverse direction; H + (aq) + HPO 4 2– (aq) → H 2 PO 4 – (aq);

22. 18.3 Salt hydrolysis Some salts doesn’t change the pH when added to a water solution, but some other ions in salts can act as acids or bases. Cations can act as acids and anions act as bases.

23. The acetate ion HAc + H 2 O H 3 O + + Ac - pK a (HAc)= 4.75 acid base Ac - + H 2 O HAc + OH - pK b (Ac - )= base acid The acetate ion is salt of a WEAK acid (acetic acid) and thus basic

24. More basic ions Ac - PO 4 3- CN - HCO 3 - CO 3 2- Salts of weak acids

25. The chloride ion HCl + H 2 O H 3 O + + Cl - pK a (HCl)= - 4 acid base Cl - + H 2 O HCl + OH - pK b (Cl - )= base acid The chloride ion is salt of a STRONG acid (hydrochloric acid) and thus so week so it is neutral (<pK w )

26. More ions with no acid/base character Na + SO 4 2- K + ClO 4 - Ca 2+ NO 3 - Cl - Derives from strong acids and bases => no acid-base activity

27. NH 3 + H 2 O NH 4 + + OH - pK b (NH 3 )= 4.74 base acid The ammonium ion NH 4 + + H 2 O H 3 O + + NH 3 pK a (NH 4 + )= acid base Ammonium ion is salt of a WEAK base (ammonia) and thus acidic

28. Metallic ions with high charge Metallic ions with high charge, e.g. Al 3+ and Fe 3+, form complexes with water: Al(H 2 O) 6 3+ and Fe(H 2 O) 6 3+. The electronegative effect of the ion weakens the O-H bond in water molecules: [Fe(H 2 O) 6 ] 3+ (aq) + H 2 O [Fe(OH)(H 2 O) 5 ] 3+ (aq) + H 3 O + (aq) An acidic solution

29. 18.4 Acid-base titrations Strong acid – Strong base Weak acid – Strong base Strong acid – Weak base

30. Strong acid – Strong base 0.1M, 10 ml 0.1M 0.1 M HCl => pH = 1 When 90% of the base been added: HCl ~0.01 => pH = 2 When 99% of the base been added: HCl ~0.001 => pH = 3 When 101% of the base been added: [OH - ] = 0.001 pH =11

31. Weak acid – strong base When strong base is added: HA +OH -  H 2 O + A - the pH gradually increase. At equivalence point all acid is consumed, pH increase rapidly. The salt of the weak acid is a weak base => pH > 7 at equivalence point. At ½ equivalence point [HA] =[A - ] => pH => pK a

32. 18.5 Indicators A weak acid/base where the colours of the protonated and ionized forms are different HIn  H + + In - Red Blue The colour depends both on pH and the pK a - value. => Different indicators change their colours at different pH

33. Indicators change colour around its pKa-value

34. How to choose indicator? If you titrate CH 3 COOH with NaOH the pH will be above 7 at the equivalence point => choose an indicator that change colour above 7 e.g. phenolphthalein (pK a =9.6), range 8.3 – 10.0. Rapid pH changes in that area. - If you titrate NH 3 with HCl the pH will be under 7 at the equivalence point => choose an indicator that change colour under 7 e.g. methyl orange(pK a = 3.7), range 3.1 – 4.4. Rapid pH changes in that area.