Topic 18- Acids and bases

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+]=[H3O+]

4. Calculate the pH in following solutions: • 1M HCl pH = -log[1] = 0 • 0.001 M HNO3 pH = -log [0.001] = 3 • 0.5 M H2SO4  2H+ pH = -log [2*0.5]= 0 • 0.15 M NaOH pOH =-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) = 101 = 10 M

6. Autoprotolysis of water H2O + H2O H3O+ + OH- Kc= [H3O+] .[OH-] [H2O] . [H2O] The concentrationofwater is not changing- it is constant Kw= K . [H2O] . [H2O] = [H3O+] .[OH-] Kw= the dissociation constant of water

7. Kw = dissociation constant of water H2O + H2O H3O+ + OH- In 25oC pure water: [H3O+] = 10-7mol/dm3 [OH-] = 10-7 mol /dm3 Kw= [H+]*[OH-] = 10-7*10-7 = 10-14mol2/dm6 -lg Kw= -lg ([H+]*[OH-]) = -lg [H+]+ -lg[OH-]=-lg10-14 pKw= pH + pOH = 14

8. The effectoftemperature on the dissociation constant Kw= 1,0 . 10 -14 vid 25 C H2O + H2O H3O+ + OH- DH> 0

9. Ka=K . [H2O] = [H3O+] . [A- ] [HA] Weak acids HA + H2O H3O+ + A- K= [H3O+] . [A- ] [HA] . [H2O] Ka = acid dissociation constant Ka=> acid strength => higher value => stronger acid. pKa = -log Ka => lower value => stronger acid

10. Someacid dissociation constants

11. Kb=K . [H2O] = [BH2+] . [OH-] [BH] Weak bases BH + H2O BH2+ + OH- K= [BH2+] . [OH-] [BH] . [H2O] Kb= base dissociation constant Kb=> base strength => higher value => stronger base pKb= -logKb => lower value => stronger base

12. Bases

13. Kwconnects Kaand Kbfor an acid/ base pair Ka* Kb= Kw = 10-14 pKa + pKb = pKw = 14 (at 25 ºC)

14. Calculate pKa of ethanoicacid, HAcWeknowthat c= 0.01M Wemeasure pH HAc+H2O H3O+ +Ac- Cstart 0.1 0 0 Ceq0.1- 10-pH10-pH 10-pH Ka = [H3O+]*[Ac-] / [HAc] = 10-pH* 10-pH/0.1- 10-pH= pKa= -log Ka=

15. Calculate pH in 0,1 M ethanoicacid, HAc HAc H3O+ +Ac- Cstart 0.1 0 0 pKa = 4.75 Ceq 0.1-X X X (see CDB) Ka = [H3O+]*[Ac-] / [HAc] = 10-4.75 = X2/0.1-X ~ X2/0.1 if x is small X2 = 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. CH3COOH and CH3COO-Na+. • Mix a weak base and its conjugate acid e.g. NH3 and NH4Cl. • 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+] = Ka* [HA] /[A-] pH = pKa -log([HA] /[A-]) Youhaveto be abletoderive it yourself!

20. Exercises 1. Which combination will form a buffer solution? A. 100 cm3 of 0.10 moldm–3 hydrochloric acid with 50 cm3 of 0.10 moldm–3 sodium hydroxide. B. 100 cm3 of 0.10 moldm–3ethanoic acid with 50 cm3 of 0.10 moldm–3 sodium hydroxide. C. 50 cm3 of 0.10 moldm–3 hydrochloric acid with 100 cm3 of 0.10 moldm–3 sodium hydroxide. D. 50 cm3 of 0.10 moldm–3ethanoic acid with 100 cm3 of 0.10 moldm–3 sodium hydroxide. 2. Buffer solutions resist small changes in pH. A phosphate buffer can be made by dissolving NaH2PO4 and Na2HPO4 in water, in which NaH2PO4 produces the acidic ion and Na2HPO4 produces the conjugate base ion. 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 cm3 of 0.10 moldm–3ethanoic acid with 50 cm3 of 0.10 moldm–3 sodium hydroxide. You have both ethanoic acid and sodium ethanoate 2. (i) Acid: H2PO4–; (Conjugate) base: HPO42–;H2PO4–(aq) H+(aq) + HPO42–(aq); strong base/OH– replaced by weak base (H2PO42–, and effect minimized) /strong base reacts with acid of buffer / equilibrium in (i) shifts in forwarddirection; OH–(aq) + H2PO4–(aq) → H2O(l) + HPO42–(aq); (iii) strong acid/H+ replaced by weak acid (H2PO4–, and effect minimized) /strong acid reacts with base of buffer / equilibrium in (i) shifts inreverse direction;H+(aq) + HPO42–(aq) → H2PO4–(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. • Cationscan act as acids and anions act as bases.

23. Ac- + H2O HAc + OH- pKb(Ac-)= base acid The acetate ion HAc + H2O H3O+ + Ac- pKa(HAc)= 4.75 acidbase The acetate ion is salt of a WEAK acid (acetic acid) and thus basic

24. Morebasicions • Ac- PO43- • CN- HCO3- • CO32- • Salts ofweakacids

25. Cl- + H2O HCl + OH- pKb(Cl-)= base acid The chloride ion HCl + H2O H3O+ + Cl- pKa(HCl)= - 4 acidbase The chloride ion is salt of a STRONGacid (hydrochloric acid) and thus so week so it is neutral (<pKw)

26. Moreionswith no acid/basecharacter • Na+ SO42- • K+ ClO4- • Ca2+ • NO3- Cl- • Derives from strong acids and bases => no acid-base activity

27. NH4+ + H2O H3O+ + NH3pKa(NH4+)= acidbase The ammonium ion NH3 + H2O NH4+ + OH- pKb(NH3)= 4.74 base acid 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. Al3+ and Fe3+, form complexes with water: • Al(H2O)63+ and Fe(H2O)63+. The electronegative effect of the ion weakens the O-H bond in water molecules: [Fe(H2O)6]3+(aq) +H2O [Fe(OH)(H2O)5]3+(aq)+ H3O+(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 base0.1M, 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- H2O + 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 => pKa

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 pKa-value. => Different indicators change their colours at different pH

33. Indicators change colour around its pKa-value

34. Howtochooseindicator? • If you titrate CH3COOH with NaOH the pH will be above 7 at the equivalence point => choose an indicator that change colour above 7 e.g. phenolphthalein (pKa =9.6), range 8.3 – 10.0. Rapid pH changes in that area. - If you titrate NH3 with HCl the pH will be under 7 at the equivalence point => choose an indicator that change colour under 7 e.g. methyl orange(pKa = 3.7), range 3.1 – 4.4. Rapid pH changes in that area.