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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.

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Presentation on theme: "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."— Presentation transcript:

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

3 The relation between oxonium ions and hydroxide ions

4 18.1 Calculations involving acids and bases
pH = -log[H3O+] pOH = -log[OH-]] [H3O+] = 10-pH [OH-] = 10-pOH pH + pOH = 14 [H+]=[H3O+]

5 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 = = 13.18

6 Calculate the [H3O+] in following solutions
pH = 5,5 [H3O+] = = 3.2*10-6 pH =- 1 [H3O+] =10-(-1) = 101 = 10 M

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

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

9 The effect of temperature on the water dissociation constant
Temperature (◦C) Kw pKw 0, 14,96 15 0, 14,35 25 1, 14,00 50 5, 13,26 100 12,29 Kw= 1, vid 25 C H2O + H2O H3O+ + OH- DH> 0

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

11 Some acid dissociation constants
Ka pKa Hydrochloric acid HCl 104 -4 Sulphuric acid H2SO4 103 -3 Nitric acid HNO3 1 -1 Oxalic acid H2C2O4 0.06 1.25 Phosphoric acid H3PO4 0.007 2.15 Salicylic acid C7H6O3 0.001 2.97 Citric acid C6H8O7 7.10-4 3.13 Ascorbic acid C6H4O2(OH)4 1.10-4 4.17 Acetic (etanoic) acid HAc CH3COOH 4.75 Carbonic acid H2CO3 6.37 Ammonium ion NH4+ 9.26

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

13 Bases Kb pKb Litium hydroxide LiOH 2.3 -0.36 Sodium hydroxide NaOH
0.63 0.2 Ammonia NH3 4.74 Hydrogen carbonate ion HCO3- 7.62 Acetate ion CH3COO- 9.25 Nitrate ion NO3- 10-15 15 Chloride ion Cl- 10-18 18

14 Kw connects Ka and Kb for a corresponding acid/ base pair, such as CH3COOH/CH3COO-
Ka * Kb = Kw = 10-14 pKa + pKb = pKw = 14 (at 25 ºC)

15 Calculate pKa of ethanoic acid, HAc (CH3COOH) We know that c= 0
Calculate pKa of ethanoic acid, HAc (CH3COOH) We know that c= 0.01M We measure pH HAc+H2O H3O+ +Ac- Cstart Ceq pH pH 10-pH Ka = [H3O+]*[Ac-] / [HAc] = 10-pH* 10-pH / pH= pKa= -log Ka =

16 Calculate pH in 0,1 M ethanoic acid, HAc
HAc H3O+ +Ac- Cstart pKa = 4.75 Ceq X X X (see CDB) Ka = [H3O+]*[Ac-] / [HAc] = = X2/0.1-X ~ X2/ if x is small X2 = 0.1* X = (0.1* )½ pH = -log [X] = -log[(0.1* )½] = 2.88

17 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

18 A Buffer: a mixture of a weak acid and its corresponding base
The equilibrium: HA(aq) +H2O H3O+(aq) + A-(aq) If strong acid is added (fully dissociated, contains mainly H3O+) => reaction goes to the left (Le Chatelier’s principle) => little change in pH (-log [H3O+])

19 A Buffer: a mixture of a weak acid and its corresponding base
The equilibrium: HA(aq) +H2O H3O+(aq) + A-(aq) If a strong base is added => OH- reacts with H3O+ to form water => equilibrium goes to the right => Restore the [H3O+] => little change in pH

20 How to prepare a buffer:
Mix a weak acid and its corresponding (conjugate) base e.g. CH3COOH and CH3COO-Na+. Mix a weak base and its corresponding (conjugate) e.g. NH3 and NH4Cl. Add strong base to an excess of weak acid. Add strong acid to an excess of weak base.

21 Add strong base to an excess of weak acid (HA)
The reaction of HA and water: HA(aq) +H2O H3O+(aq) + A-(aq) The reaction of HA and strong base (NaOH): HA + NaOH → H2O + Na+A- When you add some NaOH to a HA-solution your mixture will consist of all the above particles, but in particular the weak acid and it’s corresponding base HA/A-

22 Add strong acid to an excess of weak base (BH)
The reaction of BH and water: The reaction of B and strong acid (HCl): HCl + BH → BH2+Cl- When you add some HCl to a BH-solution your mixture will consist of all the above particles, but in particular the weak base and it’s corresponding acid BH/BH2+ BH + H2O BH2+ + OH-

23 The Hydrogen ion concentration and pH of a buffer can be calculated from the expression for the acid dissociation constant: [H3O+] = Ka* [HA] /[A-] take –log on both sides -log [H3O+] = -log Ka+ -log [HA] /[A-] identify pH = pKa -log([HA] /[A-]) You have to be able to derive the equation yourself!

24 1. Which combination will form a buffer solution?
Exercises 1. Which combination will form a buffer solution? A. 100 cm3 of 0.10 mol dm–3 hydrochloric acid with 50 cm3 of 0.10 mol dm–3 sodium hydroxide. B. 100 cm3 of 0.10 mol dm–3 ethanoic acid with 50 cm3 of 0.10 mol dm–3 sodium hydroxide. C. 50 cm3 of 0.10 mol dm–3 hydrochloric acid with 100 cm3 of 0.10 mol dm–3 sodium hydroxide. D. 50 cm3 of 0.10 mol dm–3 ethanoic acid with 100 cm3 of 0.10 mol dm–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.

25 1. Which combination will form a buffer solution?
Answers 1. Which combination will form a buffer solution? B. 100 cm3 of 0.10 mol dm–3 ethanoic acid with 50 cm3 of 0.10 mol dm–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 forward direction; 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 in reverse direction; H+(aq) + HPO42–(aq) → H2PO4–(aq);

26 18.3 Salt hydrolysis The positive and negative ion in a salt can be neutral or act as an acid or a base Cations (positive ions) can act as acids and anions (negative ions) can act as bases

27 The acetate (ethanoate) ion
HAc + H2O H3O+ + Ac pKa(HAc)= 4.75 acid base Ac- + H2O HAc + OH pKb(Ac-)= 9.25 base acid The ethanoate ion is salt of a WEAK acid (ethanoic acid) and thus basic

28 More basic ions CN- HCO3- CO32- PO43- Salts of weak acids

29 The chloride ion and thus so week so it is neutral (>pKw)
HCl + H2O H3O+ + Cl pKa(HCl)= - 4 acid base Cl- + H2O HCl + OH pKb(Cl-)= 18 base acid The chloride ion is salt of a STRONG acid (hydrochloric acid) and thus so week so it is neutral (>pKw)

30 More ions with no acid/base character
Na SO42- K ClO4- Ca2+ NO Cl- Derives from strong acids and bases => no acid-base activity

31 The ammonium ion Ammonium ion is salt of a WEAK base (ammonia)
NH3 + H2O NH4+ + OH pKb(NH3)= 4.74 base acid NH4+ + H2O H3O+ + NH pKa(NH4+)= acid base Ammonium ion is salt of a WEAK base (ammonia) and thus acidic

32 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

33 18.4 Acid-base titrations Strong acid – Strong base
Weak acid – Strong base Strong acid – Weak base

34 Strong acid – Strong base
HCl + NaOH NaCl + H2O Start with 10 ml 0.1 M HCl (pH=1). Titrate with 0.1M NaOH 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-] = pH =11

35 Weak acid – strong base CH3COOH + NaOH CH3COONa + H2O
When strong base is added 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

36 Titration simulations at

37 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

38 Structures of BTB at different pH
Increasing [OH-]/pH

39 Indicators change colour around their pKa-values

40 How to choose indicator?
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 – 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.

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