Acids and Bases shodor/unchem/basic/ab/ chemtutor/acid.htm

1. Acids and Baseshttp://www.shodor.org/unchem/basic/ab/ http://www.chemtutor.com/acid.htm

2. Strong Acids Strong acids completely dissociate in water, HCl(aq) =>H+1 (aq) + Cl-1 (aq) 1moldm-3 1moldm-3 1moldm-3 HCl - hydrochloric acid HNO3 - nitric acid H2SO4 - sulfuric acid HBr - hydrobromic acid HI - hydroiodic acid HClO4 - perchloric acid Weak Acids (Carboxylic Acids) A weak acid only partially dissociates in water CH3COOH(aq) =>H+1 (aq) + CH3COO- (aq) 1moldm-3 x moldm-3 x moldm-3 Examples of weak acids include hydrofluoric acid, HF, and acetic acid, CH3COOH.

3. Strong Bases • The hydroxides of the Group I and Group II metals usually are considered to be strong bases. • Ionize completely • NaOH(aq) => Na+(aq) + OH- (aq) • LiOH - lithium hydroxide • NaOH - sodium hydroxide • KOH - potassium hydroxide • RbOH - rubidium hydroxide • Ca(OH)2 - calcium hydroxide • Ba(OH)2 - barium hydroxide • Weak Bases( AMINES) • Examples of weak bases include ammonia, NH3, • Ionize partially NH3 + H2 O=> NH4+1 + OH- • Methylamine and diethylamine, (CH3CH2)2NH.

4. Which is not a strong acid? A. Nitric acid B. Sulfuric acid C. Carbonic acid D. Hydrochloric acid

5. Properties of Acids (page 145) 1.Produce H+ (as H3O+) ions in water HCl (aq) => H+ (aq) + Cl+ (aq) 2.Taste sour 3.Corrode metals: Zn(s) + HCl(aq)=> ZnCl2 (aq) + H2 (g) 4.Electrolytes 5.React with bases to form a salt and water: HCl + NaOH => NaCl + H2 O 6.pH is less than 7 7.Turns blue litmus paper to red 8. React with carbonates and bicarbonates to produce carbon dioxide. CaCO3 (s) + 2HCl(aq) => CaCl2 (aq) + H2 O(l) + CO2 (g)

6. Properties of Bases (page 146) • Generally produce OH- ions in water: • NaOH => Na+ + OH- • Taste bitter, chalky,soapy,slippery • Are electrolytes • Displacement of ammonia from ammonium salts • React with acids to form salts and water • pH greater than 7 • Turns red litmus paper to blue

7. Which substance, when dissolved in water, to give a 0.1 mol dm− solution, has the highest pH? • A. HCl • B. NaCl • C. NH3 • D. NaOH

8. Definitions • Arhenius • Bronsted Lowry • Lewis

9. Arrhenius Definition Arrhenius Acid - Substances in water that increase the concentration of hydrogen ions (H+). HCl(g) => H+1 (aq) + Cl-1 (aq) Base - Substances in water that increase concentration of hydroxide ions (OH-). NaOH(s) => Na+1 (aq) + OH-1 (aq)

10. Bronsted-Lowry Definition Acids are species that donate a proton (H+). HNO3 (aq) + H2O(l) => NO3-(aq) + H3O+(aq) NO3- is called the conjugate base of the acid HNO3, and H3O+ is the conjugate acid of the base H2O

11. Bases are species that accept a proton. NH3 (aq) + H2O(l) => NH4+(aq) + OH-(aq) • NH3 is a base and H2O is acting as an acid. NH4+ is the conjugate acid of the base NH3, and OH- is the conjugate base of the acid H2O. • A compound that can act as either an acid or a base, such as the H2O in the above examples, is called amphoteric

12. Conjugate Acid Base Pairs • Conjugate Base - The species remaining after an acid has transferred its proton. • Conjugate Acid - The species produced after base has accepted a proton.

13. Bronsted-Lowry Acid Base Systems • Acid Base Conjugate AcidConjugate Base • HCl + H2O  H3O+ + Cl- • H2PO4- + H2O  H3O+ +HPO42- • NH4+ + H2O  H3O+ + NH3 • Base Acid Conjugate AcidConjugate Base :NH3 + H2O  NH4++ OH- • PO43- + H2O  HPO42- + OH-

14. Lewis Definition • A Lewis acid is defined to be any species that accepts lone pair electrons. • A Lewis base is any species that donates lone pair electrons. • H+ + : OH-1 => H2O

15. Auto Ionization of Water http://www.wwnorton.com/college/chemistry/gilbert2/tutorials/interface.asp?chapter=chapter_16&folder=self_ionization • self-ionization of water • H2O (l) => H+ (aq) + OH− (aq) or • 2 H2O (l) => H3O+ (aq) + OH− (aq) hydronium ion • It is an example of autoprotolysis, and relies on the amphoteric nature of water.

16. Kw The autoionization of water in equilibrium: H2O(l)  H+(aq) + OH-(aq) ΔH> 0 The equilibrium constant expression for this reaction is given by: Keq = [H+] [OH-]/[H2 O] Kc x [H2O]= Kw = 1.0 x 10-14 ion product for water The value for Kw is for room temperature, 25 °C, and 1.0 atm The equilibrium constant expression applies not only to pure (distilled) water but to any aqueous solution. It can be used to calculate either [H+] or [OH-] provided one of them is known.

17. Kw and T: http://mmsphyschem.com/autoIon.htm H2O(l)  H+(aq) + OH-(aq) ΔH> 0 T Kw 0 1.14 x 10-15 5 1.85 x 10-15 10 2.92 x 10-15 15 4.53 x 10-15 20 6.81 x 10-15 25 1.01 x 10-14 • The ionization of water is endo so, • as the temperature increases,so does the Kw

18. Formulas pH = -log [H] pKa = -log Ka pOH = -log [OH] pKb = -log Kb pH + pOH = 14 pKb + pKa = 14 [H] [OH] = 1 x 10 -14 = Ka x Kb = Kw Kw = 1.0 x 10-14 pKw = 14 pKw = pH + pOH

19. pH Formulas

21. Lime(calcium hydroxide) is added to a lake to neutralize the effects of acid rain. The pH value of the lake water rises from 4 to 7. What is the change in concentration of H+ in the lake water? • A. An increase by a factor of 3 • B. An increase by a factor of 1000 • C. A decrease by a factor of 3 • D. A decrease by a factor of 1000

22. The pH Scale

23. Calculating the pH pH = - log [H3O+] Example 1: If [H3O+] = 1 X 10-10pH = - log 1 X 10-10 pH = - (- 10) pH = 10 Example 2: If [H3O+] = 1.8 X 10-5pH = - log 1.8 X 10-5 pH = - (- 4.74) pH = 4.74

24. pH and acidity The pH values of several common substances are shown at the right. Many common foods are weak acids Some medicines and many household cleaners are bases.

25. Indicators: Substances that change color when the concentration of hydrogen changes.

26. Neutralization An acid will neutralize a base, giving a salt and water as products Examples: HCl + NaOH  NaCl + H2O H2SO4 + 2 NaOH  Na2SO4 + 2 H2O H3PO4 + 3 KOH  K3PO4 + 3 H2O 2 HCl + Ca(OH) 2 CaCl2 + 2 H2O

27. Neutralization Problems # moles acid = # moles base If an acid and a base combine in a 1 to 1 ratio, then the volume of the acid multiplied by the concentration of the acid is equal to the volume of the base multiplied by the concentration of the base Vacid M acid = V base M base

28. Neutralization Problems Example 1: HCl + KOH  KCl + H2O If 15.00 cm3 of 0.500 M HCl exactly neutralizes 24.00 cm3 of KOH solution, what is the concentration of the KOH solution?

29. Neutralization Problems Whenever an acid and a base do not combine in a 1 to 1 ratio, a mole factor must be added to the neutralization equation n Vacid C acid = V base C base The mole factor (n) is the number of times the moles the acid side of the above equation must be multiplied so as to equal the base side. (or vice versa) Example H2SO4 + 2 NaOH  Na2SO4 + 2 H2O moles base = 2 x moles acid

30. Neutralization Problems Example 3: H3PO4 + 3 KOH  K3PO4 + 3 H2O If 30.00 cm3 of 0.300 M KOH exactly neutralizes 15.00 cm3 of H3PO4 solution, what is the concentration of the H3PO4 solution? Solution:

31. Neutralization Problems Example 2:Sulfuric acid reacts with sodium hydroxide according to the following reaction: H2SO4 + 2 NaOH  Na2SO4 + 2 H2O If 20.00 cm3 of 0.400 M H2SO4 exactly neutralizes 32.00 cm3 of NaOH solution, what is the concentration of the NaOH solution? Solution: In this case the mole factor is 2 and it goes on the acid side, since the mole ratio of acid to base is 1 to 2. Therefore 2 Vacid Cacid = Vbase Cbase 2 (20.00 cm3 )(0.400 M) = (32.00 cm3 ) Cbase Cbase = (2) (20.00 cm3 )(0.400 M) (32.00 cm3 ) Cbase = 0.500 M

32. Neutralization Problems Example 4: 2 HCl + Ca(OH)2 CaCl2 + 2 H2O If 25.00 cm3 of 0.400 M HCl exactly neutralizes 20.00 cm3 of Ca(OH)2 solution, what is the concentration of the Ca(OH)2 solution?

33. YEAR 2

34. Acid Base Dissociation • Acid-base reactions are equilibrium processes. • The relationship between the relative concentrations of the reactants and products is a constant for a given temperature. It is known as the Acid or Base Dissociation Constant: Ka and Kb • The stronger the acid or base, the larger the value of the dissociation constant.

36. Acid Strength • Strong Acid - Transfers all of its protons to water; - Completely ionized; - Strong electrolyte; - The conjugate base is weaker and has a negligible tendency to be protonated. • Weak Acid - Transfers only a fraction of its protons to water; • - Partly ionized; - Weak electrolyte; - The conjugate base is stronger, readily accepting protons from water • As acid strength decreases, base strength increases. • The stronger the acid, the weaker its conjugate base • The weaker the acid, the stronger its conjugate base

37. Acid Dissociation Constants Dissociation constants for some weak acids

38. Base Strength • Strong Base - all molecules accept a proton; - completely ionizes; - strong electrolyte; - conjugate acid is very weak, negligible tendency to donate protons. • Weak Base - fraction of molecules accept proton; - partly ionized; - weak electrolyte; - the conjugate acid is stronger. It more readily donates protons. • As base strength decreases, acid strength increases. • The stronger the base, the weaker its conjugate acid. • The weaker the base the stronger its conjugate acid.

39. Weak Acid Equilibria A weak acid is only partially ionized. Both the ion form and the unionized form exist at equilibrium HA + H2O  H3O+ + A- The acid equilibrium constant is Ka = [H3O+ ] [A-] [HA] Ka values are relatively small for most weak acids. The greatest part of the weak acid is in the unionized form

40. Weak Acid Equilibrium Constants Sample problem .A certain weak acid dissociates in water as follows: HF + H2O  H3O+ + F- If the initial concentration of HF is 1.5 M and the equilibrium concentration of H3O+ is 0.0014 M. Calculate Ka for this acid

41. Weak Base Equilibria Weak bases, like weak acids, are partially ionized. The degree to which ionization occurs depends on the value of the base dissociation constant General form: B + H2O  BH+ + OH- Kb = [BH+][OH-] [B] Example NH3 + H2O  NH4+ + OH- Kb = [NH4+][ OH-] [NH3]

42. Weak Base Equilibrium Constants Sample problem .A certain weak base dissociates in water as follows: B + H2O  BH+ + OH- If the initial concentration of B is 1.2 M and the equilibrium concentration of OH- is 0.0011 M. Calculate Kb for this base Solution Kb = [BH+ ] [OH-] [B] I C E Substituting [B] 1.2 -x 1.2-x Kb = (0.0011)2 = 1.01 x 10-6 [OH-] 0 +x x 1.1989 [BH+ ] 0 +x x x = 0.0011 1.2-x = 1.1989

43. Weak Acid Equilibria Concentration Problems Problem 1.A certain weak acid dissociates in water as follows: HA + H2O  H3O+ + A- The Ka for this acid is 2.0 x 10-6. Calculate the [HA] [A-], [H3O+] and pH of a 2.0 M solution Solution Ka = [H3O+ ] [A-] = 2.0 x 10-6 [HA] I C E Substituting [HA] 2.0 -x 2.0-x Ka = x2 = 2.0 x 10-6 [A-] 0 +x x 2.0-x [H3O+ ] 0 +x x If x <<< 2.0 it can be dropped from the denominator The x2 = (2.0 x10-6)(2.0) = 4.0 x10-6 x = 2.0 x 10-3 [A-] = [H3O+ ] = 2.0 x10-3 [HA] = 2.0 - 0.002= 1.998 pH = - log [H3O+ ] =-log (2.0 x 10-3) = 2.7

44. Weak Acid Equilibria Concentration Problems Problem 2. Acetic acid is a weak acid that dissociates in water as follows: CH3COOH + H2O  H3O+ + CH3COO- The Ka for this acid is 1.8 x 10-5. Calculate the [CH3COOH],[CH3COO-] [H3O+] and pH of a 0.100 M solution Solution Ka = [H3O+ ] [CH3COO-]= 1.8 x 10-5 [CH3COOH] I C E Substituting [CH3COOH] 0.100 -x 0.100-x Ka = x2 = 1.8 x 10-5 [CH3COO- ] 0 +x x 0.100-x [H3O+] 0 +x x If x <<< 0.100 it can be dropped from the denominator The x2 = (1.8 x10-5)(0.100) = 1.8 x10-6 x = 1.3 x 10-3 [CH3COO--] = [H3O+ ] = 1.3 x10-3 [CH3COOH] = 0.100 - 0.0013 = 0.0987 pH = - log [H3O+ ] =-log (1.3 x10-3) = 2.88

45. Weak Base Equilibria Example1. Ammonia dissociates in water according to the following equilibrium NH3 + H2O  NH4+ + OH- Kb = [NH4+][ OH-] = 1.8 x 10-5 [NH3] Calculate the concentration of [NH4+][ OH-] [NH3 ]and the pH of a 2.0M solution. I C E Substituting [NH3] 2.0 -x 2.0-x Kb = x2 = 1.8x 10-5 [OH-] 0 +x x 2.0-x [NH4+] 0 +x x If x <<< 2.0 it can be dropped from the denominator The x2 = (1.8 x10-5)(2.0) = 3.6 x10-5 x = 6.0 x 10-3 [OH-] = [NH4+] = 6.0 x10-3 [NH3] = 2.0- 0.006= 1.994 pOH = - log [OH-] =-log (6.0 x10-3) = 2.22 pH = 14-pOH = 14-2.22 = 11.78

46. Amphoteric Solutions • A chemical compound able to react with both an acid or a base is amphoteric.    • Water is amphoteric. The two acid-base couples of water are H3O+/H2O and H2O/OH-It behaves sometimes like an acid, for example • And sometimes like a base : • Hydrogen carbonate ion HCO3- is also amphoteric, it belongs to the two acid-base couples H2CO3/HCO3- and HCO3-/CO32-

47. Common Ion Effect The common ion effect is a consequence of Le Chatelier’s Principle When the salt with the anion (i.e. the conjugate base) of a weak acid is added to that acid, • It reverses the dissociation of the acid. • Lowers the percent dissociation of the acid. A similar process happens when the salt with the cation (i.e, conjugate acid) is added to a weak base. These solutions are known as Buffer Solutions.

48. Buffers • A buffer solution is na aqueous solution that resists a change in pH when a small amount of acid, base or water is added to it. Acidic Buffers: Weak acid and its salt Can be prepared by using excess of a weak acid and a strong base . NaOH(aq) + CH3 COOH(aq) => CH3 COONa(aq) + CH3 COOH(aq) + H2 O (l) salt excess weak acid CH3COOH (aq) => CH3COO- (aq) + H+ (aq) Ka = 1.74 x 10-5 A CB pH = pKa + log [ CB] / [ A ] Henderson Hasselbach Equation

49. Formulas pH = -log [H] pKa = -log Ka pOH = -log [OH] pKb = -log Kb pH + pOH = 14 pKb + pKa = 14 [H] [OH] = 1 x 10 -14 = Ka x Kb Kw = 1.0 x 10-14 pKw = 14

50. 1. Given 30 cm3 of 0.1 M ethanoic acid and 10 cm3 of 0.1 M NaOH, find the pH of the buffer solution. CH3COOH (aq) => CH3COO- (aq) + H+ (aq) Ka = 1.74 x 10-5 A CB pKa = -log (1.74 x 10-5 )=4.76 n CH3COOH = 0.1 x 0.03 = 0.003 n NaOH = 0.1 x 0.01 = 0.001 excess CH3COOH after titration = 0.002 pH = pKa + log [ CB] / [ A ] pH = 4.76 + log(0.001/0.002)= 4.5