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Arrhenius, Bronsted-Lowry, & Lewis Models of Acids and Bases and Solving Acid/Base Problems Chapter 14 Spring 2010.

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Presentation on theme: "Arrhenius, Bronsted-Lowry, & Lewis Models of Acids and Bases and Solving Acid/Base Problems Chapter 14 Spring 2010."— Presentation transcript:

1 Arrhenius, Bronsted-Lowry, & Lewis Models of Acids and Bases and Solving Acid/Base Problems Chapter 14 Spring 2010

2 Arrhenius Acids and Bases Arrhenius acids form hydrogen ions in aqueous solution with Arrhenius bases forming hydroxide ions

3 Bronsted-Lowry Acids and Bases In the Brønsted-Lowry definition, acids, referred to as "Brønsted-Lowry acids" donate H+ to a reaction and "Brønsted- Lowry bases" accept H+ ions from a reaction

4 Conjugate acid: substance formed when a base gains a hydrogen ion. Considered an acid because it can lose a hydrogen ion to reform the base. Conjugate Acids

5 In aqueous solution, the chemical reaction involved is of the form –HX + H2O X− + H3O+ Within the Brønsted-Lowry (protonic) theory of acids and bases, a conjugate acid is the acid member, HX, of a pair of two compounds that transform into each other by gain or loss of a proton. A conjugate acid can also be seen as the chemical substance that releases, or donates, a proton in the forward chemical reaction, hence, the term acid.

6 Conjugate Base In aqueous solution, the chemical reaction involved is of the form –HX + H2O X− + H3O+ The base produced, X−, is called the conjugate base, and it absorbs, or gains, a proton in the backward chemical reaction.

7 Conjugate Base conjugate base: substance formed when an acid loses a hydrogen ion. Considered a base because it can gain a hydrogen ion to reform the acid.

8 Explain acid strength in terms of equilibrium Acid strength is related to the type of acid that is in a solution; hydronium ions Diprotic acids: acids with two hydrogen atoms; can dissociate multiple times, resulting in more hydronium molecules Organic acids: acids that are also organic compounds; many are carboxylic acids, which don’t dissociate completely in water (relate to equilibrium)

9 How is acid a water and a base? A “bronsted lowry acid/base” because it can both donate and accept a hydrogen ion.

10 pH Scale The pH scale measures how acidic or basic a substance is The pH scale ranges from 0 to 14 A pH of 7 is neutral A pH less than 7 is acidic A pH greater than 7 is basic.


12 How to calculate pH The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value. Two ways to calculate: –Using hydrogen ions –Using hydroxide ions

13 Using Hydrogen Ions to Calculate pH Calculate the concentration of hydrogen (H+) ions by dividing the molecules of hydrogen ions by the volume, in liters, of the solution. Take the negative log of this number. The result should be between zero and 14, and this is the pH. Example: If the hydrogen concentration is 0.01, the negative log is 2, or the pH. The stronger the acid, the more corrosive the solution.

14 Using Hydroxide Concentration to Calculate pH The pH of a solution can also be determined by finding the pOH. Determine the concentration of the hydroxide ions by dividing the molecules of hydroxide by the volume of the solution. Take the negative log of the concentration to get the pOH. Then subtract this number from 14 to get the pH. Example: If the OH- concentration of a solution was , take the negative log of and you get five. This is the pOH. Subtract five from 14 and you get nine.

15 How to Calculate pOH To calculate the pOH of a solution you need to know the concentration of the hydroxide ion in moles per liter (molarity). The pOH is then calculated using the expression:molarity pOH = - log [OH-] Example: What is the pOH of a solution that has a hydroxide ion concentration of 4.82 x 10-5 M? pOH = - log [4.82 x 10-5] = - ( ) = 4.32

16 Calculating the Hydroxide Ion Concentration from pOH The hydroxide ion concentration can be found from the pOH by the reverse mathematical operation employed to find the pOH. [ OH- ] = 10-pOH or [OH-] = antilog ( - pOH) Example: What is the hydroxide ion concentration in a solution that has a pOH of 5.70? 5.70 = - log [OH-] = log[OH-] [OH-] = = 2.00 x 10-6 M On a calculator calculate , or "inverse" log (- 5.70).

17 A strong acid is defined as one that dissociates completely in water for every mole of acid added, one mole of free H+ is present in the solution pH = -log10 [H3O+] Ex M of HCl, pH = -log (0.01) = 2 How to Calculate pH of strong acids

18 Examples (These acids are stronger than H3O+):  HCl - hydrochloric acid  HBr - hydrobromic acid  HI - hydroiodic acid  H2SO4 - sulfuric acid (first proton)  HNO3 - nitric acid  HClO4 - perchloric acid  HClO3 - chloric acid How to Calculate pH of strong acids

19 To determine the pH of a weak acid solution you must know two things: you need the concentration of the acid in the solution, and you need the Ka of the acid (or equivalently the pKa, which you can use to determine the Ka).

20 First, we must write the equation of the acid dissociation in water. Let's use a generic acid that we'll call "HA." When it dissociates, it will form H+ and A- (A- is called the conjugate base of the acid HA). HA + H2O --> H3O+ + A-

21 Because this is a weak acid, this reaction will go to some equilibrium value, and this will be described by the equilibrium constant, Ka. The equilibrium product for this reaction is found by taking the concentration of the products and dividing them by the concentration of the reactants (with the concentration of each specie raised to its respective coefficient in the balanced reaction). So at equilibrium, we have: Ka = [H3O+] [A-] ÷ [HA] where the square brackets mean concentration (for instance "[N]" means "the concentration of N"). Notice that H2O is NOT included in the equilibrium product even though it is a reactant because pure substances are not included.

22 If you are given the pKa instead of the Ka, use this formula to find the Ka: Ka = -log10 pKa

23 pH = -log10 (x)

24 Ka is the dissociation constant for an acid in water Kb is the dissociation constant for a base in water Kw is the dissociation constant for water reacting with itself KaxKb = Kw

25 The hypocholorite ion(OCl-) is a strong oxidizing agent often found in household bleaches and disinfectants. It is also the active ingredient that forms when swimming pool water is treated with chlorine. In addition to its oxidizing abilities, the hypochlorite ion has a relatively high affinity for protons (it is a much stronger base than Cl-, for example) and forms the weakly acidic hypochlorons acid (HOCl, Ka = 3.5x10^-8). Calculate the pH of a M aqueous solution of hypochlorous acid. The pH of Weak Acids

26 Step 1 We list the major species. Since HOCl is a weak acid and remains mostly undissociated, the major species in a M HOCl solution are HOCl and H 2 0

27 Step 2 Both species can produce H+ HOCl(aq)   H + (aq) + OCL - (aq) K a = 3.5 x H20(l)  H + (aq) + OH - (aq) K w = 1.0 x

28 Step 3 Since HOCL is a significantly stronger acid than H 2 O, it will dominate in the production of H +.

29 Step 4 We therefore use the following equilibrium expression: K a = 3.5 x = [H + ][OCL-]/[HOCL]

30 Step 5 The initial concentrations appropriate for this equilibrium are [HOCL] 0 = M [OCl - ] 0 = 0 [H + ] 0 ≈ 0 (We neglect the contribution from H 2 0.)

31 Step 6 Since the system will reach equilibrium by the dissociation of HOCl, let x be the amount of HOCl (in mol/L) that dissociates in reaching equilibrium.

32 Step 7 The equilibrium concentrations in terms of x are [HOCL] = [HOCL] 0 – x = – x [OCL - ] = [OCL - ] 0 + x = 0 + x = x [H + ] = [H + ] 0 + x ≈ 0 + x = x

33 Step 8 Substituting these concentrations into the equilibrium into the equilibrium expression gives K a = 3.5 x = ((x)(x)) / (0.100 – x)

34 Step 9 Since Ka is so small, we can expect a small value for x. Thus we make the approximation [HA]0 – x = [HA]0, or – x ≈ 0.100, which leads to the expression Ka = 3.5 x 10 -8

35 Compare Lewis acids to Arrhenius acids and Bronsted-Lowry acids Lewis acid: a compound that accepts a pair of electrons from a Lewis base Arrhenius acid: a compound that dissociates in aqueous solution to form hydronium ions Bronsted-Lowry acid: a compound that donates a proton to a Bronsted-Lowry base, which accepts that proton.

36 Explain at 25 degrees C & 1 atm, F 2 is a gas while I 2 is a solid I 2 is heavier than F 2 Larger, more dispersed electron clouds London dispersion forces become stronger

37 How to calculate the pH of a strong base Use pOH, which measures the concentration of OH- ions Derived from pH pOH = -log[OH-] pOH + pH = 14

38 Example What is the pH of a 0.05 M solution of Potassium Hydroxide? pOH = - log (0.05) pOH = -(-1.3) = 1.3 pH = 14 - pOH pH = pH = 12.7 The pH of a 0.05 M solution of Potassium Hydroxide is 12.7.

39 Which salts produce neutral solutions? Why? When NaCl is dissolved in water, the products are NaOH and HCl NaOH and HCl themselves are completely dissociated No OH- or H3O+ ions produced, so solution remains neutral

40 Bronsted-Lowry Bases v. Arrhenius Bases Arrhenius Bases: dissociates in an aqueous solution to form hydroxide ions –Neutralization reaction: acid + base  salt + water Bronsted-Lowry Bases: accept a proton (hydrogen ion) from a Bronsted-Lowry Acid –React to form a conjugate acid and conjugate base

41 m eow The melting point of Naf is 993C, whereas the melting point of CsCl is 645C Both are ionic compounds with the same charges on the anions and the cations. The ionic radius of Na is smaller than the ionic radius of Cs and the ionic radius of F is smaller than the ionic radius of Cl. The ionic centers are closer in NaF than in CsCl. Melting occurs when the attraction between the cation and the anion are overcome due to thermal motion. Since the lattice eneregy is inversly proportional to the distance between the ion centers (Coulomb’s Law), the compound with the smaller ions will have the stronger attractions and the higher melting point.

42 ICl4 and BF4 The shape of ICl4 is square planar due to the bonds of the Iodine atom to 4 chlorine atoms. The shape of BF4 is tetrahedral because of the presence of more valence electrons, which bend the shape and make it tetrahedral instead of square planar even though like ICl4, there are 4 bonds to the surrounding atoms.

43 Explain why Ammonia is soluble in water while Phosphine is only moderately soluble? NH3 can participate in H-bonding with water, because of the high electronegativity value of N, the H bonded to it, and the lone pair of electrons on the N. P has the hydrogens, and the lone pair of electrons, but it does not have a high electronegativity value, so it does not H- bond to water as much.

44 Definition the ratio of the amount of a substance that is dissociated at equilibrium to the initial concentration of the substance in a solution

45 Example: The pH of a molar solution of unknown monoprotic acid was measured and found to be Calculate the percentage dissociation of this acid. 1.% = [HA] dissociated /[HA] initial * 100 = 2.[HA] dissociated: pH = -log[H + ] [H + ] = 10 -pH = = / M = * 100 = %

46 Arrhenius and Bronsted-Lowry Bases Arrhenius: anything that forms hydroxide ions in solution Bronsted-Lowry: anything that accepts protons The strong bases: compounds that have a K a of 13 or greater, e.g. KOH, Ba(OH) 2, CsOH, NaOH, Sr(OH) 2, Ca(OH) 2, LiOH, RbOH, Mg(OH) 2

47 How to calculate the pH of a strong base Use pOH, which measures the concentration of OH- ions Derived from pH pOH = -log[OH-] pOH + pH = 14

48 Example What is the pH of a 0.05 M solution of Potassium Hydroxide? pOH = - log (0.05) pOH = -(-1.3) = 1.3 pH = 14 - pOH pH = pH = 12.7 The pH of a 0.05 M solution of Potassium Hydroxide is 12.7.

49 Explain the value and significance of the amine group of chemicals. Amine - A compound containing an amino group (-NH2). Example: R-NH2

50 Amines are organic compounds formed from ammonia (NH3 by the replacement of one or more hydrogen atoms by hydrocarbon groups (unlike the amides where the hydrogen in ammonia is replaced by organic acid groups). Primary amines have one hydrogen atom replaced by a hydrocarbon group, secondary amines have two hydrogen atoms replaced, and tertiary amines have all three hydrogen atoms replaced. Blah

51 Primary amines also are manufactured by a reaction known as the Hofmann degradation which takes amides as the starting point. The amide is heated with bromine, and the resulting mixture is then treated with excess sodium hydroxide to give the required primary amine.Reduction of amides will also yield amines, as will reduction of alkyl cyanides and nitroalkenes. Secondary amines are manufactured by heating a primary amine with an alkyl halide, although care must be used to ensure that excess alkyl halide is not used. Tertiary amines are manufactured by heating an alcoholic solution of ammonia with excess alkyl halide. More Blah

52 Polyprotic acids are able to donate more than one proton (H+) per acid molecule, in contrast to monoprotic acids that only donate one proton per molecule.

53 Equilibrium Each acid equilibrium has a separate equilibrium equation. Each equation, in turn, has a different constant. To distinguish these acid equilibrium constants, they are numbered 1 through 3. Ka1 is the constant for the first proton ionization, Ka2 is the second, etc. The acid equilibria are expressed by the three equilibrium equation:

54 Two key features of polyprotic acids are that they lose their protons in a stepwise manner and that each proton is characterized by a different pK a. The factors contributing to the pK a of each acidic proton in a polyprotic species are the same factors that determine the relative acidity of monoprotic acids--the dominant factor is strength of the acid-H bond. Consider, for example, the triprotic acid H3PO4 shown in : As each proton is lost from phosphoric acid, the phosphorous becomes more electron rich, and less electron withdrawing. Therefore, the loss of each proton strengthens the O-H bond and increases the pK a of the phosphate species. This trend is evident in the pK a data given in. In general, it is true that K a1, K a2, K a3, and so on, for polyprotic acids. As you may have guessed, calculating the pH of a polyprotic acid solution is not as simple as it is for monoprotic acids. In fact, it is quite a messy problem. However, that mess can be quickly cleaned up by making the assumption, as we did for a mixture of acids, that only the strongest acid (i.e. only the first dissociation) has a significant effect on the pH.

55 Acid and Base Definitions In the Lowry-Brønsted model: "A Brønsted acid is a proton donor, and a Brønsted base is a proton acceptor (abstractor)". In the Lewis model: "A Lewis acid is an electron-pair acceptor, and Lewis base is an electron-pair donor ". The two theories can be reconciled by recognizing that the proton, H+, is a unique and versatile Lewis acid that is the agent of Brønsted acidity.

56 Acid/Base Definitions All Brønsted acids are proton/Lewis base complexes. The transfer of H+ between Lewis bases equates with Brønsted acidity. While a Brønsted acid is an H+ donor, the proton, H+, is a Lewis acid. All Lewis bases can be protonated. It follows that the ability of a species to complex a proton defines that species as being both a Brønsted base and a Lewis base. Any species able to complex with a Lewis base is a Lewis acid.

57 Acid/Base Definitions The Lewis theory suggests that acids react with bases to share a pair of electrons, with no change in the oxidation numbers of any atoms. Many chemical reactions can be sorted into one or the other of these classes. Either electrons are transferred from one atom to another, or the atoms come together to share a pair of electrons. The principal advantage of the Lewis theory is the way it expands the number of acids and therefore the number of acid-base reactions. In the Lewis theory, an acid is any ion or molecule that can accept a pair of nonbonding valence electrons. In the preceding section, we concluded that Al 3+ ions form bonds to six water molecules to give a complex ion. Al 3+ (aq) + 6 H 2 O(l) Al(H 2 O) 6 3+ (aq)

58 Acid/Base Definitions This is an example of a Lewis acid-base reaction. The Lewis structure of water suggests that this molecule has nonbonding pairs of valence electrons and can therefore act as a Lewis base. The electron configuration of the Al 3+ ion suggests that this ion has empty 3s, 3p, and 3d orbitals that can be used to hold pairs of nonbonding electrons donated by neighboring water molecules. Al 3+ = [Ne] 3s0 3p0 3d0 Thus, the Al(H 2 O) 6 3+ ion is formed when an Al 3+ ion acting as a Lewis acid picks up six pairs of electrons from neighboring water molecules acting as Lewis bases to give an acid-base complex, or complex ion.

59 Which salts produce neutral solutions? Why? When NaCl is dissolved in water, the products are NaOH and HCl NaOH and HCl themselves are completely dissociated No OH- or H3O+ ions produced, so solution remains neutral


61 Which salts produce basic solutions?






67 Lewis Bases More general theory, encompasses much of previous theories Covers many reactions that don’t involve Bronsted-Lowry acids/bases Donate a pair of electrons Lewis bases react with Lewis acids to create a new, covalently bonded adduct Ex. BF 3 (Lewis acid) + NH 3 (Lewis base)  BF 3 NH 3 (new B-N bond)

68 Dissociation of Acid in Water  HCN(aq) + H 2 O(l)  H 3 O + (aq) + CN - (aq) K a = 6.2 x  Since HCN is a weak acid, CN - appears to be a strong base, showing a very high affinity for H + compared to H 2 O with which it is competing.  However, we must also look at the reaction in which cyanide ion reacts with water.

69 Base Strength  CN - (aq) + H 2 O(l)  HCN(aq) + OH - (aq)  K b = K w / K a  1.0 x / 6.2 x = 1.6 x  K b = 1.6 x  Now, CN - appears to be a weak base because the K b value is only 1.6 x  In this reaction, CN - is competing with OH - for H +, instead of competing with H 2 O as it did in the previous reaction.  Relative base strength: OH - > CN - > H2O

70 A SALT IS A… Salt is conjugate of either an acid or a base (sometimes both) according to Bronsted- Lowry

71 THESE PRODUCE ACID SOL. NH 4 NO 3 (s) → NH 4+ (aq) + NO 3– (aq) NH 4+ can donate an H +, so NH 4+ is an acid and NH 3 is its conjugate base. NO 3– can accept an H +, so NO 3– is a base and HNO 3 is its conjugate acid. In water NH 4+ would be NH 4+ + H 2 O → NH 3 + H 3 O + This reaction occurs, since NH 4+ is a weak acid and NH 3 is a weak base Result is to increase the H 3 O + concentration, so the solution becomes acidic.

72 ANOTHER ONE: ALCL 3 AlCl 3 (s) → Al 3+ (aq) + 3 Cl – (aq) Al 3+ ions don’t exist in water, they become [Al(H 2 O) 6 ] 3+ complex ions. The aluminum complex ion can act like an acid and donate an H + to become [Al(H 2 O) 5 (OH)] 2+. The Cl – ion can accept an H +, so Cl – is a base and HCl is its conjugate acid (strong acid). No reaction occurs between Cl – and water because HCl is a strong acid [Al(H 2 O) 6 ] 3+ + H 2 O → [Al(H 2 O) 5 (OH)] 2+ + H 3 O + Therefore, when AlCl 3 is added to water, the result is to increase the H 3 O + concentration, so the solution becomes acidic.

73 KBR DOES NOT QUALIFY BECAUSE… KBr (s) → K + (aq) + Br – (aq) K + is not a conjugate acid or a base. Br – can accept an H +, so Br – is a base and HBr is its conjugate acid. Br – + H 2 O → HBr + OH – But this reaction does not occur to any reasonable extent because HBr is a strong acid, so the equilibrium in the reaction lies very strongly to the left. Since no new H + or OH – are produced, the resulting KBr (aq) solution remains as neutral as the original water solvent.

74 For BINARY ACIDS (ex. H-X) In general, the larger the atom X, the stronger the acid. For larger atoms of X, the e¯ cloud is more diffuse. H-X bond breaks EASILY than in a smaller atom of X. In general, the strength of an acid increases as electronegativity of X increases

75 Strength of OXY-ACIDS For Same Structure, but different central atom: –Ex. H – O – X : H– O – Cl > H– O – Br > H – O – I –The ability of X to withdraw e¯ density from O-H increases with increasing electronegativity of X. e.g. HClO3 > HBrO3 For Same Central Atom, but with different number of Oxygen-atoms –The acid strength increases with the increase in number of Oxygen-atoms, i.e. with increase –in oxidation number of the central atom.

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