Arrhenius, Bronsted-Lowry, & Lewis Models of Acids and Bases and Solving Acid/Base Problems

1. Arrhenius, Bronsted-Lowry, & Lewis Models of Acids and Basesand 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 Acids • 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.

5. Conjugate Acids • 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 0.00001, take the negative log of 0.00001 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: • 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) = 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-] -5.70 = log[OH-] [OH-] = 10-5.70 = 2.00 x 10-6 M • On a calculator calculate 10-5.70, or "inverse" log (- 5.70).

17. How to Calculate pH of strong acids • 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. 0.01 M of HCl, pH = -log (0.01) = 2

18. How to Calculate pH of strong acids • 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

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 pH of Weak Acids • 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 0.100 M aqueous solution of hypochlorous acid.

26. Step 1 • We list the major species. Since HOCl is a weak acid and remains mostly undissociated, the major species in a 0.100 M HOCl solution are • HOCl and H20

27. Step 2 • Both species can produce H+ • HOCl(aq)   H+(aq) + OCL-(aq) Ka = 3.5 x 10-8 • H20(l)H+(aq) + OH-(aq) Kw = 1.0 x 10-14

28. Step 3 • Since HOCL is a significantly stronger acid than H2O, it will dominate in the production of H+.

29. Step 4 • We therefore use the following equilibrium expression: • Ka = 3.5 x 10-8 = [H+][OCL-]/[HOCL]

30. Step 5 • The initial concentrations appropriate for this equilibrium are • [HOCL]0 = 0.100 M • [OCl-]0 = 0 • [H+]0≈ 0 (We neglect the contribution from H20.)

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 = 0.100 – 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 • Ka = 3.5 x 10-8 = ((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 0.100 – 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, F2 is a gas while I2 is a solid • I2 is heavier than F2 • 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 - pOHpH = 14 - 1.3pH = 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. meow • 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 0.009300 molar solution of unknown monoprotic acid was measured and found to be 2.990. Calculate the percentage dissociation of this acid. • % = [HA] dissociated /[HA] initial * 100 = • [HA] dissociated: pH = -log[H+] • [H+] = 10-pH = 10-2.990 = 0.001023293 • 0.001023293/0.009300 M = • 0.110031505 * 100 = • 11.00%

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 Ka 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 - pOHpH = 14 - 1.3pH = 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. Blah • 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.