1 / 124

Acids, Bases, and Salts

Acids, Bases, and Salts. Chapters 4.3, 16 and 17. Classification of Acids and Bases is one of the oldest in History. The medieval Alchemists first used the terms “acid”, “alkali”, and “salt”. Acids were probably the most easily recognized chemical because of their sour taste.

merrill
Download Presentation

Acids, Bases, and Salts

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Acids, Bases, and Salts Chapters 4.3, 16 and 17

  2. Classification of Acids and Bases is one of the oldest in History. The medieval Alchemists first used the terms “acid”, “alkali”, and “salt”. Acids were probably the most easily recognized chemical because of their sour taste.

  3. Properties of Acids • Have a sour taste • Change the color of some plant pigments • Dissolve or corrode certain metals and minerals; damages skin -“Corrosive” • Produce gas bubbles when combined with minerals. • Dissolve in water to form aqueous solutions. • Ionize in water to produce H+ ions

  4. Examples of Acids • HCl hydrochloric acid (stomach acid) • H2SO4 sulfuric acid (battery acid) • HNO3 nitric acid (used to make fertilizer) • HBr hydrobromic acid • HI hydroiodic acid • HClO3 chloric acid • HClO4 perchloric acid There are 7 strong Acids. (all other acids are weak. . .) HC2H3O2 acetic acid H2CO3 carbonic acid

  5. Properties of Bases • Have a bitter, chalky taste • Have a soapy, slippery feeling • Change the color of some plant pigments • Can burn and irritate skin - “caustic” • Bases can destroy the properties of acids when mixed in proper proportions. • Dissolve in water to form aqueous solution; accept H+ ions and produce OH- ions

  6. Examples of Bases • NaOH sodium hydroxide (lye) • LiOH • KOH • RbOH • CsOH • Ca(OH)2 • Sr(OH)2 • Ba(OH)2 Group IA metal hydroxides Heavy Group IIA metal hydroxides ALL other bases are weak: NH3

  7. Electrolytes and Nonelectrolytes: • Strong acids and bases completely ionize in solution making them strong electrolytes. • Weak acids and bases only partly ionize making them weak electrolytes in solution. • Acids and Bases are the only molecular compounds that can act as electrolytes.

  8. Arrhenius Definition • Acids produce hydrogen ions in aqueous solution. HCl + H2O ---> H+(aq) + Cl- (aq) • Bases produce hydroxide ions when dissolved in water. NaOH + H2O ---> Na+ (aq) + OH- (aq) • Limited to aqueous solutions. • Arrhenius limits us to one kind of base. • NH3 ammonia could not be an Arrhenius base.

  9. Bronsted-Lowry Definitions • An acid is a H+ (proton) donor and a base is a H+ (proton) acceptor. • Acids and bases always come in pairs. • Acid + Base Conjugate + Conjugate acid base • HCl is an acid. • When it dissolves in water it gives its proton to water. • HCl(g) + H2O(l) H3O+ + Cl- H+ donor H+ acceptor conjugate conjugate acid base

  10. Bronsted-Lowry expands the list • When ammonia is dissolved in water - NH3 + H2O NH4+ + OH- H+ acceptor H+ donor conjugate conjugate acid base The conjugate acid is what results when the base accepts the H+ The conjugate base is what remains after the acid has donated the H+ BASE ACID

  11. Competing Pairs NH3 + H2O NH4+ + OH- H+ acceptor H+ donor conjugate conjugate base acid acid base • This is an equilibrium. • Competitionfor H+between NH3 and OH • The stronger base controls direction. • If OH is a stronger base it takes the H+,forms water. • Equilibrium moves to left. • If NH3 is stronger, it takes the H+,forms NH4+ • Equilibrium moves to right.

  12. Relative strength of Acids and Bases: • The stronger the acid or base, the weaker its conjugate pair. • Strong acids completely transfer their protons to water; therefore, the conjugate bases have a tendency to be protonated. • Weak acids only partly dissociate; therefore, the conjugate bases show a slight ability to remove protons from water. • Substances with negligible acidity, contain hydrogen, but do not demonstrate any acidic behavior; their conjugate bases are strong bases.

  13. Lewis Acids and Bases • Most general definition. Gives us many things which act as acid-bases. • Acids are electron pair acceptors. • Bases are electron pair donors. F H B F :N H F H Lewis Acids involve molecules which have an incomplete octet. Lewis Bases have an unshared pair of electrons.

  14. Lewis Acids and Bases • Boron triflouride wants more electrons. • It acts as a Lewis Acid. F H B F :N H F H Ammonia, NH3 donates a pair of electrons. It acts as a Lewis base.

  15. Lewis Acids and Bases • BF3 is Lewis acid • NH3 is a Lewis base. H F F H B N F H Lewis Acid Lewis Base

  16. Water H+ + OH- ---> H2O Lewis Base Lewis Acid • • H+ [ O - H ]- • • • • Donates a pair of electrons

  17. A New View of Water:Autoionization • Water is amphoteric: it behaves as both an acid and a base • In any sample of water, to a very small degree, water self-ionizes: H2O ---> H+ + OH actually, 2 H2O H3O+ + OH

  18. Kw • Since there is an equilibrium established 2 H2O H3O+ + OH we can write an equilibrium expression: [H3O+] [OH] Keq = [H2O] 2 We don’t include water in the equilibrium expression because the amount is huge and constant Kw = [H3O+] [OH] Kw - “Ion Product Constant for water”

  19. Kw • Kw = [H3O+] [OH] For any aqueous solution at 25C, Kw = 1.0 x 10 14(a very small amt. of ionic activity!) This means: Kw = [H3O+] [OH] 1.0 x 10 14 = [H3O+] [OH] [H3O+] = [OH] = 1.0x 10 7 [H3O+] is also [H+] So, any Acidic Solution: [H+] > [OH] And any Basic Solution: [H+] < [OH]

  20. Acid vs. Basic Solutions • Remember water behaves as both an acid and a base. • 2H2O(l) H3O+(aq) + OH-(aq) • KW= [H3O+][OH-] = [H+][OH-] • At 25ºC, KW = 1.0 x10-14 • In EVERY aqueous solution. Therefore, • Neutral solution [H+] = [OH-]= 1.0 x10-7 • Acidic solution [H+] > [OH-] • Basic solution [H+] < [OH-]

  21. pH • pH= -log[H+] • Used because [H+] is usually very small, always less than 1 M • As pH decreases, [H+] increases exponentially • Helpful to see a graph of y =  log [x]

  22. pH =  log [H+]y =  log [x] This is where pH is concerned. pH [H+] (1,0)

  23. pH • [H+] = 1.0 x 10-8 M [H+] = 1.0 x 103 M pH= 8.00 pH = 3.00 • Sig figs - only the digits after the decimal place of a pH are significant - only 2 sig. figs in pH = 8.00 p stands for “ log” Also. . . pOH = -log[OH-] pKw = -log Kw

  24. [H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14 pH 0 1 3 5 7 9 11 13 14 14 13 11 9 7 5 3 1 0 pOH 10-14 10-13 10-11 10-9 Basic 10-7 10-5 10-3 10-1 100 [OH-] Acidic Neutral Basic

  25. Relationships KW = [H+][OH-] -log KW = -log([H+][OH-]) -log KW = -log[H+]+ -log[OH-] • So, pKW = pH + pOH and since, KW = 1.0 x10-14 • Then, 14.00 = pH + pOH • [H+], [OH-], pH and pOH; given any one of these we can find the other three.

  26. Calculations with pH • Remember every solution has both a pH and a pOH. • Remember the chemistry, but don’t try to memorize there is no one way to do this.

  27. Example #1 • A solution has [H+] = 1.5 x106 M • Determine the pH, pOH, and [OH]. pH = log[H+] = log [1.5 x 106 M] = 5.82 14.00 = pH + pOH 14.00 = 5.82 = pOH pOH = 8.18 Or, take the antilog of 8.18. Kw = [H+] [OH] 1.0 x 1014 = 1.5 x 106 M [OH] [OH] = 6.6 x 10 9 M

  28. Example #2 • A solution has a pOH = 4.92 • Calculate the pH, [H+], and [OH]. pH = 14.00  pOH pH = 14.00  4.92 pH = 9.08 Take the anti-log of 9.08 [H+] = 8.3 x 1010 M Kw = [H+] [OH] 1 x 10 14 = [8.3 x 1010 M] [OH] [OH] = 1.2 x 105 M

  29. Acid and Base dissociation constants Ka and Kb

  30. Ka -“Acid dissociation constant” General equation for an acid: • HA(aq) + H2O(l) H3O+(aq) + A-(aq) • Ka = [H3O+][A-] [HA] • H3O+ is often written H+ ignoring the water in equation (it is implied).

  31. Ka -Acid dissociation constant • Ignoring the water: • HA(aq) H+(aq) + A-(aq) • Ka = [H+][A-] [HA] • We can write the expression for any acid.

  32. Strong Acids • HCl, HNO3, H2SO4, HBr, HI, HClO4,HClO3 • Strong acids dissociate completely. • HCl ---> H+ (aq) + Cl(aq) • Ka = [H+][Cl-] [HCl] • Strong acids favor products. • Equilibrium far to right. Ka is very large. • Conjugate base must be weak.

  33. Weak Acids • Weak acids only partly dissociate. • Weak acids favors reactants. • HF (aq) H+(aq) + F (aq) • Ka = [H+][F-] [HF] • Ka will be small. • ALWAYS WRITE THE MAJOR SPECIES. • It will be an equilibrium problem from the start. • Rest is just like last chapter - an equilibrium problem!

  34. Strong acids Kais very large [H+] is equal to [HA] A- is a weaker base than water Weak acids Kais small [H+] <<< [HA] A- is a stronger base than water Summary

  35. Types of Acids • Oxyacids - Proton is attached to the oxygen of an ion. Example: nitrous acid, HNO2 • Organic acids contain the Carboxyl group -COOH with the H attached to O example: Benzoic acid, C6H5COOH • Polyprotic Acids- more than 1 acidic hydrogen (diprotic, triprotic).

  36. Polyprotic acids • Always dissociate stepwise. • The first H+ comes off much easier than the second. • Ka for the first step is much bigger than Ka for the second. • Denoted Ka1, Ka2, Ka3 • Example: Carbonic Acid • H2CO3 H+ + HCO3- Ka1= 4.3 x 10-7 • HCO3- H+ + CO3-2 Ka2= 4.3 x 10-10 Almost all of the H+ comes from the first step, so successive steps are often ignored except for sulfuric acid.

  37. Sulfuric acid is special • H2SO4(aq) ----> H+(aq) + HSO4 (aq) • In first step it is a strong acid. • HSO4 (aq)H+(aq) + SO42 (aq) • 2nd step it’s weak: Ka2 = 1.2 x 10-2 • However, 2nd step too large Ka to ignore.

  38. Ka Problems • Problem: What is the pH of a 0.0658M HCl solution? • Always write the major species: HCl --> H+ + Cl • Strong acids completely dissociate! • [H+] = [HCl] HCl ---> H+(aq) + Cl (aq) [H+] = [HCl] [H+] = 0.0658 M pH = log [0.0658 M] = 1.18

  39. Try this. • Calculate the pH of 2.0 M acetic acid HC2H3O2. Then find the pOH and[OH-]. Weak acids only partly dissociate. HC2H3O2 H+ + C2H3O2- Ka = [H+][C2H3O2-] [HC2H3O2] Look up Ka = 1.8 x 105 1.8 x 105 = [ x ][ x ] [ 2.0 M ] pH = 2.2 pOH = 11.8 [OH] = 1.6 x 1012 M X2 = 1.8 x 10-5 x 2.0 [H+] = X = 0.0060 M

  40. Bases • Strong Bases: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2 and Ba(OH)2. • The OH-is a strong base. • Hydroxides of the alkali metals are strong bases because they dissociate completely when dissolved. • The hydroxides of alkaline earths Ca(OH)2 etc. are strong dibasic bases, but they don’t dissolve well in water. • Again, strong bases favor products. Kb is very large, [OH-] = [B] • Calculations just like strong acids.

  41. Bases without OH- • Bases are proton acceptors. • NH3 + H2O NH4+ + OH- • It is the lone pair on nitrogen that accepts the proton. • Many weak bases contain N

  42. Weak Bases • NH3 + H2O NH4+ + OH- • General Equation: • B(aq) + H2O(l) BH+(aq) + OH- (aq) Kb = [BH+][OH- ] [B] Kb “base dissociation constant”

  43. Strength of Bases • Hydroxides are strong. • Others are weak. • Smaller the Kbweaker the base.

  44. Kb problem • Calculate the pH of a 0.25 M NH3. • Always write the equation and the species! NH3 + H2O NH4+ + OH- Kb = [NH4+][OH- ] [NH3] Look up Kb = 1.8 x 105 1.8 x 105 = [ x ][ x ] [NH3] (1.8 x 105 ) (0.25 M) = x2 14.00 = pH + pOH pH = 11.33 x = 0.0021 M = [OH-] pOH = 2.67

  45. Percent Dissociation • The amount of the acid (or base) [HA] that has dissociated [x] divided by the acid’s initial conentration, [HA0], then multiplied by 100%. • Percent Dissociation = (x/[HA0]) x 100% • When making assumptions in equilibrium concentrations, it is best to test the assumption by making sure that the percent dissociation is less than or equal to 5%.

  46. Example • The percent dissociation of an acid, HA, which is 0.100M is 2.5%. Calculate the Ka of the acid. • (x/.100) x 100% = 2.5% • x = 2.5 x 10-3M • Ka = (2.5x10-3)2 / (.100 – 2.5x10-3) • Ka = 6.4 x 10-5

  47. A mixture of Weak Acids/Bases • The process is the same. • Determine the major species. • The stronger will predominate. • If one acid has a relatively highier Ka value, it will be the focus of the solution. • Calculate the pH of a mixture: 1.20 M HF (Ka = 7.2 x 10-4) and 3.4 M HOC6H5 (Ka = 1.6 x 10-10) • HF dominates so the pH of the mixture will be based on it.

  48. Relationship of Ka and Kb NH3 + H2O NH4+ + OH- NH4+ + H2O NH3 + H3O+ Kb = [NH4+] [OH-] Ka = [NH3] [H+] [NH3] [NH4+] if added together, NH3 + H2O NH4+ + OH + NH4+ + H2O NH3 + H3O+ = 2 H2O H3O+ + OH- • we get the autoionization of water equation

  49. Relationship of Ka and Kb • Ka x Kb = ( [NH3] [H+] ) ( [NH4+] [OH-] ) [NH4+] [NH3] • Ka x Kb = [H+] [OH-] = Kw = 1.0 x 10-14 • So, Ka x Kb = Kw

  50. Relationship of Ka and Kb • Ka x Kb = Kw= 1.0 x 1014 As the strength of an acid increases (larger Ka), the strength of it’s conjugate base must decrease (smaller Kb). • We can now calculate Kb for any base if we know Ka for conjugate acid and vice versa. Ka x Kb = Kw = 1.0 x 1014 Taking the – log of both sides and simplifying . . . • pKa + pKb = pKw = 14.00

More Related