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Acids and Bases

Acids and Bases. They’re Everywhere. Definitions. Arrhenius Acids Acids produce hydrogen ions in solution Bases produce hydroxide ions in solution Only have one type of base, hydroxide First description of what an acid/base was. -1. +1. +. . Definitions. Bronsted-Lowry Model

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Acids and Bases

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  1. Acids and Bases They’re Everywhere

  2. Definitions • Arrhenius Acids • Acids produce hydrogen ions in solution • Bases produce hydroxide ions in solution • Only have one type of base, hydroxide • First description of what an acid/base was

  3. -1 +1 +  Definitions • Bronsted-Lowry Model • Acid is a proton donor • Base is a proton acceptor H---Cl + H---O---H  (H---O---H)+1 + Cl-1 H Water is the base (proton acceptor) HClis the acid (proton donor)

  4. Conjugate Pairs • The reaction that occurs is really an equilibrium • The H3O+1 is called the hydronium ion H---Cl + H---O---H  (H---O---H)+1 + Cl-1 H HCl + H2O Acid Base H3O+1 + + Cl-1 Acid Base These are conjugate acid base pairs. Water is a base on one side and a acid on the other, this is called amphoteric

  5. Quick Practice • Strong acid HNO3 + H2O ↔ H3O+ + NO3- • What are the conjugate pairs? • Weak acid HF + H2O ↔ H3O+ + F- • What are the conjugate pairs?

  6. Weak Acids • Weak acids partially dissociate in water: • HA(aq)  H+(aq) + A- (aq) • Bronsted-Lowry: HA + H2O  H3O+ + A- • What is Kc for this dissociation? • Kc in a weak acid case is better known as Ka, or the acid-dissociation constant. • The larger the value of Ka, the stronger the acid.

  7. Acid Dissociation Constant • HA + H2O  H3O+1 + A- • Generalized expression • The equilibrium expression for this process Ka = [H3O+1 ] [A-] = acid dissociation constant [HA] Note: the [water] is a constant (concentration of a pure solid or pure liquid is not included in an equilibrium expression) Water does play an important part in the dissociation of the acid!

  8. Acid Strength • HA(aq) + H2O  H3O+1(aq) + A-(aq) • A strong acid lies to the right Ka >>1 • [H+1]  [HA]0 • A weak acid lies to the left Ka<<1 • [H+1] << [HA]0 • Typical Ka (dissociation) constants for weak acids • HF 7.2 x 10-7 • HC2H3O2 1.8 x 10-5

  9. Conjugate Pairs and Strength • HA(aq) + H2O  H3O+1(aq) + A-(aq) • Strong acids have weak conjugate bases • Strong bases have weak conjugate acids http://www.chem.ubc.ca/courseware/pH/index.html

  10. Weak Acid

  11. Acid Strength • In an acid, the strength of the bond between the acidic hydrogen and the other atom (H-X) determines how strong the acid is. • In general, the strength of an H-X bond weakens as atoms get bigger. • So, going down a group, the strength of an acid increases. • HF < HCl < HBr < HI

  12. Acid Strength • Going across a row, bond strengths don’t change all that much. So, bond polarity is the major factor – the more polar the bond, the stronger the acid • Period 2: CH4 < NH3 < H2O < HF

  13. Acid-Base Behavior and Chemical Structure Binary Acids

  14. Acid-Base Behavior and Chemical Structure Factors That Affect Acid Strength Consider H-X. For this substance to be an acid we need: • H-X bond to be polar with H+ and X- (if X is a metal then the bond polarity is H-, X+ and the substance is a base). • the H-X bond must be weak enough to be broken, • the conjugate base, X-, must be stable.

  15. Acid-Base Behavior and Chemical Structure • Binary Acids • Acid strength increases across a period and down a group. • Conversely, base strength decreases across a period and down a group. • What differences in atomic structure account for these variations? • The kernel charge increases across a period. Therefore, the nonmetallic element has a stronger pull on the shared electron pair and H is more easily ionized. • As you go down a group, the strength of the H-X bond weakens as the X has more shells and its size increases. Therefore, H is more easily ionized

  16. Acid-Base Behavior and Chemical Structure • Oxyacids • Oxyacids contain O-H bonds. • All oxyacids have the general structure Y-O-H. • The strength of the acid depends on Y and the atoms attached to Y. • If Y is a metal (low electronegativity), then the substances are bases. • If Y has intermediate electronegativity (e.g. I, EN = 2.5), the electrons are between Y and O and the substance is a weak oxyacid.

  17. Acid-Base Behavior and Chemical Structure • Oxyacids • If Y has a large electronegativity (e.g. Cl, EN = 3.0), the electrons are located closer to Y than O and the O-H bond is polarized to lose H+. • The number of O atoms attached to Y increase the O-H bond polarity and the strength of the acid increases (e.g. HOCl is a weaker acid than HClO2 which is weaker than HClO3 which is weaker than HClO4 which is a strong acid).

  18. Acid-Base Behavior and Chemical Structure Oxyacids

  19. Carboxylic Acids • On a similar note, carboxylic acids contain –OH groups, but are acids, because of the additional attached oxygen “aldehyde” group on the final carbon in the chain. • Carboxylic acids are also stabilized by resonance once the hydrogen goes away.

  20. B-l Acids and Bases • Weak acids and Bases • Know the difference between a Bronsted-Lowry acid/base and an Arrhenius acid and base

  21. Strong Acid • Example 0.10 M HNO3 • H2O  H3O+ + OH- • There are two sources of H+ • The nitric acid and water • Since [H+] >>[OH-] in 0.1M nitric • Autoionization of water is insignificant • All the H+ is from HNO3 • [H+] = 0.10M pH = -log 0.1 = 1.0

  22. Strong Acid • The acid contributes all the [H+] • Example 1.0 x 10-10 HCl • The [H+] from autoionization (1 x 10-7M) is much higher. pH = - log 1 x 10-7 = 7

  23. Weak Acids • Treat like any equilibrium problem • What is pH of a 1.00M HF solution • Kc = 7.2 x 10 –4 • Kw = 1.0 x 10-14 • Since the Kc is so much bigger than the Kw, • HF is the major source of H+

  24. pH of 1.0 M HF Solution HF + H2O  H3O+ + F- I 1.0 0(1 x 10-7)0 C -x +x +x E 1.0 – x x x Ka = 7.2 x 10-4 = [F- ] [H+] = x • x = x2 [HF] 1 – x 1 Assume x is small (5% rule) X = 2.7 x 10-2 = [H+]

  25. Check for 5% Rule • Ka = x2 x2 [HA]o – x [HA] Ka • [HA]  x2 (Ka • [HA])1/2 x X x 100  5% [HA]0 • X = 2.7 x 10-2 x 100  5% 1

  26. Weak Acid Equilibrium • List the Major species in solution • Don’t forget water as an acid source! • Choose the species that can produce H+ • Write balanced equation • Using the K values, choose dominate source H+ • Write the equilibrium expression for dominate H+ • Do “ICE” and solve using 5% rule • Verify 5% • Calculate [H+ ] and pH

  27. Polyprotic Acids • When acids are polyprotic, like the triprotic H3PO4, where all three protons are weak acids, different Ka values are used. • H3PO4 H2PO4- + H+Ka1 = 7.1 x 10-3 • H2PO4-  HPO42- + H+ Ka2 = 6.3 x 10-8 • HPO42-  PO43- + H+Ka3= 4.5 x 10-13 • If Ka1 is more than 103 larger than Ka2, you can ignore Ka2 and treat it like a monoprotic acid.

  28. Calculate pH of 0.100 M HOCl • Ka = 3.5 x 10-8 • You Calculate the pH • Pg 675 if you need book • pH = 4.23

  29. pH of Weak Acid Mixture • Calculate the pH of a mixture of 1.00 M HCN, 5.00 M HNO2 and the equilibrium concentration of [CN-1] • HCN Ka = 6.2 x 10-10 • HNO2 Ka = 4.0 x 10-4 • H20 Kw = 1 x 10-14 How do you approach this?

  30. Calculate pH • Ka = 4.0 x 10-4 = [H+][NO2-] [HNO2] HNO2 H+ + NO2- I 5.00 0 0 C -x +x +x E 5.00 – x x x

  31. Calculate pH • Ka = 4.0 x 10-4 = x2 = x2 5.00 - x 5 x = [H+] = 4.5 x 10-2 M pH = - log [H+] = 1.35 Now calculate [CN-], you now [HCN] and [H+]

  32. Calculate [CN-] • Ka = 6.2 x 10 –10 6.2 x 10 –10 = [CN-][H+] = [CN-][4.5 x 10-2] [CN] 1.00 Solve for [CN-] = 1.4 x 10-8 M

  33. Percent Dissociation • % dissociation = [amount disassociated] x 100 [initial concentration] In the HF example [H+] = 1.27 x 10-2 M x 100 = 1.27% [HF] 1.00 M

  34. 1.00 M HC2H3O2 Left side of room Write on board 0.100 M HC2H3O2 Right side of room Write on board Calculate the Percent Dissociation

  35. Percent Dissociation • For solutions of weak acids: • the more dilute the solution • The greater the percent dissociation General Proof Suppose have acid HA, with [HA]0 Dilute it to 1/10 th initial concentration Q = (x/10)(x/10) = x2 = 1/10 Ka [HA]/10 10 [HA] Since Q < Ka, the reaction moves to the right And you get a greater percent dissociation

  36. Ka from % Dissociation • Lactic acid is 3.7% dissociated @ 0.100 M • HC3H5O  H+ + C3H5O- • Ka = [H+][C3H5O -] [HC3H5O] 3.7% = x x 100 x = 3.7 x 10 -3 [HC3H5O] Ka = [H+][C3H5O -] = (3.7 x 10 –3) (3.7 x 10 –3) [HC3H5O] 0.100 Ka = 1.4 x 10 -4

  37. [H+] [Bz-] [HBz] Weak acid equilibria Example Determine the pH of a 0.10 M benzoic acid solution at 25 oC if Ka = 6.5 x 10-5 HBz(aq) + H2O(l) H+(aq) + Bz-(aq) • The first step is to write the equilibrium expression. • Ka =

  38. Weak acid equilibria HBz H+Bz- Initial conc., M 0.10 0.00 0.00 Change, DM -x+x +x Eq. Conc., M 0.10 - x xx [H+] = [Bz-] = x We’ll assume that [Bz-] and [H+] are negligible compared to [HBz], since the value of the Ka<< [HBz]. (6.5 x 10-5 << 0.10 M)

  39. x2 0.10 Weak acid equilibria Solve the equilibrium equation in terms of x Ka = 6.5 x 10-5 = x = (6.5 x 10-5 )(0.10) = 0.00255 M H+ pH = - log (0.0025 M) = 2.6

  40. x2 0.10-x -b + b2 - 4ac 2a Weak acid example Now, lets go for the exact solution Earlier, we found that for 0.10 M benzoic acid Ka = 6.5 x 10-5 =  X2 + 6.5 x 10-5 X - 6.5 x 10-6 = 0 Use the quadratic equation to solve for x. X =

  41. Weak acid example -6.5 x 10-5 + [(6.5 x 10-5)2 +4 x 6.5 x 10-6]1/2 2 X = X = 0.00252 M H+ pH = - log (0.00252 M) = 2.599 versus pH = - log (0.00255 M) = 2.593 In this case, there is no significant difference between our two answers. If the Ka value is more than 2 powers of 10 different than the [acid], you can ignore the change in [acid].

  42. [OH-] [BH+] [B] Dissociation of bases, Kb The ionization of a weak base can also be expressed as an equilibrium. B (aq)+ H2O(l) BH+(aq) +OH-(aq) The strength of a weak base is related to its equilibrium constant, Kb. Kb =

  43. [OH-] [HBz] [Bz-] Weak base equilibria Example The Bz-(aq) formed in the benzoic acid solution is a weak conjugate base. Determine the pH of a 0.10 M sodium benzoate solution NaBz(aq), at 25 oC Bz-(aq) + H2O(l) HBz(aq) + OH-(aq) The Na+(aq) are spectator ions, and are not part of the equilibrium expression. Kb =

  44. Weak base equilibria Example The Kb value is related to the Ka value by the equation Ka x Kb = Kw = 1.0 x 10-14 [H+] [Bz-] [OH-] [HBz] = [H+] [OH-] [HBz] [Bz-] Kb= Kw /Ka = 1.0 x 10-14 / 6.5 x 10-5 = 1.5 x 10-10

  45. Weak base equilibria Bz - OH- HBz Initial conc., M 0.10 0.00 0.00 Change, DM -x+x +x Eq. Conc., M 0.10 - x x x [OH-] = [HBz] = x We’ll assume that [HBz] and [OH-] are negligible compared to [Bz -], since the value of the Ka << [Bz -]. (1.5 x 10-10 << 0.10 M)

  46. x2 0.10 Weak base equilibria Solve the equilibrium equation in terms of x Kb = 1.5 x 10-10 = x = (1.5 x 10-10 )(0.10) = 3.9 x 10-6 M pOH = - log (3.9 x 10-6 M) = 5.4 pH = 14 - pOH = 8.6

  47. Relationship Between Ka and Kb • What happens when you multiply Ka and Kb together? • Ka x Kb = [H+][OH-] = Kw = 1.0 x 10-14 • And, just like pH + pOH = 14.00 for strong acids/bases at standard temperature… • pKa + pKb = pKw = 14.00

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