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Chapter 14 Acids and Bases

Chapter 14 Acids and Bases. Arrhenius: 1 st to Define Acids. Acids  H + (aq) Bases  OH - (aq) 3 Problems with definition: Limited to aqueous solutions Only one kind of base: OH - NH 3 could not be base. Brønsted-Lowry Definitions. Acid proton (H + ) donor Base proton acceptor

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Chapter 14 Acids and Bases

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  1. Chapter 14 Acids and Bases

  2. Arrhenius: 1st to Define Acids • Acids H+(aq) • Bases  OH-(aq) • 3 Problems with definition: • Limited to aqueous solutions • Only one kind of base: OH- • NH3 could not be base.

  3. Brønsted-Lowry Definitions • Acid proton (H+) donor • Base proton acceptor • Acids and bases always come in pairs. • Example H2O(l) + HCl(g)⇌H3O+ + Cl- • Water as base  Hydronium ion • Remember – when acids are added to water the process is Dissociation

  4. Bronsted Lowry • General Reaction: • HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq) • Acid Base Conjugate acidConjugate base • There are 2 conjugate acid-base pairs • HA & A- • H2O & H3O+ • This is equilibrium • Competitionfor H+between H2O and A- • The stronger base controls direction.

  5. Acid Dissociation Constant Ka • HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq) • The equilibrium expression is: • Ka is the acid dissociation constant • Ignore the water in equation because it’s (l) and its concentration remains constant.

  6. Acid Dissociation Constant Ka • H3O+ is often written H+ • HA(aq) ⇆ H+(aq) + A-(aq) • We can write the expression for any acid. • Strong acids dissociate completely. • Equilibrium far to right. • Conjugate base must be weak.

  7. Write The Dissociation Reaction • Example Hydrochloric acid: HCl(aq)⇔ H+(aq) + Cl-(aq) Your Turn Write the Reactions: • Acetic Acid HC2H3O2 • NH4+ • (C6H5NH3+) Which H? • [Al(H2O)6]+3

  8. Acid Strength Table 14.1 p.627 Strength of acid is defined by the equilibrium position of its dissociation reactions.

  9. Strong Acids Strong SolubleBases • Hydrochloric Acid • Hydrobromic Acid • Hydroiodic Acid • Nitric Acid • Perchloric Acid • Chloric Acid • Sulfuric Acid • Lithium Hydroxide • Sodium Hydroxide • Potassium Hydroxide • Rubidium Hydroxide • Cesium Hydroxide • Calcium Hydroxide • Strontium Hydroxide • Barium Hydroxide KNOW THESE

  10. Rules for Writing and Naming Acidsreview p.67 • All inorganic acid formulas start with H • Binary Acids • Two elements or H plus non-oxy polyatomic anion • Name comes from root of anion • Begins w/ “hydro-” & ends “-ic” • Ex. HCl: hydrochloric acid • Ex. HCN: hydrocyanic acid

  11. Rules for Writing and Naming Acids • Oxyacids • Anion contains Oxygen • Named from root name of anion • Use suffix “ic” or “-ous” • Anion end –ate Acid becomes -ic acid • Anion end –ite Acid becomes -ous acid • Ex. Nitrate⇒ HNO3 is Nitric Acid • Ex. Nitrite⇒ HNO2 is Nitrous Acid

  12. Other Types of Acids • Table 14.2 Common Monoprotic Acids – acids that donate only 1 acidic proton • Polyprotic Acids- more than 1 acidic proton donated (diprotic, triprotic) • Organic acids contain the Carboxyl group COOH with the H attached to O

  13. Water as an Acid & Base • Amphoteric: Behave As Acid & Base, water most common • Example: Auto-ionization of Water 2H2O(l)⇆ H3O+(aq) + OH-(aq) • Leads to equilibrium expression: Kw= [H3O+][OH-]=[H+][OH-] KW is dissociation constant for water At 25oC KW = 1.0 X 10-14 therefore [H+] = [OH-] = 1.0 X 10-7M

  14. Meaning of Kw • In any aqueous solution, we know KW = 1.0X10-14 • There are 3 possible situations: • Neutral solution [H+] = [OH-]= 1.0 x10-7 • Acidic solution [H+] > [OH-] • Basic solution [H+] < [OH-] • In each case: KW = 1.0X10-14

  15. Calculate [H+] or [OH-] Calculate [H+] for the following solution at 25oC and state whether the solution is neutral, acidic or basic: 1.0X10-5M OH- Solution: KW=[H+] [OH-]=1.0X10-14 , since [OH-] = 1.0X10-5M, solving for [H+] gives: Since: 1.0X10-5 [OH-] > 1.0X10-9 [H+] the solution is basic • Your Turn: • 1.0 X 10-7 M OH- • 10.0 M H+

  16. The pH Scale • pH= -log[H+] • Use pH scale because [H+] is usually very small Sig Figs Rule: Number of decimal places inlog = number of sig. figs. in original number • [H+] = 1.0 x 10-8 pH= 8.00 • Similar for other quantities: pOH= -log[OH-] pK = -log K 2 decimals 2 sig figs

  17. Calculating pH & pOH • Given 1.0 X 10-3 M OH- at 25oC • Calculate pH & pOH • Solution: Your turn: #42 p. 674

  18. Lots of Relationships • pH = -log[H+] • pOH = -log[OH-] • pK = -log K • KW = [H+][OH-] • log KW = log [H+]+ log[OH-] • -log KW = -log[H+] - log[OH-] • pKW = pH + pOH • Since KW = 1.0 x10-14 • thenpKW = -log (1.0 x10-14 )=14.00 • pH + pOH = 14.00 Given any one of these [H+], [OH-] ,pH, and pOH we can find the other three.

  19. [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

  20. Calculating pH, pOH, [H+], & [OH-] The pH of a sample of blood was measured to be 7.41 at 25oC. Calculate the pOH, [H+], and [OH-] for the sample. Since pH + pOH = 14.00 pOH = 14.00 – pH = 14.00 – 7.41 = 6.59 Since pH = -log[H+] 7.41 = -log[H+] take the antilog of -7.41=3.9x10-8 Similarly for pOH = -log[OH-] 6.59 = -log[OH-] take the antilog of -6.59=2.6x10-7 Your turn: #44 p. 674

  21. Calculate pH of Strong Acid (SA) Solutions Strong Acids: HCl, HBr, HI, HNO3, H2SO4, HClO4 • When working with a SA solution, focus on solution components • Determine which are significant and which can be ignored ALWAYS WRITE THE MAJOR SPECIES Calculate pH of 1.0M HCl solution There will be no HCl in solution because all SA completely dissociate Major species: H+, Cl-, H2O Because want pH, focus on species that produce H+ Consider H+ from H2O, the dissociation of HCl drives H+ contribution from water to left, so H+ from H2O is negligible [H+] is 1.0 and pH = -log[1.0] = 0

  22. Calculate pH of Strong Acid Solutions Calculate pH of 0.10 M HNO3 solution • Solution: • Since HNO3 is strong acid, Major species: H+, NO3-, H2O, OH- • [HNO3] virtually 0 because SA • [OH-] very small because H2O(l) [H+](aq) + [OH-](aq) is driven to left by H+ from acid (Le Chatelier’s Principle) • Source of H+ : HNO3, and H2O, negligible from H2O • Leaves HNO3- as source of H+ so: • [H+]=0.10 M and pH = -log(0.10) = 1.00 Your turn: #48 p.674

  23. Calculating pH of Weak Acid • Calculate the pH of a 1.00 M solution of HF with a Ka=7.2 X 10-4 • List major species: HF and H2O • Decide what is the major source of H+ because we are trying to calculate pH • Compare • Ka= 7.2 X 10-4 • Kw = 1.0 X 10-14

  24. Calculating pH of Weak Acid • Write equilibrium expression: • Be nICE and do ICE

  25. Calculating pH of Weak Acid • Substitute equilibrium concentrations into expression • We approximate [HF] to 1.00 M which allows us to simplify the calculation: • Solve for x = 2.7 X 10-2

  26. Validation of Answer • How valid is the approximation of [HF] = 1.00 M? • Test for validity of approximation: • Compare the size of x and [HA]0 • If the expression • Then the approximation is considered valid • In problem • Therefore the approximation is acceptable. • Now can plug it in and find pH

  27. Solving Weak Acid Equilibrium Problems • List major species • Choose the species that can produce H+ & write balanced equation for reactions producing H+ • Using Ka or Kw decide which equilibrium will dominate production of H+ • Write equilibrium expression for dominate equilibrium. • Do an ICE Table: • List the initial conc. of species participating in dominate equilibrium • Define change needed to achieve equilibrium in terms of x • Initial conc. + Change = Equilibrium Concentration • Substitute Equilibrium Concentration into expression #4 • Solve for x • Validate approximation for x • Calculate [H+] & pH

  28. Calculating pH of Weak Acid • Sample Exercise 14.8 p. 638 • Lets study this one together • Pay close attention to comments between equations and calculations. • Ask yourself why they do what they do? • You will use these steps in ALL problems in next couple of chapters, so learn them well! Your turn # 53 p.675

  29. Percent Dissociation aka Percent Ionization Weak acid: percent dissociation increases as acid becomes more dilute. Calculate the percent dissociation of acetic acid (Ka=1.8x10-5) for a 1.00M solution & 0.100 M solution Solution: • List major species: • HC2H3O2 & H2O • Choose species that produces most H+ : HC2H3O2& write equation: HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

  30. Percent Dissociation • Write equilibrium expression • Create ICE chart & fill in • Plug equilibrium conc. into equilibrium expression & solve for X (make appropriate approximations) • X = 4.2 X 10-3 M

  31. Percent Dissociation • Calc. percent dissociation: • You calculate % dissociation for 0.100 M solution. General Trend: For weak acid solutions Fig 14.10 p. 643 • [acid] and [H+] decrease as % dissociation increases • % dissociation increase as acid is diluted Calculating Ka from percent dissociation is illustrated in Sample Exercise 14.11 on p.643 STUDY this sample & do #65 p.675

  32. The pH of Weak Acid Mixture • The process is the same. STUDYSample Exercise 14.9 on page 639 • Determine the major species. • Key Point : Compare K values for all species that produce H+, the larger K will predominate. • Key Point: Remember there is only one kind of H+ in the solution and it does not matter from which acid it originates so you use the value in all [H+] • Your Turn: #62 p. 675

  33. POP QUIZ • A 0.0560 g sample of acetic acid is added to enough water to make 50.00 mL of solution. Calculate the [H+], [C2H3O2-], [HC2H3O2], and the pH at equilibrium. Ka for acetic acid is 1.8x10-5 • Show all setups and work

  34. Commercial Interruption • http://www.chemtutor.com/ • It’s free • It’s good • It comes w/ NSF recommendation • Use it!

  35. Bases • Remember the OH-list of strong bases. • 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. Not very soluble. • Used as antacids because [OH-] can’t build up.

  36. Bases without OH- • Bases are proton acceptors (Brønsted-Lowry). • Example: NH3 + H2O ⇆ NH4+ + OH- • NH3 is a base. It is the lone pair on nitrogen that accepts the proton, H2O is acid • General reaction between base (B) & water B(aq) + H2O(l) ⇆ BH+(aq) + OH- (aq) • Equilibrium constant

  37. pH of Weak Base Calculate the pH for a 15.0M solution of NH3 (Kb=1.8x10-5) • List Major Species: NH3 & H2O (Which one dominates the production of OH-?) • Use Kb and Kw values to decide. • Write simple dissociation reaction • Write equilibrium expression • Build an ICE table • Substitute equilibrium concentrations into the equilibrium expression and solve for x

  38. pH of Weak Base • Apply 5% validation rule • Plug the value of x back into Kw=[H+][OH-] and solve for [H+] • Find pH Appendix Five on pages A22-23 have 3 tables with Ka and Kb values for common acids and bases.

  39. Polyprotic Acids 14.7 • Polyprotic acids furnish more than 1 proton • Always dissociate stepwise i.e. 1 proton at a time • The first H+ comes off much easier than the second. • Denoted Ka1, Ka2, Ka3 • Ka1 is much bigger than Ka2 • Ka1 > Ka2 > Ka3 for typical weak acid

  40. Polyprotic acid • H2CO3⇆ H+ + HCO3- Ka1= 4.3 x 10-7 • HCO3-⇆H+ + CO3-2 Ka2= 4.3 x 10-10 • Base (HCO3- )in first step is acid in second. • In calculations we can normally ignore the second dissociation because the other steps make very small contributions to [H+]

  41. Calculate the Concentration Calculate the pH of a 5.0 M H3PO4 solution and the equilibrium concentration of the species H3PO4, H2PO4-, HPO4-2, and PO4-3 • Use table 14.4 p. 651 for Ka values • List major species & write balanced equation • Use Ka & Kw to decide dominant equilibrium • Write equilibrium expression • Create ICE table • Solve for x • Repeat for each successive species

  42. Acid-Base Properties of Salts 14.8 • Salt is another name for ionic compounds • Cations of strong bases(Na+ & K+) have no effect on [H+] • Anion of strong acids(Cl- & NO3-) have no effect on [H+] • Thus these salts NaCl, KNO3 have no effect on [H+] • Aqueous solutions of these salts are neutral (pH = 7) • IF the salt has an anion that produces a strong conjugate base, the solution is basic • Example: Na(C2H3O2)

  43. Basic Salts • For any salt solution • That has a neutral cation (cation from strong base) • And an anion which is the conjugate base of a weak acid • the solution will be basic • For any weak acid and it’s conjugate base, If you know one (Ka or Kb ) you can calculate the other because you always know KW Ka x Kb = KW = 1.0x10-14

  44. Salts as Weak Base • The anion of a weak acid is a weak base. • Calculate the pH of a 0.30 M NaF solution KaHF = 7.2 x 10-4 • List major species (which one dominates) • Write equilibrium reaction • Kb expression from Ka x Kb = Kw • Create ICE table • Solve for x and use 5% rule to validate approx. • Calculate pH

  45. Acidic salts • Salts in which anion is not a base and cation is the conjugate acid of weak base produces acidic solution • The same development as bases leads to • Ka x Kb = KW • Calculate the pH of a 0.10 M NH4Cl , Kb of NH3 1.8 x 10-5 • A 2nd type of acidic salts are those of highly charged metal ions, Example Al+3

  46. Oxides of the Representative Elements

  47. Qualitative PredictionsSalts w/ Acidic & Basic Properties • Ka > Kb acidic • Ka < Kb basic • Ka = Kb Neutral

  48. Effect of Structure on Acid-base Properties 14.9 • Any molecule with an H in it is a potential acid. • The stronger the X-H bond the less acidic Table 14.7 • The more polar the X-H bond the stronger the acid (use electronegativity). • The more polar H-O-X bond -stronger acid.

  49. Strength of Oxyacids • The more oxygen hooked to the central atom, the more acidic the hydrogen. HClO4 > HClO3 > HClO2 > HClO • Remember that the H is attached to an oxygen atom. See Ka values on Table 14.8 • The oxygens are electronegative & pull electrons away from hydrogen allowing it to dissociate more completely

  50. Cl O H Strength of oxyacids Electron Density

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