1 / 93

Acids & Bases

Acids & Bases. Unit 13. Overview. Acid/base properties Definitions Arrhenius Bronsted-Lowry Lewis Acid-Base Reactions Neutralization Sulfides Carbonates pH/ pOH Scale Acid/base strength Factor affecting K a , K b , K w Percent ionization. Vocabulary Polyprotic Acids

Download Presentation

Acids & Bases

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 Unit 13

  2. Overview • Acid/base properties • Definitions • Arrhenius • Bronsted-Lowry • Lewis • Acid-Base Reactions • Neutralization • Sulfides • Carbonates • pH/pOH Scale • Acid/base strength • Factor affecting • Ka, Kb, Kw • Percent ionization • Vocabulary • Polyprotic Acids • Amphoteric • Anhydrides • Acids/Bases & Salts • Determine acidity • Calculations • Common Ion Effect • Buffers • Henderson-Hasselbalch • Titration • Indicators • 4 types of curves

  3. Acids • Sour taste • React with active metals to produce hydrogen gas • Change the color of acid-base indicators • React with bases to produce salt and water • Conduct an electric current (electrolytes) • Turn litmus paper red

  4. Common Acids • Sulfuric Acid • Car batteries; production of metals, paints, dyes, detergents • Nitric Acid • Explosives, pharmaceuticals, rubber, plastics, dyes • Phosphoric acid • Soda, fertilizers, animal feed, detergents • Hydrochloric Acid • Stomach acid, cleaning metals, found in hardware stores (muriatic acid) • Acetic Acid • Vinegar, food supplements, fungicide • Citric Acid • Fruit juices

  5. Acids • Binary acids • Contain only two different elements • Name as “hydro - ic acid” • Example: HCl (hydrochloric acid) • Oxyacids • Acid consisting of hydrogen and a polyatomic anion that contains oxygen (oxyanion) • To name, drop ending of polyatomic ion and ad “- ic acid” • Example: HNO3 (nitric acid)

  6. Bases • Taste bitter • Feel slippery • Change the color of acid-base indicators • React with bases to produce salt and water • Conduct an electric current (electrolytes) • Turn litmus paper blue

  7. Common Bases • Ammonium hydroxide • Household cleaners, window cleaner • Ammonia • (Gas) inhalant to revive unconscious person • Sodium bicarbonate (baking soda) • Acid neutralizers in acid spills • Antacids for upset stomachs • Sodium hydroxide • Drain cleaner (drano), oven cleaner, production of soap • Magnesium hydroxide • Antacids, milk of magnesia, laxatives

  8. Definition: Arrhenius • Acid • Substance that ionizes in water and produces H+ ions • Example: HCl H+ + Cl- • Base • Substance that ionizes in water and produces OH- ions • Example: NaOH Na+ + OH-

  9. Definition: Bronsted-Lowry • Acid • Substance that is capable of donating a proton (H+ ion) • Base • Substance that is capable of accepting a proton (H+ ion)

  10. Examples: Bronsted-Lowry HC2H3O2 + H2O ↔ C2H3O2- + H3O+ Acids: HC2H3O2 and H3O+ Bases: H2O and C2H3O2- NH3 + H2O ↔ NH4+ + OH- Acids: H2O and NH4+ Bases: NH3 and OH- Notice that water can act as an acid or a base

  11. Bronsted-Lowry • Conjugate Pair – a BL acid/base pair(one with H+ and one without H+) • Examples: • HC2H3O2 and C2H3O2- • H3O+ and H2O • H2O and OH- • NH4+ and NH3

  12. Bronsted-Lowry • The more easily a substance gives up a proton, the less easily the conjugate base accepts a proton (and vice versa) • The stronger the acid, the weaker the conjugate base • The stronger the base, the weaker the conjugate acid

  13. Acid Base Reactions • Neutralization • Salt + water • Sulfides • Salt + sulfide gas • Carbonates • Salt + CO2 + H2O

  14. Neutralization • Solution of an acid and solution of a base are mixed • Products have no characteristics of either the acid or the base • Acid + Base (metal hydroxide)  salt + water • Salt comes from cation of base and anion of acid HY + XOH  XY + H2O HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

  15. Sulfides • Acid reacts with a sulfide • Gaseous product (H2S) has a foul odor (rotten eggs) • Acid + metal sulfide  salt + hydrogen sulfide • Salt comes from cation of sulfide and anion of acid HY + XS  XY + H2S HCl(aq) + Na2S(aq)  NaCl(aq) + H2S(g)

  16. Carbonates • Carbonates and bicarbonates react with acids HY + XHCO3 XY + H2CO3 H2CO3 is not stable so breaks into H2O and CO2 Then HY + XHCO3 XY + H2O + CO2 HCl(aq) + NaHCO3(aq)  NaCl(aq) + H2CO3 (aq) HCl(aq) + NaHCO3(aq)  NaCl(aq) + H2O(l) + CO2(g)

  17. The pH Scale • pH scale: measures concentration of hydrogen ions in solution pH = -log[H+]therefore [H+] = 10-pH • Example: What is the pH of a solution with a [H+] of 1.4×10-5? pH = -log[1.4×10-5] =4.9

  18. pH Scale Acids: pH < 7 Neutral: pH = 7 Bases: pH > 7 Increasing [H+] means decreasing pH Increasing pH means decreasing [H+]

  19. The pOH Scale • pOH scale: measures concentration of hydroxide ions in solution pOH = -log[OH-]therefore [OH-] = 10-pOH • Example: What is the [OH-] of a solution with a pOH of 6.2? [OH-] = 10-6.2= 6.3×10-7

  20. Comparing pH and pOH pH + pOH = 14 • An acid has a pH of 4, what is the pOH? 4 + pOH = 14 pOH = 10

  21. Measuring pH • pH meter • Electrodes measure [H+] • Acid-base indicators • Change color in presence of acid or base (or certain pH ranges) • Litmus paper, phenolphthalein, cabbage juice, methyl orange, thymol blue…

  22. Strong Acids and Bases • Completely ionize in solution (strong electrolytes) • If acid/base is not on this list, it is a weak acid/base

  23. Weak Acids and Bases • Do not completely ionize in water (weak electrolytes) • Common weak acids: • HF, acids with -COOH group • Common weak bases: • NH3

  24. Factors Affecting Acid Strength • Electronegativity of element bonded to H • Binary acids • More electronegative bond = stronger acid • Example: HCl stronger than HBr • Bond Strength • Stronger bonds do not allow hydrogen to dissociate as easily • Reason why HF is not a strong acid (F is most electronegative, but H-F bond is strongest bond) • Stability of Conjugate base • More stable the conjugate base, the stronger the acid

  25. Factors Affecting Acid Strength • For polyatomic ions, the more electronegative the nonmetal, the stronger the acid (when comparing acids with same number of O atoms) • Example: HClO3 is stronger than HBrO3 • For polyatomic ions, when nonmetal is the same, the more O atoms, the stronger the acid • Example: HClO3 is stronger than HClO2

  26. Percent Ionization • Tells us what percent of an acid (or base) is ionized in water • Helps determine the strength of an acid (or base) Percent Ionization = ×100 [H+] at equilibrium Initial Acid Concentration

  27. Percent Ionization (Example) • A 0.035 M solution of HNO2 contains 3.7×10-3 M H+(aq). Calculate the percent ionization. = = 11% This means that 11% of the acid will dissociate in water. 3.7×10-3 M 0.035 M

  28. Strong Acids HA(aq) + H2O(l)  H+(aq)+ A-(aq) acid water protonconjugate base • Strong acids dissociate completely • The dissociation is not reversible • The acid is the only significant source of H+ ions, so pH can be calculated directly from the [H+] • Example: A 0.20 M solution of HNO3 has an [H+] of 0.20 M • pH = -log[H+]

  29. Strong Bases • Strong bases dissociate completely • The dissociation is not reversible • The base is the only significant source of OH- ions, so pOH can be calculated directly from the [OH-] • Example: A 0.30 M solution of NaOH has a [OH-] of 0.30 M • pOH = -log[OH-]

  30. Equilibrium Time!

  31. Weak Acids and Bases • Do not dissociate completely • Reversible reactions • Need to use equilibrium to solve for [H+] K = [Products] [Reactants]

  32. Acid Dissociation HA(aq) + H2O(l)  H+(aq)+ A-(aq) acid water protonconjugate base • Write the equilibrium expression for the acid dissociation constant, Ka.

  33. Base Dissociation B(aq) + H2O(l) ↔BH+(aq)+ OH-(aq) base water conjugate hydroxide acid ion • Write the equilibrium expression for the base dissociation constant, Kb.

  34. Size of K • The greater the Ka, the stronger the acid • The smaller the Ka, the weaker the acid • The greater the Kb, the stronger the base • The smaller the Kb, the weaker the base

  35. Example: Weak Acid Equilibrium Problem • What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? • Step #1: Write the dissociation equation HC2H3O2 C2H3O2- + H+

  36. Example: Weak Acid Equilibrium Problem • What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? • Step #2: ICE HC2H3O2 C2H3O2- + H+ 0.50 0 0 +x +x - x x x 0.50 - x

  37. Example: Weak Acid Equilibrium Problem • What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? • Step #3: Set up the equilibrium expression If percent ionization is less than 5%, you can ignore using the quadratic.

  38. Example: Weak Acid Equilibrium Problem • What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? • Step #5: Solve for pH You can use the Kb expression to solve for pOH using the same method!

  39. Example 2: Weak Acid Equilibrium - Solving for Ka • A student prepared a 0.10 M solution of formic acid (HCOOH) and measured its pH. The pH at 25°C was found to be 2.38. Calculate the Ka for formic acid at this temperature. • Step #1: Solve for [H+] from pH [H+] = 10-2.38 = 4.2×10-3 M

  40. Example 2: Weak Acid Equilibrium - Solving for Ka • A student prepared a 0.10 M solution of formic acid (HCOOH) and measured its pH. The pH at 25°C was found to be 2.38. Calculate the Ka for formic acid at this temperature. • Step #2: Set up ICE table HCOOH(aq)  HCOO- + H+ 0.10 0 0 4.2×10-3

  41. Example 2: Weak Acid Equilibrium - Solving for Ka • A student prepared a 0.10 M solution of formic acid (HCOOH) and measured its pH. The pH at 25°C was found to be 2.38. Calculate the Ka for formic acid at this temperature. • Step #3: Use stoichiometry to complete table HCOOH(aq)  HCOO- + H+ 0.10 0 0 4.2×10-3 4.2×10-3 4.2×10-3 4.2×10-3 4.2×10-3 0.0096

  42. Example 2: Weak Acid Equilibrium - Solving for Ka • A student prepared a 0.10 M solution of formic acid (HCOOH) and measured its pH. The pH at 25°C was found to be 2.38. Calculate the Ka for formic acid at this temperature. • Step #4: Solve for Ka using equilibrium expression Ka = 1.8×10-4

  43. Rule of Thumb The larger the value of Ka, the stronger the acid

  44. (Self-) Auto-ionization of Water • According to Bronsted Lowry, H2O can act as either an acid or a base • Auto-ionization: One water molecule can donate a proton to another water molecule • Extremely rapid reaction and no molecule remains ionized for long • At room temperature 1 out of every 109 molecule are ionized at a given instant • Water is a nonelectrolyte and consists almost entirely of H2O molecules H2O(l) + H2O(l) ↔ H3O+ + OH-

  45. Auto-ionization of Water • H2O(l) + H2O(l) ↔ H3O+ + OH- • Auto-ionization of water is an equilibrium process (use Kw - ion product constant) Kw = [H3O+][OH-] Also written as Kw = [H+][OH-] • At 25°C, Kw =1.4×10-14

  46. Auto-ionization of Water 1.4×10-14 = [H+][OH-] • In basic solutions, [OH-] > [H+] • In acidic solutions, [H+] > [OH-] • In neutral solutions, [H+] = [OH-]

  47. Auto-ionization of Water 1.4×10-14 = [H+][OH-] • If the concentration of one ion is known, you can solve for the concentration of the other ion Example: Calculate the concentration of H+ in a solution in which the concentration of OH- is 0.010M. 1.4×10-14 = [H+][0.010] [H+] = 1.0×10-12 M

  48. Relating pKw to pKa and pKb • Acid or base dissociation constants are sometimes expressed as pKa and Kb. • pKa = –logKa • pKb = -logKb pKw = 14 = pKa+ pKb

  49. Polyprotic Acids • Acids with more than one ionizable H+ ion • The acid-dissociation constants are Ka1, Ka2, etc… • The first proton is most easily removed • As protons are removed, it becomes more and more difficult to remove protons • Ka1>Ka2>Ka3…. H2SO3(aq) ↔ H+(aq) + HSO3-(aq) Ka1 = 1.7×10-2 HSO3-(aq) ↔ H+(aq) + SO3-2(aq) Ka2 = 6.4×10-8

  50. Polyprotic Acids • To calculate the overall K for the reaction, treat it as a multi-step equilibrium Overall reaction… H2SO3(aq) ↔ H+(aq) + HSO3-(aq) Ka1 = 1.7×10-2 HSO3-(aq) ↔ H+(aq) + SO3-2(aq) Ka2 = 6.4×10-8 H2SO3(aq) ↔ 2H+(aq) + SO3-2(aq) Ka = Ka1 × Ka2 Ka = (1.7×10-2)(6.4×10-8) = 1.1×10-9

More Related