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Chapter 14: Acids and Bases

Chapter 14: Acids and Bases

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Chapter 14: Acids and Bases

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  1. Chapter 14: Acids and Bases • Acids • Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. • React with certain metals to produce hydrogen gas • React with carbonates and bicarbonates to produce carbon dioxide gas • Bases • Have a bitter taste, feel slippery, • Many soaps contain bases

  2. Arrhenius Definition • Acids produce hydrogen ions in aqueous solution. • Bases produce hydroxide ions when dissolved in water. • Limits to aqueous solutions. • Only one kind of base. • NH3 ammonia could not be an Arrhenius base.

  3. Bronsted-Lowry Definitions • An acid is an proton (H+) donor and a base is a proton acceptor. • Acids and bases always come in pairs. • HCl is an acid. • When it dissolves in water it gives its proton to water. • HCl(g) + H2O(l) ↔ H3O+ + Cl- • Water is a base -makes hydronium ion

  4. Pairs • The general reaction that occurs when an acid is dissolved in water can best be represented by the following equation: HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) Acid + Base ↔ Conjugate acid + Conjugate base This is an equilibrium

  5. In fact… HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) …there is a competition for the H+between H2O and A- • The stronger base controls the direction of the shift. • If H20 is stronger than A- SHIFT RIGHT • If A- is stronger than H20 SHIFT LEFT

  6. So what? • What is the equilibrium expression for the general equation? • Called the acid dissociation constant.

  7. Because [H2O] is constant • It is NOT included in the equilibrium expression…making Ka the same form as that for simple dissociation into acids. HA(aq) ↔ H+(aq) + A-(aq) BUT Don’t forget that water has an important role in causing the acid to ionize!!

  8. Try this!! • Write the simple dissociation reaction for each of the following acids…(omit water) • Perchloric acid (HClO4) • The ammonium ion (NH4+) • Hydrofluoric acid (HF)

  9. 14.2 Acid Strength Acid strength is defined by the equilibrium position of its dissociation reaction. Strong acids Ka is large [H+] ~ [HA] A- is a weaker base than water Equilibrium lies far to the right HA completely dissociates Weak acids Ka is small [H+] <<< [HA] A- is a stronger base than water Equilibrium lies far to the left HA still present at equilibrium

  10. Opposites • The weaker the acid, the stronger the conjugate base. • Water is not strong enough to attract the H+ • The stronger the acid, the weaker the conjugate base • Water wins the struggle for the H+

  11. Types of acids • Monoprotic – one acidic proton • Polyprotic Acids- more than 1 acidic hydrogen (diprotic, triprotic). • Sulfuric, Phosphoric • Oxyacids - Proton is attached to the oxygen of an ion. • Organic acids contain the Carboxyl group -COOH with the H attached to O • Generally very weak.

  12. Amphoteric • A substance that can act as both an acid and a base is said to be amphoteric • Water is the most common!!! • Seen by its ability to autoionize (transfer a proton from one water molecule to another to produce a hydronium ion and the hydroxide ion) 2H2O(l) ↔ H3O+(aq) + OH-(aq)

  13. 2H2O(l) ↔ H3O+(aq) + OH-(aq) • Ion product constant/dissociation constant The product of the molar concentrations of H+ and OH- ions at a particular temperature. At 25oC Kw=1.0x10-14

  14. For ANY aqueous solution @ 25oC • Neutral solution [H+] = [OH-]= 1.0 x10-7 • Acidic solution [H+] > [OH-] • Basic solution [H+] < [OH-]

  15. = [OH-] = Kw 1 x 10-14 1.3 [H+] • What is the concentration of OH- ions in a HCl solution whose hydrogen ion concentration is 1.3 M? Kw = [H+][OH-] = 1.0 x 10-14 [H+] = 1.3 M = 7.7 x 10-15M

  16. Acetic acid (HC2H3O2) and HCN are both weak acids. Acetic acid is a stronger acid than HCN. HC2H3O2Ka = 1.8 x 10-5 HCN Ka = 6.2 x 10-10 Calculate the Kb values for C2H3O2− CN−

  17. = [H+] = Kw 1 x 10-14 2.0x10-2 [OH-] Try this one • Calculate [H+] in a solution in which [OH-] = 2.0x10-2M. Is the solution acidic, neutral, or basic? Kw = [H+][OH-] = 1.0 x 10-14 [OH-] = 2.0x10-2M = 5x10-13M

  18. 14.3 pH Scale • pH = pouvoirhydrogene (Fr.) “hydrogen power” • pH scale provides a convenient way to represent solution acidity • definition: pH = -log[H+] • in neutral pH = -log(1 x 10-7) = 7.0 • The number of decimal places for a log = number of sig figs in the original number.

  19. Other important relationshipsgiven any one of these we can find the other three. pOH= -log[OH-] The relationship between pH and pOH from the Kw Kw= [H+] [OH-] = 1.0x10-14 Take the –log of both sides -log([H+] [OH-]) = -log(1.0x10-14) -log[H+] + -log[OH-] = 14 pH + pOH = 14

  20. Calculating pH and pOH • If a solution has a [H+] of .0035M what is the pH? • [OH-]? • pOH? OR

  21. Can you… …find the [H+] from the pH? Sure, it is the inverse operation. Inverse log(-pH) = [H+] So if the pH is 7.0 Inverse log (-7) = 1x10-7

  22. How about these? • The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater? pH = -log[H+] Inverse log(-pH) = [H+] = 10-4.82 = 1.5 x 10-5M [H+] = 10-pH

  23. The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood? There is more than one way to complete this problem.

  24. Now for the fun stuff  • Remember aqueous solutions contain many components and doing the quantitative descriptions of the equilibria can be tough if we don’t keep the following strategies in mind: • Think chemistry: focus on the components (major ions in the solution) • Be systematic • Be flexible: there’s no ONE way to do these • Be patient: break the problem into steps • Be confident: YOU CAN DO IT!!

  25. Thinking about acid-base problems • What are the major species in solution? • What is the dominant reaction that will take place? • Is it an equilibrium reaction or a reaction that will go essentially to completion? • React all major species until you are left with an equilibrium reaction. • Solve for the pH.

  26. Let’s look at a bunch of problems Consider an aqueous solution of 2.0 x 10-3MHCl What are the major species in solution? H+Cl- H2O

  27. Consider a 0.80 M aqueous solution of the weak acid HCN (Ka = 6.2 x 10-10). What are the major species in solution? HCN H2O THINK!!! Why aren’t H+ or CN- major species?

  28. Which reaction controls the pH? Explain. HCN(aq) + H2O H3O+ (aq) + CN-(aq) Ka = 6.2 x 10-10 H2O + H2O H3O+(aq) + OH-(aq) Kw = 1.0 x 10-14 • Since Ka is > Kw the first reaction is the dominant primary reaction

  29. Common strong acids • Remember if it is a strong acid, it is said to completely dissociate in water. HCl HNO3 H2SO4 HClO4 [H+] = [HA] [OH-] is going to be small because of equilibrium If [HA] < 10-7 water contributes H+

  30. 14.4 Calculating the pH of Strong Acids • Now that we know how to find the major species and what controls the pH, let’s actually calculate the pH of some reactions. Calculate the pH of 0.6 M HCl(aq) What do we know? HCl = strong acid = complete dissociation pH = -log(0.6)

  31. What is the pH of a 2 x 10-3 M HNO3 solution? • HNO3 is a strong acid – 100% dissociation HNO3(aq) + H2O (l) ↔ H3O+(aq) + NO3-(aq) Since H+ comes from the acid pH = -log(2x10-3 )= 2.7

  32. 14.5 Calculating pH of weak acid solutions Calculate the pH of a 1.00M solution of HF (Ka = 7.2 x 10-4) Let’s proceed the same way we did with strong acids. • Start by writing the major species in the solution HF and H2O

  33. Then… • Decide which of the major species furnishes the H+ HF ↔ H+ + F- Ka = 7.2x10-4 H2O ↔ H+ + OH- Ka = 1x10-7 Although HF is a weak acid, you can see that it is stronger than water . Therefore, the first reaction is the dominant reaction.

  34. Here’s where things get different Create an ICE diagram of what we know about the components.

  35. Ka = 7.2x10-4 • Set up your equilibrium expression • We know that HF will only slightly dissociate and Ka of HF is so small compared to 1.00, the “x” in the denominator won’t affect this expression.

  36. From Ka to pH

  37. Let’s recap the steps to typical equilibrium problems with a weak acid. • List the major species in the solution • Choose the species that can produce H+, and write balanced equations for the reactions. • Using the values of the equilibrium constants for the reactions you have written, decide which will dominate in producing H+

  38. Write the equilibrium expression for the dominant equilibrium • Do an ICE diagram • Substitute the equilibrium concentrations into the expression • Solve for x (assuming that [HA]0-x≈[HA]0) • Verify the approximation is < ±5% • Calculate [H+] and pH

  39. Let’s try a few more problems with weak acids • Calculate the pH of a 0.036M nitrous acid (HNO2) solution (Ka = 4.5x10-4) Major species? HNO2 and H2O Both produce H+ compare Ka values: HNO2 is stronger Ice diagram?

  40. Assuming x is really small and we omit it… x = 4.0 x 10-3M Test the approximation: This is more than 5% so we must go back and leave x in and do the quadratic formula

  41. Since x can’t be negative, we know the concentration of x is 3.8 x 10-3M.

  42. What if… • …I’m given the pH and need to find Ka? The pH of a 0.10M solution of formic acid (HCOOH) is 2.39 What is the Ka of the acid? Major species? HCOOH and H2O Find [H+] from pH

  43. Look at the changes (ICE) • Now find Ka:

  44. Mixtures of Acids • The process is the same. • Determine the major species. • The stronger acid will predominate. • The one with the bigger Ka if concentrations are comparable

  45. Calculate the pH of a mixture 2.00M formic acid (HCOOH, Ka = 1.77 x 10-4) and 1.50M hydrobromous acid (HOBr, Ka= 2.06 x 10-9). • What are the concentrations of both the hypobromite ion (OBr-) and hydroxide ion at equilibrium?

  46. Major species (sources of H+) • HCOOH ↔ H+ + COOH- Ka= 1.77 x 10-4 • HOBr ↔ H+ + OBr- Ka = 2.06 x 10-9 • H2O ↔ H+ + OH- Ka = 1.0 x 10-7 • Which is the dominant reaction? • Largest Ka is HCOOH therefore it is the supplier of the H+ • Now What? ICE diagram

  47. Assuming x is “negligible”

  48. But…I’m looking for concentrations of OBr- and OH- • What do I know? • HCOOH ↔ H+ + COOH- Ka= 1.77 x 10-4 • HOBr ↔ H+ + OBr- Ka = 2.06 x 10-9 • H2O ↔ H+ + OH- Kw = 1.0 x 10-14 • 1.5M Hydrobromous acid • So… And THEN…

  49. How about this… • How would you prepare 1600mL of a pH = 1.5 solution using concentrated (12M) HCl?