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Chapter 14 Acids and Bases

Chapter 14 Acids and Bases. 2006, Prentice hall. CHAPTER OUTLINE. GENRAL PROPERTIES OF ACIDS & BASES. Many common substances in our daily lives are acids and bases.

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Chapter 14 Acids and Bases

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  1. Chapter 14 Acids and Bases 2006, Prentice hall

  2. CHAPTER OUTLINE

  3. GENRAL PROPERTIESOF ACIDS & BASES • Many common substances in our daily lives are acids and bases. • Oranges, lemons and vinegar are examples of acids. In addition, our stomachs contain acids that help digest foods. • Antacid tablets taken for heartburn and ammonia cleaning solutions are examples of bases.

  4. GENRAL PROPERTIESOF ACIDS • General properties associated with acids include the following: • sour taste • change color of litmus from blue to red • react with metals to produce H2 gas • react with bases to produce salt & water

  5. Hydrofluoric acid Structure of Acids • binary acids have acid hydrogens attached to a nonmetal atom • HCl, HF

  6. Structure of Acids • oxy acids have acid hydrogens attached to an oxygen atom • H2SO4, HNO3

  7. Structure of Acids • carboxylic acids have COOH group • HC2H3O2, H3C6H5O3 • only the first H in the formula is acidic • the H is on the COOH

  8. GENRAL PROPERTIESOF BASES • General properties associated with bases include the following: • bitter taste • slippery, soapy feeling • change color of litmus from red to blue • react with acids to produce salt & water

  9. Structure of Bases • most ionic bases contain OH ions • NaOH, Ca(OH)2 • some contain CO32- ions • CaCO3 NaHCO3 • molecular bases contain structures that react with H+ • mostly amine groups

  10. amine groups Ammonia H – N – H H + H+ H H – N+ – H H Ammonium ion

  11. ARRHENIUSACIDS & BASES According to the Arrhenius definition, • Acids are substances that produce hydronium ions (H3O+) in aqueous solution. • The most common definition of acids and bases was formulated by the Swedish chemist Svante Arrhenius in 1884. HCl(g) + H2O (l) H3O+ (aq) + Cl– (aq) HCl(g) H+(aq) + Cl– (aq) Commonly written as H2O

  12. ARRHENIUSACIDS & BASES According to the Arrhenius definition, • Bases are substances that produce hydroxide ion (OH-) in aqueous solution. NaOH(s) Na+(aq) + OH–(aq) NH3 (aq) + H2O (l) NH4+ (aq) + OH – (aq) H2O

  13. BRØNSTED-LOWRYACIDS & BASES • The Arrhenius definition of acids and bases is limited to aqueous solutions. • A broader definition of acids and bases was developed by Brønsted and Lowry in the early 20th century. • According to Brønsted-Lowry definition, an acid is a proton donor, and a base is a proton acceptor. A substance that can act as a Brønsted-Lowry acid and base (such as water) is called amphiprotic. Base Acid Acid Base NH3 (aq) + H2O (l) → NH4+ (aq) + OH– (aq) HCl(g) + H2O (l) → H3O+ (aq) + Cl– (aq)

  14. BRØNSTED-LOWRYACIDS & BASES • In Brønsted-Lowry definition, any pair of molecules or ions that can be inter-converted by transfer of a proton is called conjugate acid-base pair. HCl(g) + H2O (l) → H3O+ (aq) + Cl– (aq) Conjugate acid Conjugate base Acid Base

  15. BRØNSTED-LOWRYACIDS & BASES NH3 (aq) + H2O (l) → NH4+ (aq) + OH– (aq) Base Acid Conjugate acid Conjugate base

  16. Example 1: Identify the conjugate acid-base pairs for each reaction shown below: H2O + Cl HCl + OH Conjugate acid Acid Base Conjugate base

  17. Example 1: Identify the conjugate acid-base pairs for each reaction shown below: C6H5OH + C2H5O C6H5O + C2H5OH Conjugate base Conjugate acid Acid Base

  18. Example 2: Write the formula for the conjugate acid for each base shown: HS + H+ H2S NH3 + H+ NH4+ CO32 + H+ HCO3

  19. Example 3: Write the formula for the conjugate base for each acid shown: HI - H+ I CH3OH - H+ CH3O HNO3 - H+ NO3

  20. ACID & BASESTRENGTH • According to the Arrhenius definition, the strength of acids and bases is based on the amount of their ionization in water. • Strong acids and bases are those that ionize completely in water. • Strong acids and bases are strong electrolytes. 100 100 100 1M 1M 1M

  21. ACID & BASESTRENGTH • Weak acids and bases are those that ionize partially in water. • Weak acids and bases are weak electrolytes. 100 ~1 ~1 1M ~0.01M ~0.01M

  22. IONIZATION OFSTRONG vs. WEAK ACIDS Ionizes completely Ionizes partially

  23. COMMON ACIDS

  24. COMMON BASES

  25. COMPARISON OFACIDS & BASES

  26. Titration • using reaction stoichiometry to determine the concentration of an unknown solution • Titrant (unknown solution) added from a buret • indicators are chemicals added to help determine when a reaction is complete • the endpoint of the titration occurs when the reaction is complete

  27. Titration

  28. Titration The base solution is the titrant in the buret. As the base is added to the acid, the H+ reacts with the OH– to form water. But there is still excess acid present so the color does not change. At the titration’s endpoint, just enough base has been added to neutralize all the acid. At this point the indicator changes color.

  29. Example 1 • The titration of 10.00 mLHCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Find: concentration HCl, M • Collect Needed Equations and Conversion Factors: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 1 mole HCl = 1 mole NaOH 0.100 M NaOH0.100 mol NaOH 1 L sol’n

  30. = 1.25 x 10-3 mol HCl molarity = 0.125 M HCl

  31. IONIZATIONOF WATER • Water can act both as an acid and a base. • In pure water, one water molecule donates a proton to another water molecule to produce ions. H2O + H2O  H3O+ + OH– Conjugate acid Acid Base Conjugate base

  32. IONIZATIONOF WATER • In pure water, the transfer of protons between water molecules produces equal numbers of H3O+ and OH– ions. • The number of ions produced in pure water is very small, as indicated below: • When the concentrations of H3O+ and OH– are multiplied together, the ion-product constant (Kw) is formed. • All aqueous solutions have H3O+ and OH– ions. • An increase in the concentration of one of the ions will cause an equilibrium shift that causes a decrease in the other one. [H3O+] = [OH–] = 1.0 x 10-7 M Kw=[H3O+][OH–] =[1.0x10-7][1.0x10-7]=1.0x10-14

  33. ACIDIC & BASICSOLUTIONS • When [H3O+] and [OH–] are equal in a solution, it is neutral. • When [H3O+] is greater than [OH–] in a solution, it is acidic. • For example, if [H3O+] is 1.0 x 10–4 M, then [OH–] would be 1.0 x 10–10 M.

  34. ACIDIC & BASICSOLUTIONS • When [OH-] is greater than [H3O+] in a solution, it is basic. • For example, if [OH-] is 1.0 x 10–6 M, then [H3O+] would be 1.0 x 10–8 M.

  35. ACIDIC & BASICSOLUTIONS Basic Acidic Neutral [H3O+]>[OH-] [H3O+]=[OH-] [H3O+]<[OH-]

  36. pH SCALE • The acidity of a solution is commonly measured on a pH scale. • The pH scale ranges from 0-14, where acidic solutions are less than 7 and basic solutions are greater than 7. pH = -log [H3O+]

  37. pH SCALE Acidic solutions pH < 7 H3O+ > 1x10-7 Neutral solutions pH = 7 H3O+ = 1x10-7 Basic solutions pH > 7 H3O+ < 1x10-7

  38. Example 1: The [H3O+] of a liquid detergent is 1.4x10–9 M. Calculate its pH. Solution is basic pH = -log [H3O+] = -log [1.4x10-9] = -(-8.85) 2 significant figures pH = 8.85 The number of decimal places in a logarithm is equal to the number of significant figures in the measurement.

  39. Example 2: The pH of black coffee is 5.3. Calculate its [H3O+]. Solution is acidic [H3O+] = antilog (-pH) = 10 –pH = 10 -5.3 1 significant figure [H3O+] = 5 x 10-6

  40. Example 3: The [H3O+] of a solution is 3.5 x 10–3 M. Calculate its pH. Solution is acidic pH = -log [H3O+] = -log [3.5x10-3] = -(-2.46) 2 significant figures pH = 2.46

  41. Example 4: The pH of tomato juice is 4.1. Calculate its [H3O+]. Solution is acidic [H3O+] = antilog (-pH) = 10 –pH = 10 -4.1 1 significant figure [H3O+] = 8 x 10-5

  42. 2 sig figs Example 5: The [OH] of a cleaning solution is 1.0 x 105 M. What is the pH of this solution? Solution is basic Kw=[H3O+][OH–] [H3O+] = = 1.0 x 109 M pH = log[H3O+] = log(1.0x109) = 9.00

  43. Example 6: The pH of a solution is 11.50. Calculate the [H3O+] for this solution. Solution is basic [H3O+] = antilog (-pH) = 10 –pH = 10 -11.50 [H3O+] = 3.2 x 10-12

  44. THE END

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