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Chapter 19 - Neutralization

Chapter 19 - Neutralization. Section 19.1 Neutralization Reactions. OBJECTIVES: Explain how acid-base titration is used to calculate the concentration of an acid or a base. Section 19.1 Neutralization Reactions. OBJECTIVES: Explain the concept of equivalence in neutralization reactions.

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Chapter 19 - Neutralization

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  1. Chapter 19 - Neutralization

  2. Section 19.1Neutralization Reactions • OBJECTIVES: • Explain how acid-base titration is used to calculate the concentration of an acid or a base

  3. Section 19.1Neutralization Reactions • OBJECTIVES: • Explain the concept of equivalence in neutralization reactions.

  4. Acid-Base Reactions • Acid + Base Water + Salt • Properties related to every day: • antacids depend on neutralization • farmers use it to control soil pH • formation of cave stalactites • human body kidney stones

  5. Acid-Base Reactions • Neutralization Reaction - a reaction in which an acid and a base react in an aqueous solution to produce a salt and water: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H2SO4(aq) + 2KOH(aq)K2SO4(aq) + 2 H2O(l) • Table 19.1, page 458 lists some salts

  6. Titration • Titration is the process of adding a known amount of solution of known concentration to determine the concentration of another solution • Remember? - a balanced equation is a mole ratio • Sample Problem 19-1, page 460

  7. Titration • The concentration of acid (or base) in solution can be determined by performing a neutralization reaction • An indicator is used to show when neutralization has occurred • Often use phenolphthalein- colorless in neutral and acid; turns pink in base Simulation Simulation Worksheet

  8. Steps - Neutralization reaction 1. A measured volume of acid of unknown concentration is added to a flask 2. Several drops of indicator added 3. A base of known concentration is slowly added, until the indicator changes color-measure the volume • Figure 19.4, page 461

  9. Neutralization • The solution of known concentration is called the standard solution • added by using a buret • Continue adding until the indicator changes color • called the “end point” of the titration • Sample Problem 19-2, page 461

  10. Equivalents • One mole of hydrogen ions reacts with one mole of hydroxide ions • does not mean that 1 mol of any acid will neutralize 1 mol of any base • because some acids and bases can produce more than 1 mole of hydrogen or hydroxide ions • example: H2SO4(aq)2H+ + SO42-

  11. Equivalents • Made simpler by the existence of a unit called an equivalent • One equivalent (equiv) is the amount of acid (or base) that will give 1 mol of hydrogen (or hydroxide) ions • 1 mol HCl = 1 equiv HCl • 1 mol H2SO4 = 2 equiv H2SO4

  12. Equivalents • In any neutralization reaction, the equivalents of acid must equal the equivalents of base • How many equivalents of base are in 2 mol Ca(OH)2? • The mass of one equivalent is it’s gram equivalent mass (will be less than or equal to the formula mass): HCl = 36.5 g/mol; H2SO4 = 49.0 g/mol

  13. Equivalents • Sample Problem 19-3, page 462 • Sample Problem 19-4, page 462

  14. Normality (N) • Useful to know the Molarity of acids and bases • Often more useful to know how many equivalents of acid or base a solution contains • Normality (N) of a solution is the concentration expressed as number of equivalents per Liter

  15. Normality (N) • Normality (N) = equiv/L • equiv = Volume(L) x N; • and also know: N=M x eq; • M = N / eq • Sample Problem 21-5, page 621 • Diluting solutions of known Normality: N1 x V1 = N2 x V2 • N1 and V1 are initial solutions • N2 and V2 are final solutions

  16. Normality (N) • Titration calculations often done more easily using normality instead of molarity • In a titration, the point of neutralization is called the equivalence point • the number of equivalents of acid and base are equal

  17. Normality (N) • Doing titrations with normality use: NA x VA = NB x VB • Sample Problem 19-6, page 464 • Sample Problem 19-7, page 464 • Sample Problem 19-8, page 464

  18. Section 19.2 Salts in Solution • OBJECTIVES: • Demonstrate with equations how buffers resist changes in pH

  19. Section 19.2 Salts in Solution • OBJECTIVES: • Calculate the solubility product constant (Ksp) of a slightly soluble salt

  20. Salt Hydrolysis • A salt: • comes from the anion of an acid (Cl-) • comes from the cation of a base (Na+) • formed from a neutralization reaction • some neutral; others acidic or basic • Salt hydrolysis- salt reacts with water to produce acid or base solution

  21. Salt Hydrolysis • Hydrolyzing salts usually made from: • strong acid + weak base, or • weak acid + strong base • Strong refers to the degree of ionization (100%) What pH will result from the above combinations?

  22. Salt Hydrolysis • To see if the resulting salt is acidic or basic, check the “parent” acid and base that formed it: NaCl HCl + NaOH NH4OH H2SO4 + NH4OH CH3COOK CH3COOH + KOH

  23. Strong Acids HCl HClO4 H2SO4 HI HNO3 HBr Strong Bases Mg(OH)2 NaOH Ca(OH)2 KOH To determine if a salt is made From a combination which is: acid/base weak/weak weak/strong strong/strong strong/weak Na Cl OH H

  24. Lab 42 : Salt Hydrolysis Universal Indicator Colors

  25. Buffers • Buffers are solutions in which the pH remains relatively constant when small amounts of acid or base are added • made from a pair of chemicals • a weak acid and one of it’s salts; • HA / A- • or a weak base and one of it’s salts • NH3 / NH4+ Simulation

  26. Buffers • A buffer system is better able to resist changes in pH than pure water • Since it is a pair of chemicals: • one chemical neutralizes any acid added, while the other chemical would neutralize any additional base • they make each other in the process!

  27. Unbuffered reaction between and acid an base Buffered solution and reaction of an acid with a base HA / A- HCl + NaOH  NaCl + HOH HCl + A- HA + Cl- pH pH Add strong acid Add strong acid

  28. Buffers • Example: Ethanoic (acetic) acid and sodium ethanoate (also called sodium acetate) • HC2H302 / NaC2H302 becomes • HC2H302 / C2H302 1- • Weak acid weakbase • The buffer capacity is the amount of acid or base that can be added before a significant change in pH

  29. Buffers • Buffers that are crucial to maintain the pH of human blood: • carbonic acid - hydrogen carbonate H2CO3 / HCO3- 2. dihydrogen phosphate - monohydrogen phoshate H2PO4- / HPO42- • Table 19.2, page 469 has some important buffer systems • Sample Problem 19-9, page 468

  30. Calculating ksp or Solubility Product Constant Copy Example 10/11 pg 470/471 into your notes What does a high Ksp mean? What does a low Ksp mean?

  31. Solubility Product Constant • Salts differ in their solubilities • Table 19.3, page 470 • Most “insoluble” salts will actually dissolve to some extent in water • said to be slightly, or sparingly, soluble in water

  32. Solubility Product Constant • Consider: AgCl(s) • The “equilibrium expression” is: Ag+(aq) + Cl-(aq) [ Ag+ ] x [ Cl- ] Keq = [ AgCl ]

  33. Solubility Product Constant • But, the [ AgCl ] is constant as long as some undissolved solid is present • Thus, a new constant is developed, and is called the “solubility product constant” (Ksp): Keq x [ AgCl ] = [ Ag+ ] x [ Cl- ] = Ksp

  34. Solubility Product Constant • Values of solubility product constants are given for some sparingly soluble salts in Table 19.4, page 471 • Although most compounds of Ba are toxic, BaSO4 is so insoluble that it is used in gastrointestinal examinations by doctors! - p.632

  35. Solubility Product Constant • To solve problems: • a) write equation, • b) write expression, and • c) fill in values using x for unknowns • Sample Problem 21-10, page 634 • Sample Problem 21-11, page 634

  36. Common Ion Effect • A “common ion” is an ion that is common to both salts in solution • example: You have a solution of lead (II) chromate. You now add some lead (II) nitrate to the solution. • The lead (II) ion is the common ion

  37. Pb2+ + CrO42- PbCrO4 Add Pb(NO3)2 • This causes a shift in equilibrium (due to Le Chatelier’s principle), and is called the common ion effect shift PbCrO4 Pb2+ + CrO42-

  38. Common Ion Effect • Sample ProblemThe Ksp of silver iodide is 8.3x10-17. What is the iodide concentration of a 1.00L saturated solution of AgI to which 0.020 mol of AgNO3 is added? • 1. Write the equilibrium equation. AgI(s) Ag1+ + I1- 2. Write the Ksp expressionKsp= [Ag1+]1 [I1-]1 = 8.3x10-17 Ksp= (x)1 (x)1 = 8.3x10-17 Ksp = (x + 0.020)1 (x)1 = 8.3x10-17 Ksp = (0.020) (x) = 8.3x10-17 x= 4.2x10-15 Note: X is So small that it can be ignored • The [ ] of iodide ion is 4.2x10-15 M

  39. The solubility product constant (Ksp) can be used to predict whether a precipitate will form or not in a reaction • if the calculated ion-product concentration is greater than the known Ksp, a precipitate will form

  40. ksp Sample Problem: A student prepares a solution by combining 0.025 mol CaCl2 with 0.015 molPb(NO3)2 in a 1 L container. Will a precipitate form? WHAT WE KNOW!!! CaCl2+ Pb(NO3)2 Ca(NO3)2(aq) +PbCl2(s) X ppt 0.025 mol/L 0.015 mol/L PbCl2 Pb2+ + 2 Cl1- ksp = [Pb2+]1 [Cl1-]2 The values we need to solve the problem!

  41. 0.050 mol Cl- L Pb(NO3)2 Pb2+ + 2NO3- 0.015 mol/L 0.015 mol Pb2+ 1L • Calculate the concentration of the ion’s used to make the precipitate. CaCl2 Ca 2+ 2Cl- + 0.025 mol/L 2 mol Cl- 0.025 mol CaCl2 1L 1 mol CaCl2 0.015 mol Pb(NO3)2 1 mol Pb2+ 1L 1 mol Pb(NO3)2

  42. 2) Calculate the Ksp for the precipitate PbCl2 Pb2+ + 2Cl- Ksp = [Pb2+]1 [Cl-]2 Ksp = ( 0.015 M)1 ( 0.050 M) 2 Ksp for PbCl2 = 3.75 x 10 -5 Calculated Ksp 3.75 x 10 -5 is > Known Ksp 1.7 x 10 -5 for PbCl2 See pg 471 Table 19-4 for known values Calc > Known = PPT Known> Calc = No PPT Known = Calc = No PPT

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