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Chapter 21 “Neutralization”

Chapter 21 “Neutralization”. Milbank High School. Section 21.1 Neutralization Reactions. OBJECTIVES: Explain how acid-base titration is used to calculate the concentration of an acid or a base. Section 21.1 Neutralization Reactions. OBJECTIVES:

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Chapter 21 “Neutralization”

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  1. Chapter 21“Neutralization” Milbank High School

  2. Section 21.1Neutralization Reactions • OBJECTIVES: • Explain how acid-base titration is used to calculate the concentration of an acid or a base.

  3. Section 21.1Neutralization 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 from insoluble salts

  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 21.1, page 614 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 21-1, page 616

  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

  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 21.4, page 617

  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 21-2, page 618

  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 21-3, page 620 • Sample Problem 21-4, page 620

  14. Normality (N) • It is useful for us 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 21-6, page 623 • Sample Problem 21-7, page 623 • Sample Problem 21-8, page 624

  18. Section 21.2Salts in Solution • OBJECTIVES: • Demonstrate with equations how buffers resist changes in pH.

  19. Section 21.2Salts in Solution • OBJECTIVES: • Calculate the solubility product constant (Ksp) of a slightly soluble salt.

  20. Salt Hydrolysis • A salt is an ionic compound that: • comes from the anion of an acid • comes from the cation of a base • is formed from a neutralization reaction • some neutral; others acidic or basic • “Salt hydrolysis” - a salt that reacts with water to produce acid or base

  21. Salt Hydrolysis • Hydrolyzing salts usually from: • strong acid + weak base, or • weak acid + strong base • Strong refers to the degree of ionization • How do you know if it’s strong? • Refer to handout provided

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

  23. 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; or a weak base and one of it’s salts

  24. 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 • AND, they produce each other in the process!!!

  25. Buffers • Example: Ethanoic (acetic) acid and sodium ethanoate (also called sodium acetate) • Examples on page 628 of these • The buffer capacity is the amount of acid or base that can be added before a significant change in pH

  26. Buffers • Buffers that are crucial to maintain the pH of human blood: 1. carbonic acid (H2CO3) & hydrogen carbonate (HCO31-) 2. dihydrogen phosphate (H2PO41-) & monohydrogen phoshate (HPO42-) • Table 21.2, page 629 has some important buffer systems • Sample Problem 21-9, page 630

  27. Solubility Product Constant • Salts differ in their solubilities • Table 21.3, page 631 • Most “insoluble” salts will actually dissolve to some extent in water • said to be slightly, or sparingly, soluble in water

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

  29. 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

  30. Solubility Product Constant • Values of solubility product constants are given for some common slightly soluble salts in Table 21.4, page 632 • Although most compounds of Ba are toxic, BaSO4 is so insoluble that it is used in gastrointestinal examinations by doctors! - p.632

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

  32. 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 is a common ion • This causes a shift in equilibrium (due to Le Chatelier’s principle), and is called the common ion effect

  33. Common Ion Effect • Sample Problem 21-12, page 636 • The solubility product constant (Ksp) can be used to predict whether a precipitate will form or not: • if the ion-product concentration is greater than the allowed Ksp, then a precipitate will form • Sample Problem 21-13, page 637

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