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Chapter 19 Section 3

Chapter 19 Section 3. Acid-Base Titration. Neutralization Reaction. Reaction between an acid and a base HA + BOH → H 2 O + BA HA = general form of acid BOH = general form of base (Ex) HCl + NaOH → (Ex) H 2 SO 4 + KOH → Always produce water and a salt

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Chapter 19 Section 3

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  1. Chapter 19 Section 3 Acid-Base Titration

  2. Neutralization Reaction • Reaction between an acid and a base HA + BOH → H2O + BA HA = general form of acid BOH = general form of base (Ex) HCl + NaOH→ (Ex) H2SO4 + KOH → • Always produce water and a salt • Salt is NOT always table salt

  3. Neutralization Problems • Two ways to solve • Stoichiometrically • Using the relationship between mol of H3O+ and mol of OH- * mol of H3O+ = mol of OH- * mol of H3O+ = Macid x Vacid in liters x (# of H+) * mol of OH− = Mbase x Vbase in liters x (# of OH− )

  4. Example What is the concentration of a hydrochloric acid solution if 35.00 cm3 of it are exactly neutralized by 14.8 cm3 of a 0.500 M sodium hydroxide?

  5. Titration • Titration (10 min) • The gradual addition of a solution (=titrant) to another solution • The titrant’s concentration and volume used are known • The titrant is called standard solution • Purpose: to determine the concentration of a solution

  6. Reading a buret

  7. Titration Curves Adding a base to an acid

  8. Adding an acid to a base

  9. Based on the concept of equivalence point • Equivalence point is when the neutralization reaction is complete • At equivalence point, mol H+ = mol OH- • An indicator tells the equivalence point • End point is when the color of the indicator changes • A well-chosen indicator has its end point close to the equivalence point

  10. Equivalence Point Calculations (1) At equivalence point, mol H+ = mol OH- • nMaVa = mMbVb • n = the subscript after H+ • m = the subscript after OH- (Ex) H2SO4: n = 2 Al(OH)3: m = 3 (2) Solve stoichiometrically from a balanced chemical equation

  11. Example 1 • What is the molarity of NaOH if 20.0 mL of the solution is neutralized by 17.4 mL of 1.00 M H3PO4 solution? • Answer: 2.61 M

  12. Example 2 • What is the molarity of NaOH if 20.0 mL of the solution is neutralized by 17.4 mL of 1.00 M H3PO4 solution? • Answer: 2.61 M

  13. Buffer • A mixture of an acid and a base – usually a conjugate pair (Ex) HC2H3O2 and C2H3O2- • What is the buffer for? • To maintain the pH of the solution about constant when an acid or a base is added

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