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III. Titration (p. 493 - 503)

Ch. 15 & 16 - Acids & Bases. III. Titration (p. 493 - 503). A. Neutralization. Chemical reaction between an acid and a base. Products are a salt (ionic compound) and water. A. Neutralization. ACID + BASE  SALT + WATER. HCl + NaOH  NaCl + H 2 O. strong. strong. neutral.

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III. Titration (p. 493 - 503)

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  1. Ch. 15 & 16 - Acids & Bases III. Titration(p. 493 - 503) C. Johannesson

  2. A. Neutralization • Chemical reaction between an acid and a base. • Products are a salt (ionic compound) and water. C. Johannesson

  3. A. Neutralization ACID + BASE  SALT + WATER HCl + NaOH  NaCl + H2O strong strong neutral HC2H3O2 + NaOH  NaC2H3O2 + H2O weak strong basic • Salts can be neutral, acidic, or basic. • Neutralization does not mean pH = 7. C. Johannesson

  4. standard solution unknown solution B. Titration • Titration • Analytical method in which a standard solution is used to determine the concentration of an unknown solution. C. Johannesson

  5. B. Titration • Equivalence point (endpoint) • Point at which equal amounts of H3O+ and OH- have been added. • Determined by… • indicator color change • dramatic change in pH C. Johannesson

  6. B. Titration moles H3O+ = moles OH- MVn = MVn M: Molarity V: volume n: # of H+ ions in the acid or OH- ions in the base C. Johannesson

  7. B. Titration • 42.5 mL of 1.3M KOH are required to neutralize 50.0 mL of H2SO4. Find the molarity of H2SO4. H3O+ M = ? V = 50.0 mL n = 2 OH- M = 1.3M V = 42.5 mL n = 1 MV# = MV# M(50.0mL)(2) =(1.3M)(42.5mL)(1) M = 0.55M H2SO4 C. Johannesson

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