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

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Presentation on theme: "Ch. 15 & 16 - Acids & Bases III. Titration (p. 493 - 503)"— Presentation transcript:

1 Ch. 15 & 16 - Acids & Bases III. Titration (p )

2 A. Neutralization Chemical reaction between an acid and a base.
Products are a salt (ionic compound) and water. Neutralization occurs when hydronium ions and hydroxide ions are supplied in equal numbers by reactants.

3 ACID + BASE  SALT + WATER
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.

4 B. Titration standard solution unknown solution Because acids and bases react, the progressive addition of an acid to a base can be used to compare the concentrations of the acid and base.

5 Titration Titration: analytical method in which a standard solution is used to determine the concentration of another solution. Standard solution: one for which the concentration is known

6 B. Titration Equivalence point (endpoint)
Point at which equal amounts of H3O+ and OH- have been added. End point Point in a titration at which an indicator changes color.

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

8 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

9 A 15. 5 mL sample of 0. 215 M KOH solution is mixed with
A 15.5 mL sample of M KOH solution is mixed with .645 M HCl, in a titration experiment. Calculate the volume of the hydrochloric acid solution. Answer: L


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