Presentation is loading. Please wait.

Presentation is loading. Please wait.

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

Similar presentations


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

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

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

3 A. Neutralization ACID + BASE  SALT + WATER HCl + NaOH  NaCl + H 2 O HC 2 H 3 O 2 + NaOH  NaC 2 H 3 O 2 + H 2 O Salts can be neutral, acidic, or basic. Neutralization does not mean pH = 7. weak strong neutral basic

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

5  Equivalence point (endpoint) Point at which equal amounts of H 3 O + and OH - have been added. Determined by… indicator color change B. Titration dramatic change in pH

6 B. Titration moles H 3 O + = 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

7 B. Titration  42.5 mL of 1.3M KOH are required to neutralize 50.0 mL of H 2 SO 4. Find the molarity of H 2 SO 4. H3O+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 H 2 SO 4


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

Similar presentations


Ads by Google