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C. Johannesson III. Titration Ch. 14 & 15 - Acids & Bases.

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Presentation on theme: "C. Johannesson III. Titration Ch. 14 & 15 - Acids & Bases."— Presentation transcript:

1 C. Johannesson III. Titration Ch. 14 & 15 - Acids & Bases

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

3 C. Johannesson 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 C. Johannesson 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 C. Johannesson  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 C. Johannesson 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 C. Johannesson 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


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