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

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

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

2 Neutralization Neutralization is a chemical reaction between an acid and a base. The products are a salt (any ionic compound) and water. Acid + Base Salt + Water Ex. HCl + NaOH NaCl + H 2 O

3 Neutralization Continued… Neutralization does not necessarily mean pH = 7 Salts can be neutral, acidic, or basic depending on the strength of the acid and base involved in the reaction. Ex. HCl + NaOH NaCl + H 2 O strongstrong neutral Ex. HC 2 H 3 O 2 + NaOHNaC 2 H 3 O 2 + H 2 O weak strong basic

4 Indicators Acid-base indicators are compounds whose colors are sensitive to pH. - the color of the indicator changes as pH changes - indicators are typically made from weak acids or bases Acid-base indicators are good for determining approximate pH. If exact pH is needed, a pH meter can be used.

5 Titration Titration is an analytical method in which a standard solution is used to determine the concentration of an unknown solution. The equivalence point tells you when equal amounts of H 3 O + and OH - are in the solution. - you know you have reached the equivalence point when there is a dramatic change in pH or a color change in the indicator. The point at which the indicator changes color is called the endpoint.

6 Calculations moles H 3 O + = moles OH - M * V * n = M * V * n M = molarity (moles/L) V = volume (L) n = # of H + ions in the acid OR # of OH - ions in the base

7 Practice 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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