1. Acid-Base Reactions

2. Neutralization acid + base salt + water HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (l) H + + Cl - + Na + + OH - Na + + Cl - + H 2 O (l) H + + OH - H 2 O (l)

3. Titration Titrant or Standard Solution (known concentration) Analyte (unknown concentration) stopcock Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point End point—the point which the indicator used in titration changes color

4. Before the endpoint Past the endpoint (overshot) At the Endpoint **Indicator used: phenolphthalein (colorless in an acid, and pink in a base)

5. 9.50 mL of 0.500 M NaOH was needed to neutralize 5.89 mL of nitric acid. Calculate the molarity of the acid. Example 1 9.50mL NaOH 0.00475 mol HNO 3 1000 mL 1 L 1000 mL 5.89 mL HNO 3 0.500 mol NaOH = = 1 L NaOH 0.806 M HNO 3 1 mol NaOH 1 mol HNO 3 1 L 0.806 mol HNO 3 L 0.00475 mol HNO 3 HNO 3 (aq) + NaOH (aq)  H 2 O (l) + NaNO 3 (aq) Turn moles of titrant into moles of analyte (because we are neutralizing and therefore at the equivalence point) then divide the moles of analyte by L used (in problem) to solve for concentration (M) of analyte.

6. Alternate way to solve Example 1 9.50 mL of 0.500 M NaOH was needed to neutralize 5.89 mL of nitric acid. Calculate the molarity of the acid. HNO 3 (aq) + NaOH (aq)  H 2 O (l) + NaNO 3 (aq) ** at the equivalence point mol of acid = mol of base (if 1:1 mol ratio of acid to base) ? = 0.806 M HNO 3

7. In a titration, 43.21 mL of a 0.03020 M RbOH solution is required to exactly neutralize 30.00 mL of sulfuric acid. What is the molarity of the sulfuric acid solution? Example 2 H 2 SO 4 (aq) + 2RbOH (aq)  2H 2 O (l) + Rb 2 SO 4 (aq) 43.21mL RbOH 1000 mL 1 L 0.03020 mol RbOH 1 L NaOH 2 mol RbOH 1 mol H 2 SO 4 0.02175 mol H 2 SO 4 L or 0.02175 M H 2 SO 4 0.0006525 mol H 2 SO 4 1000 mL 30.00 mL H 2 SO 4 1 L = 0.0006525 mol H 2 SO 4 =