Acid-Base Reactions

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)

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

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

9.50 mL of M NaOH was needed to neutralize 5.89 mL of nitric acid. Calculate the molarity of the acid. Example mL NaOH mol HNO mL 1 L 1000 mL 5.89 mL HNO mol NaOH = = 1 L NaOH M HNO 3 1 mol NaOH 1 mol HNO 3 1 L mol HNO 3 L 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.

Alternate way to solve Example mL of 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) ? = M HNO 3

In a titration, mL of a M RbOH solution is required to exactly neutralize 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 mol RbOH 1 L NaOH 2 mol RbOH 1 mol H 2 SO mol H 2 SO 4 L or M H 2 SO mol H 2 SO mL mL H 2 SO 4 1 L = mol H 2 SO 4 =