1. TITRATION Titration of a strong acid with a strong base ENDPOINT = POINT OF NEUTRALIZATION = EQUIVALENCE POINT At the end point for the titration of a strong acid with a strong base, the moles of acid (H + ) equals the moles of base (OH - ) to produce the neutral species water (H 2 O). If the mole ratio in the balanced chemical equation is 1:1 then the following equation can be used. MOLES OF ACID = MOLES OF BASE n acid = n base Since M=n/V M A V A = M B V B

2. TITRATION 1. Suppose 75.00 mL of hydrochloric acid was required to neutralize 22.50 mLof 0.52 M NaOH. What is the molarity of the acid? HCl + NaOH  H 2 O + NaCl M a V a = M b V b rearranges to M a = M b V b / V a so M a = (0.52 M) (22.50 mL) / (75.00 mL) = 0.16 M = 0.16 M Now you try: 2. If 37.12 mL of 0.843 M HNO 3 neutralized 40.50 mL of KOH, what is the molarity of the base? M b = 0.773 mol/L

3. TITRATION Titration of a strong acid with a strong base ENDPOINT = POINT OF NEUTRALIZATION = EQUIVALENCE POINT At the end point for the titration of a strong acid with a strong base, the moles of acid (H + ) equals the moles of base (OH - ) to produce the neutral species water (H 2 O). If the mole ratio in the balanced chemical equation is NOT 1:1 then you must rely on the mole relationship and handle the problem like any other stoichiometry problem. MOLES OF ACID = MOLES OF BASE n acid = n base

4. TITRATION 1. If 37.12 mL of 0.543 M LiOH neutralized 40.50 mL of H 2 SO 4, what is the molarity of the acid? 2 LiOH + H 2 SO 4  Li 2 SO 4 + 2 H 2 O First calculate the moles of base: 0.03712 L LiOH (0.543 mol/1 L) = 0.0202 mol LiOH Next calculate the moles of acid: 0.0202 mol LiOH (1 mol H 2 SO 4 / 2 mol LiOH)= 0.0101 mol H 2 SO 4 : Last calculate the Molarity: M a = n/V = 0.010 mol H 2 SO 4 / 0.4050 L = 0.248 M M a = n/V = 0.010 mol H 2 SO 4 / 0.4050 L = 0.248 M 2. If 20.42 mL of Ba(OH) 2 solution was used to titrate29.26 mL of 0.430 M HCl, what is the molarity of the barium hydroxide solution? M b = 0.260 mol/L

5. PRACTICE PROBLEMS 1. How many milliliters of 1.25 M LiOH must be added to neutralize 34.7 mL of 0.389 M HNO 3 ? 2. How many mL of 0.998 M H 2 SO 4 must be added to neutralize 47.9 mL of 1.233 M KOH? 10.8 mL 29.6 mL