1. Acid-Base Titration Problems West Valley High School General Chemistry Mr. Mata

2. Titration Sample Problem #1 Calculate the concentration (molarity) of an HCl solution determined by titration to which 50 mL of the acid is added to a flask. Indicator is added, and 29.6 mL of 0.967 M NaOH are required to reach the end point. M a V a = M b V b -> M a = M b V b V a M a = ? M V a = 50 mL M b = 0.967 M V b = 29.6 mL M a = M b V b = (0.967 M) (29.6 mL) = 0.57 M V a 50 mL

3. Titration Sample Problem #2 How much 3.5 M NaOH is required to neutralize 75 mL of 12 M HCl? M a V a = M b V b -> V b = M a V a M b M a = 12 M V a = 75 mL M b = 3.5 M V b = ? mL V b = M a V a = (12 M) (75 mL) = 257.1 mL M b 3.5 M

4. Titration Sample Problem #3 An 80 mL sample of KOH was titrated to the endpoint with 39 mL of 6.5 M HCl. What was the molarity of the KOH? M a V a = M b V b -> M b = M a V a V b M a = 6.5 M V a = 39 mL M b = ? M V b = 80 mL M b = M a V a = (6.5 M) (39 mL) = 3.2 M V b 80 mL

5. Titration Sample Problem #4 How much 2.8 M Ca(OH) 2 is necessary to titrate 44 mL of 0.75 M H 2 SO 4 solution to the endpoint? M a V a = M b V b -> V b = M a V a M b M a = 0.75 M V a = 44 mL M b = 2.8 M V b = ? mL V b = M a V a = (0.75 M) (44 mL) = 11.8 mL M b 2.8 M

6. Practice Titration Problem #1 Calculate the molarity of an HCl solution to which 150 mL of acid is added to a flask. Indicator is added, and 80 mL of 12 M NaOH are used to reach the end point. M a V a = M b V b -> M a = M b V b V a M a = ? M V a = 150 mL M b = 12 M V b = 80 mL M a = M b V b = (12 M) (80 mL) = 6.4 M V a 150 mL

7. Practice Titration Problem #2 How much 8 M NaOH is required to neutralize 500 mL of 3 M HCl? M a V a = M b V b -> V b = M a V a M b M a = 3 M V a = 500 mL M b = 8 M V b = ? mL V b = M a V a = (3 M) (500 mL) = 187.5 mL M b 8 M

8. Practice Titration Problem #3 A 200 mL sample of KOH was titrated to the endpoint with 60 mL of 10 M HCl. What was the molarity of the KOH? M a V a = M b V b -> M b = M a V a V b M a = 10 M V a = 60 mL M b = ? M V b = 200 mL M b = M a V a = (10 M) (60 mL) = 3.0 M V b 200 mL