1. The Chemistry of Acids and Bases

2. Acids and Bases: Titration At the conclusion of our time together, you should be able to: 1. Do a titration of an acid with a base 2. Calculate the equivalence point (neutralization point) for a titration problem 3. Define normality

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4. ACID-BASE REACTIONS Titrations H 2 C 2 O 4 (aq) + 2 NaOH(aq) ---> acid base acid base Na 2 C 2 O 4 (aq) + 2 H 2 O(liq) Carry out this reaction using TITRATION. Oxalic Acid, H 2 C 2 O 4

5. Setup For Titrating An Acid With A Base

6. TitrationTitration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base) This is called the NEUTRALIZATION point. This is called the NEUTRALIZATION point. The equivalence point has been reached

7. The Formula for Titration Use: Use: M 1 * E * V 1 = M 2 * E * V 2 M = molarity of acid and base E = number of H+ or OH- per mole of acid or base M x E = Normality V = volume

8. Various Protic Titration #1: Standardize a solution of NaOH — i.e., accurately determine its concentration. 32.7 mL of 0.100 M NaOH is neutralized with 25.0 mL of HCl by titration to an equivalence point. What is the concentration of the acid?

9. How do you do this?? Use: Use: M 1 * E * V 1 = M 2 * E * V 2 0.100 M x 1 x 32.7 mL = M x 1 x 25.0 mL = 0.131 M HCl pH= -log (0.131 x 1) = 0.883

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11. #2: You have 50.0 mL of 0.250 M NaOH neutralized 20.0 mL of sulfuric acid. Determine the M of the acid? 0.250 M x 1 x 50.0 mL = M x 2 x 20.0 mL = 0.313 M H 2 SO 4 For each mole of sulfuric acid there are 2 moles of H +, therefore, pH = -log (0.313 x 2) = 0.203

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13. Acids and Bases: Titration Let’s see if you can: 1. Do a titration of an acid with a base 2. Calculate the equivalence point (neutralization point) for a titration problem 3. Define normality

14. #4 Various Protic Use: Use: M 1 * E * V 1 = M 2 * E * V 2 0.0500 M x 2 x 29.6 mL = M x 1 x 40.0 mL = 0.0740 M HNO 3 pOH= -log (0.0500 x 2) = 1.00

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16. Normality is defined as 1234567891011121314151617181920 21222324252627282930 1. M * V 2. E * V 3. M * E 4. M * E * V 5. Not listed

17. A method used to determine the concentration of an unknown acid or base. 1234567891011121314151617181920 21222324252627282930 1. Equivalence point 2. Neutralization point 3. Titration 4. Titrant 5. Normality 6. Flask 7. Buret 8. Not listed

18. The titrant is usually added to a(n) 1234567891011121314151617181920 21222324252627282930 1. Equivalence point 2. Neutralization point 3. Titration 4. Titrant 5. Normality 6. Flask 7. Buret 8. Not listed

19. The indicator is usually added to the 1234567891011121314151617181920 21222324252627282930 1. Equivalence point 2. Neutralization point 3. Titration 4. Titrant 5. Normality 6. Flask 7. Buret 8. Not listed

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21. 3 M sulfuric acid would have a normality of 1234567891011121314151617181920 21222324252627282930 1. 3 N 2. 2 N 3. 5 N 4. 6 N 5. 9 N 6. Not listed

22. 100.0 mL of 3.0 M sulfuric acid neutralizes 50.0 mL of sodium hydroxide. What is the molarity of the base? 1234567891011121314151617181920 21222324252627282930 1. 12 M 2. 6.0 M 3. 3.0 M 4. 1.5 M 5. 0.33 M 6. Not listed

23. 100.0 mL of 3.000 M nitric acid neutralizes 3.000 M of aluminum hydroxide. How many mL of the base did you use? 1234567891011121314151617181920 21222324252627282930 1. 100.0 mL 2. 50.00 mL 3. 33.33 mL 4. 16.67 mL 5. 8.333 mL 6. Not listed

24. Murphy's Laws of Science and Technology Whenever a system becomes completely defined, some fool discovers something which either abolishes the system or expands it beyond recognition.

26. The Chemistry of Acids and Bases

27. Acids and Bases: Titration At the conclusion of our time together, you should be able to: 1. Do a titration of an acid with a base 2. Calculate the equivalence point (neutralization point) for a titration problem 3. Define normality 4. Determine the percent acid with a titration problem

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29. Lemonade Titration Problem 0.100 M x 1 x 8.70 mL = M x 3 x 10.0 mL = 0.0290 M citric acid

30. Lemonade Titration Problem 0.0290 M x 0.010 L = x g 192.14 g/mol = 0.0557 g in 10 mL

31. Lemonade Titration Problem Using the density of citric acid in lemonade: 10.0 mL x 0.960 g/mL = 9.60 g lemonade in 10.0 mL of solution 0.0557 g citric acid in 10.0 mL x 100 = 9.60 g lemonade in 10.0 mL 0.580%