Acid Base Titrations

1. Experiment #7 Acid Base Titrations

2. What are we doing in this experiment? Determine the concentration of acetic acid in vinegar. To follow the ionization of phosphoric acid with the addition of NaOH and plot a titration curve.

3. What is acid-base titration? A TITRATION WHICH DEALS WITH A REACTION INVOLVING ACID AND A BASE. What is a titration? The act of adding standard solution in small quantities to the test solution till the reaction is complete is termed titration.

4. What is a standard solution? A standard solution is one whose concentration is precisely known. What is a test solution? A test solution is one whose concentration is to be estimated

5. Titration of Vinegar against NaOH Vinegar is an acetic acid solution of certain concentration So a titration of vinegar against NaOH actually means, a reaction between acetic acid and NaOH. HC2H3O2(aq) + NaOH(aq)NaC2H3O2 + H2O(l) Sodium hydroxide Sodium Acetate Acetic acid water water Base Acid salt

6. Titration of Vinegar against NaOH HC2H3O2(aq) + NaOH(aq)NaC2H3O2 + H2O(l) CH3COOH(aq) + NaOH(aq)CH3COONa + H2O(l) 1 molecule 1 molecule 1 molecule 1 molecule 1 mole 1 mole 1 mole 1 mole 1 mole CH3COOH ≡ 1 mole NaOH

7. Titration of Vinegar against NaOH In this experiment, we are trying to find the [CH3COOH] in vinegar. So we have to know the [NaOH] accurately first, before finding the [CH3COOH] in Vinegar. Finding the [NaOH] should be pretty easy right?!! Ya! Kind of …..

8. How hard is it to make 500 mL of 0.1 M NaOH? Should not be that hard right?!! Calculate the weight of NaOH (2.0 g) Weigh out the NaOH on the balance Dissolve it in 500 mL of water in a volumetric flask Hold on! We have a problem here!!

9. So what is your problem? I don’t have a problem but Mr. NaOH seems to have a problem here NaOH is hygroscopic.. What do you mean by hygroscopic? It absorbs moisture. NaOH absorbs moisture from air.

10. What if NaOH is hygroscopic? Let us say, for the problem at hand, we need 2.0 g NaOH to make a 0.1 M NaOH solution. By the time we weigh out the NaOH for our solution, it would have absorbed moisture. So the total weight of 2.0 g is not all due to NaOH. It has some contribution from the water that the NaOH absorbed.

11. What if NaOH is hygroscopic? Since the total weight of 2.0 g is not all due to NaOH, the concentration of the 0.1 M NaOH that we are trying to make is not going to be What we expect it to be. We will not know the accurate concentration of the NaOH solution, just by dissolving 2.0 g of NaOH in 500 mL of water. But the requirement for a titration is that we know the concentration of at least one of the Solutions very precisely.

12. How do we find the [NaOH] precisely? Through standardization What is standardization? It is just a technical term for doing a titration using a primary standard to find the precise concentration of a secondary standard.

13. What is a primary standard? A primary standard should possess the following qualities: (i) It must be available in very pure form (ii) It should not be affected by exposure to moisture or air. (iii) It should maintain its purity during storage. (iv) The reactions involving the primary standard should be stoichiometric and fast. (v) It should have high molecular weight.

14. C-H COOH HC COOK HC CH Which primary standard are we going to use? Potassium hydrogen phthalate, abbreviated as KHP. Remember!! KHP is not the molecular formula for Potassium hyrogen phthalate. It is just an abbreviation. So when calculating the molecular weight of KHP, do not add up the atomic weights of K, H and P. Potassium Hydrogen Phthalate, KHC8H4O4

15. Standardization KHC8H4O4(aq) + NaOH(aq) KNaC8H4O4 + H2O(l) water Base Acid salt 1 molecule 1 molecule 1 molecule 1 molecule 1 mole 1 mole 1 mole 1 mole 1 mole KHC8H4O4≡ 1 mole NaOH

16. Standardization Vfinal- Vinital= Vused (in mL) Vinitial moles of KHP = Moles of NaOH moles of KHP = MNaOH× VNaOH “ 0.1 M NaOH ” Vfinal moles of KHP = MNaOH× Vused End point: Pale Permanent Pink color 250mL 250mL 250mL KHP + H2O+ 2-3 drops of phenolphthalein

17. Titration of Vinegar vs. NaOH Vfinal- Vinital= Vused (in mL) Vinitial moles of acetic acid = Moles of NaOH moles of acetic acid = MNaOH× VNaOH “ 0.1 M NaOH ” Vfinal moles of acetic acid = MNaOH× Vused End point: Pale Permanent Pink color 250mL 250mL 250mL vinegar + H2O+ 2-3 drops of phenolphthalein

18. Hydrolysis of salts formed from strong bases and weak acids NaCH3COO (CH3COO-) NaOH + CH3COOH SB WA HA(aq) + OH-(l) H2O+ A-(aq) conjugate acid conjugate base WA SB

19. Hydrolysis of salts formed from strong bases and weak acids A-(aq) + H2O (l) HA+ OH-(aq)

20. Hydrolysis of salts formed from strong bases and weak acids Multiplying and dividing the numerator and denominator by [H3O+] Rearranging the equation (1)

21. Hydrolysis of salts formed from strong bases and weak acids Remember!! H2O (l) + H2O(l) H3O+ (aq) + OH-(aq) H2O(l) H+ (aq) + OH-(aq) (2)

22. Hydrolysis of salts formed from strong bases and weak acids Remember!! For a monoprotic weak acid (HA) dissolved in water, HA(aq) + H2O(l) H3O+(aq) + A-(aq) conjugate acid conjugate base acid base (3)

23. Hydrolysis of salts formed from strong bases and weak acids (1) (2) (3) Substituting 3 and 2 in 1

24. pH at the equivalence point HA(aq) + OH- (l) A- + H2O(aq) A-(aq) + H2O (l) HA+ OH-(aq) At equivalence point, [HA] = [OH-]

25. pH at the equivalence point Now we make 2 substitutions in the above equations

26. pH at the equivalence point At equivalence point, moles of A- moles of HA

27. pH at the equivalence point Taking log on both sides and multiplying both sides of the equation by -1

28. Titration curve of phosphoric acid, H3PO4 Phosporic acid is a triprotic acid

29. Titration curve of phosphoric acid, H3PO4