Experiment #7 Acid Base Titrations
Experiment #7 Acid Base Titrations 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. 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. 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 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) Acetic acid Acid + NaOH(aq) NaC2H3O2 + H2O(l) Sodium hydroxide Sodium Acetate Base salt water water Titration of Vinegar against NaOH HC2H3O2(aq) + CH3COOH(aq) NaOH(aq) NaC2H3O2 + H2O(l) + NaOH(aq) CH3COONa + H2O(l) 1 molecule 1 molecule 1 molecule 1 mole 1 mole 1 mole 1 mole CH3COOH ≡ 1 mole NaOH 1 molecule 1 mole 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 ….. 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!! 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. 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. 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. 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. 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. 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. C-H HC COOH HC COOK CH Potassium Hydrogen Phthalate, KHC8H4O4 Standardization KHC8H4O4(aq) + NaOH(aq) KNaC8H4O4 + H2O(l) Acid Base salt 1 molecule 1 molecule 1 molecule 1 mole 1 mole 1 mole 1 mole KHC8H4O4 ≡ 1 mole NaOH water 1 molecule 1 mole Standardization Vfinal- Vinital= Vused (in mL) Vinitial moles of KHP = Moles of NaOH “ 0.1 M NaOH ” moles of KHP = MNaOH × VNaOH moles of KHP = MNaOH × Vused Vfinal M NaOH moles of KHP V used 250mL 250mL 250mL KHP + H2O+ 2-3 drops of phenolphthalein End point: Pale Permanent Pink color Titration of Vinegar vs. NaOH Vfinal- Vinital= Vused (in mL Vinitial moles of acetic acid = Moles of NaOH “ 0.1 M NaOH ” moles of acetic acid = MNaOH × VNaOH moles of acetic acid = MNaOH × Vused Vfinal M acetic acid moles of NaOH V (L) vinegar 250mL 250mL 250mL vinegar + H2O+ 2-3 drops of phenolphthalein End point: Pale Permanent Pink color Hydrolysis of salts formed from strong bases and weak acids NaCH3COO (CH3COO-) NaOH + CH3COOH SB WA HA(aq) + OH-(l) WA SB H2O + A-(aq) conjugate conjugate acid base Hydrolysis of salts formed from strong bases and weak acids A-(aq) + H2O (l) Kb HA HA OH eqb A eqb eqb HA OH Kb A + OH-(aq) Hydrolysis of salts formed from strong bases and weak acids HA OH Kb A Multiplying and dividing the numerator and denominator by [H3O+] HA OH H O Kb 3 A H O 3 Rearranging the equation Kb HA 3 H O OH A H O 3 1 (1) Hydrolysis of salts formed from strong bases and weak acids Remember!! H3O+ (aq) H2O (l) + H2O(l) H+ (aq) H2O(l) + OH-(aq) + Ionic product of water , Kw H 3O OH Kw (25 C ) H O OH 110 3 OH-(aq) 14 (2) Hydrolysis of salts formed from strong bases and weak acids Remember!! For a monoprotic weak acid (HA) dissolved in water, H3O+(aq) + A-(aq) conjugate conjugate acid base HA(aq) + H2O(l) acid base H O A Ka 3 eqb eqb HA eqb H O A Ka 3 HA HA 1 K a H 3O A (3) Hydrolysis of salts formed from strong bases and weak acids HA H O OH (1) K A H O 1 3 b 3 Ionic product of water , Kw H 3O OH HA 1 K a H 3O A (3) Substituting 3 and 2 in 1 1 Kw Kw Kb Ka 1 Ka Kb K a K w (2) pH at the equivalence point HA(aq) + OH- (l) A- A-(aq) + H2O (l) Kb + H2O(aq) HA HA OH eqb A HA OH A eqb eqb Kb At equivalence point, [HA] = [OH-] OH A 2 Kb + OH-(aq) pH at the equivalence point OH K A 2 b Now we make 2 substitutions in the above equations Kw OH H 3O 2 Kw Kw 2 K a A H 3O H O 2 3 1 Kw Ka A Kw Kb Ka pH at the equivalence point H O 2 3 1 Kw Ka A Kw Ka H 3O A At equivalence point, moles of A- moles of HA moles of HA A CHA Vsolution at equivalence ( L) KwKa H 3O CHA pH at the equivalence point KwKa H 3O CHA Taking log on both sides and multiplying both sides of the equation by -1 KwKa Log H 3O Log C1 HA Kw Ka 2 pH Log CHA KwKa 1 pH Log 2 CHA Titration curve of phosphoric acid, H3PO4 Phosporic acid is a triprotic acid Titration curve of phosphoric acid, H3PO4
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