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III. Titrations

III. Titrations. In an acid-base titration , a basic (or acidic) solution of unknown [ ] is reacted with an acidic (or basic) solution of known [ ]. An indicator is a substance used to visualize the endpoint of the titration.

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III. Titrations

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  1. III. Titrations • In an acid-base titration, a basic (or acidic) solution of unknown [ ] is reacted with an acidic (or basic) solution of known [ ]. • An indicator is a substance used to visualize the endpoint of the titration. • At the equivalence point, the number of moles of acid equals the number of moles of base.

  2. III. Visual of a Titration

  3. III. Titration/pH Curves • A titration or pH curve is a plot of how the pH changes as the titrant is added. • It is possible to calculate the pH at any point during a titration. • Multiple pH’s can be calculated, and the results plotted to create the theoretical titration curve.

  4. III. Find the Equivalence Point! • The keys to these types of problems are writing the titration equation and finding the equivalence point of the titration. • The calculation then depends on what region of the titration curve you are in: • Before titration begins • Pre-equivalence • Equivalence point • Post-equivalence

  5. III. Illustrative Problem • Sketch the pH curve for the titration of 25.0 mL of 0.100 M HCl with 0.100 M NaOH.

  6. III. Illustrative Problem Solution • Write the titration equation. • HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) • Calculate the equivalence point. What volume of NaOH is needed to completely react with HCl?

  7. III. Illustrative Problem Solution • Calculate initial pH before titration. • Since HCl is strong, 0.100 M HCl has [H3O+] = 0.100 M, and pH = 1.000. • Calculate the pH of some points in the pre-equivalence region. • As NaOH is added, the neutralization reaction OH-(aq) + H3O+(aq) H2O(l) takes place. • We calculate the pH after addition of 5.00 mL of NaOH.

  8. III. Illustrative Problem Solution • To calculate the pH after 5.00 mL, we need to calculate initial moles of acid and the number of moles of base. • We put these moles into a reaction chart.

  9. III. Illustrative Problem Solution The H3O+ leftover is in a larger volume so we calculate its concentration.

  10. III. Illustrative Problem Solution • We do the same thing for some other points: 10.0 mL, 15.0 mL, and 20.0 mL. • Results summarized below.

  11. III. Illustrative Problem Solution • Calculate the pH at the equivalence point. • For a strong-strong titration, pH always equals 7.00 at the equivalence point! • Calculate the pH of some points in the post-equivalence region. • In this region, the pH depends on the excess OH- added.

  12. III. Illustrative Problem Solution • To find the excess added, calculate how many mL past the equivalence point have been added, convert to moles, and divide by total volume. For 30.0 mL: pH can then be found from pOH.

  13. III. Illustrative Problem Solution • Again, calculate for additional points like 35.0, 40.0, and 50.0 mL. Results summarize below.

  14. III. Illustrative Problem Solution • Now we plot the data points and sketch the pH titration curve!

  15. III. Sample Problem • A 0.0500 L sample of 0.0200 M KOH is being titrated with 0.0400 M HI. What is the pH after 10.0 mL, 25.0 mL, and 30.0 mL of the titrant have been added?

  16. III. Weak Acid/Base Titrations • The situation becomes a little more complicated when a weak acid/base is titrated with a strong base/acid. • Again, the keys are to identify the titration reaction and the equivalence point. • The method of calculating the pH will then depend on the region of the titration curve.

  17. III. Four Different Regions • For a weak acid titrated with a strong base, there are 4 regions as well: • Before titration: only HA in solution, so it’s a weak acid problem! • Pre-equivalence: a mixture of HA and A-, so it’s a buffer! • Equivalence point: only A- in solution, so it’s a weak base problem! • Post-equivalence: adding excess OH-, so it’s a dilution problem! • For a weak base titrated with a strong acid, everything is just rewritten w/ conjugates!

  18. III. Illustrative Problem • A 50.00 mL sample of 0.02000 M CH3COOH is being titrated with 0.1000 M NaOH. Calculate the pH before the titration begins, after 3.00 mL of the titrant have been added, at the equivalence point, and after 10.20 mL of the titrant have been added. Note that Ka = 1.75 x 10-5 for acetic acid.

  19. III. Illustrative Problem Solution • First, we need to write the titration eqn. • CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O(l) • Next, we calculate the equivalence point.

  20. III. Illustrative Problem Solution • Before titration begins, it’s just a weak acid problem. Solving this (with simplification), we get [H3O+] = 5.916 x 10-4, so pH = 3.228.

  21. III. Illustrative Problem Solution • After 3.00 mL of 0.1000 M NaOH have been added, we will have a mixture of CH3COOH and CH3COO- in solution. • Since it’s a buffer, we can use the Henderson-Hasselbalch eqn.

  22. III. Illustrative Problem Solution • In the H-H equation, we need a ratio of CH3COO- to CH3COOH. • After adding 3.00 mL, we are 3.00/10.00 to equivalence. We can use a relative concentration chart.

  23. III. Illustrative Problem Solution • Now we just plug the relative final row into the H-H equation.

  24. III. Illustrative Problem Solution • At the equivalence point, all CH3COOH has been converted to CH3COO-. • Initial moles of CH3COOH = moles of CH3COO- at the equivalence point, but the volume has increased. • Must calculate [CH3COO-] at equiv. pt.

  25. III. Illustrative Problem Solution • Now we solve a weak base problem. Using Kb = 5.714 x 10-10 and the simplification, x = 3.089 x 10-6. Thus, pOH = 5.5102 and pH = 8.490. Note that pH does not equal 7.00!!

  26. III. Illustrative Problem Solution • At 10.20 mL of added titrant, we are 0.20 mL past equivalence, and the pH depends only on excess OH-. Thus: Of course, this means that pOH = 3.479 and pH = 10.52.

  27. III. Sample Problem • A 25.00 mL sample of 0.08364 M pyridine is being titrated with 0.1067 M HCl. What’s the pH after 4.63 mL of the HCl has been added? Note that pyridine has a Kb of 1.69 x 10-9.

  28. III. Sample Weak/Strong Curves • Important aspects about pH curves for weak/strong titrations: • At equivalence, pH does not equal 7.00. • At ½ equivalence, pH = pKa.

  29. III. Polyprotic Acid Titration • If the Ka’s are different enough, you will see multiple equivalence points. • Since protons come off one at a time, 1st equiv. pt. refers to the 1st proton, 2nd to the 2nd, etc.

  30. III. Detecting the Equiv. Pt. • During a titration, the equivalence point can be detected with a pH meter or an indicator. • The point where the indicator changes color is called the endpoint. • An indicator is itself a weak acid that has a different color than its conjugate base.

  31. III. Phenolphthalein

  32. III. Indicators • The indicator has its own equilibrium: • HIn(aq) + H2O(l) In-(aq) + H3O+(aq) • The color of an indicator depends on the relative [ ]’s of its protonated and deprotonated forms. • If pH > pKa of HIn, color will be In-. • If pH = pKa of HIn, color will be in between. • If pH < pKa of HIn, color will be HIn.

  33. III. Selecting an Indicator

  34. IV. Solubility • In 1st semester G-chem, you memorized solubility rules and regarded compounds as either soluble or insoluble. • Reality is not as clear cut – there are degrees of solubility. • We examine solubility again from an equilibrium point of view.

  35. IV. Solubility Equilibrium • If we apply the equilibrium concept to the dissolution of CaF2(s), we get: • CaF2(s) Ca2+(aq) + 2F-(aq) • The equilibrium expression is then: • Ksp = [Ca2+][F-]2 • Ksp is the solubility product constant, and just like any other K, it tells you how far the reaction goes towards products.

  36. IV. Some Ksp Values

  37. IV. Calculating Solubility • Recall that solubility is defined as the amount of a compound that dissolves in a certain amount of liquid (g/100 g water is common). • The molar solubility is obviously the number of moles of a compound that dissolves in a liter of liquid. • Molar solubilities can easily be calculated using Ksp values.

  38. IV. Ksp Equilibrium Problems • Calculating molar solubility is essentially just another type of equilibrium problem. • You still set up an equilibrium chart and solve for an unknown. Pay attention to stoichiometry!

  39. IV. Sample Problem • Which is more soluble: calcium carbonate (Ksp = 4.96 x 10-9) or magnesium fluoride (Ksp = 5.16 x 10-11)?

  40. IV. The Common Ion Effect • The solubility of Fe(OH)2 is lower when the pH is high. Why? • Fe(OH)2(s) Fe2+(aq) + 2OH-(aq) • Le Châtelier’s Principle! • common ion effect: the solubility of an ionic compound is lowered in a solution containing a common ion than in pure water.

  41. IV. Sample Problem • Calculate the molar solubility of lead(II) chloride (Ksp = 1.2 x 10-5) in pure water and in a solution of 0.060 M NaCl.

  42. IV. pH and Solubility • As seen with Fe(OH)2, pH can have an influence on solubility. • In acidic solutions, need to consider if H3O+ will react with cation or anion. • In basic solutions, need to consider if OH- will react with cation or anion.

  43. IV. Sample Problems • Which compound, FeCO3 or PbBr2, is more soluble in acid than in base? Why? • Will copper(I) cyanide be more soluble in acid or base? Why? • In which type of solution is AgCl most soluble: acidic, basic, or neutral?

  44. IV. Precipitation • Ksp values can be used to predict when precipitation will occur. • Again, we use a Q calculation. • If Q < Ksp, solution is unsaturated. Solution dissolve additional solid. • If Q = Ksp, solution is saturated. No more solid will dissolve. • If Q > Ksp, solution is supersaturated, and precipitation is expected.

  45. IV. Sample Problems • Will a precipitate form if 100.0 mL 0.0010 M Pb(NO3)2 is mixed with 100.0 mL 0.0020 M MgSO4? • The concentration of Ag+ in a certain solution is 0.025 M. What concentration of SO42- is needed to precipitate out the Ag+? Note that Ksp = 1.2 x 10-5for silver(I) sulfate.

  46. V. Complex Ions • In aqueous solution, transition metal cations are usually hydrated. • e.g. Ag+(aq) is really Ag(H2O)2+(aq). • The Lewis acid Ag+ reacts with the Lewis base H2O. • Ag(H2O)2+(aq) is a complex ion. • A complex ion has a central metal bound to one or more ligands. • A ligand is a neutral molecule or an ion that acts as a Lewis base with the central metal.

  47. V. Formation Constants • Stronger Lewis bases will replace weaker ones in a complex ion. • e.g. Ag(H2O)2+(aq) + 2NH3(aq) Ag(NH3)2+(aq) + 2H2O(l) • For simplicity, it’s common to write Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) • Since this is an equilibrium, we can write an equilibrium expression for it.

  48. V. Formation Constants Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) • Kf is called a formation constant. • Unlike other equilibrium constants we’ve seen, Kf’s are large, indicating favorable formation of the complex ion.

  49. V. Sample Formation Constants

  50. V. Calculations w/ Kf’s • Since Kf’s are so large, calculations with them are slightly different. • We assume the equilibrium lies essentially all the way to the right. • This changes how we set up our equilibrium chart.

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