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Chapter 16

Chapter 16. From Kotz , Chemistry & Chemical Reactivity, 5th edition Illustrates the transformation of one insoluble lead compound into an even less soluble compound. Buffers, Titrations, Solubility, and other useless information. 16.1 Common Ion Effect.

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Chapter 16

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  1. Chapter 16 From Kotz, Chemistry & Chemical Reactivity, 5th edition Illustrates the transformation of one insoluble lead compound into an even less soluble compound. Buffers, Titrations, Solubility, and other useless information

  2. 16.1 Common Ion Effect • According to Le Chatelier’s principle: if the concentration of a substance involved in an equilibrium is changed, the system will adjust to accommodate the change and maintain the value of the K • Common ion effect: refers to the equilibrium disturbance involving a weak acid or base and its conjugate partner

  3. Common Ion Effect • CH3CO2H + OH- H2O + CH3CO2- • The addition of more acetate ion suppresses the ionization of the acetic acid and causes the system to shift left. • To calculate the pH of a solution containing both a weak acid and its conjugate base, the small changes in initial concentration can (usually) be ignored

  4. Common Ion Effect • Ex. Suppose a solution is 0.10M acetic acid and 0.050M sodium acetate. The Ka of acetic acid is 1.8 x 10-5. Calculate pH. • Ka = [H3O+][CH3CO2-] [CH3CO2H] 1.8 x 10-5 = [H3O+][0.050] [H3O+] = 3.6x10-5 [0.10] pH = 4.44 What is the pH of the same solution without the common ion added?

  5. Common Ion – a special case • What is the pH when 25.0mL of 0.0500M sodium hydroxide is added to 25.0mL of 0.100M lactic acid? (Ka of lactic acid is 1.4x10-4) • 0.00125 mol of NaOH is present • 0.00250 mol of lactic acid is present • All of the base combines with half of the acid. 0.00125 mol of lactate ion is produced and 0.00125 mol of lactic acid remain • This is the half equivalence point! Ka = [H+]

  6. 16.2 Buffer Solutions:Controlling pH • A buffer solution is resistant to change in pH when an acid or base is added to the solution • Buffer solutions are simply an example of the common ion effect; they consist of a weak acid and its conjugate base (the common ion) or a weak base and its conjugate acid in solution, in nearly equimolar quantities

  7. Buffers • There are two requirements for a buffer: • Two substances are needed: an acid capable of reacting with added OH- ions and a base that can consume added H3O+ ions • The acid and base must not react with each other

  8. Buffers • From a list of weak acids (or bases), one is chosen that has a pKa close to the pH value desired • The relative amounts of the weak acid and its conjugate base are adjusted to achieve exactly the pH required

  9. Some Commonly Used Buffer Systems

  10. Buffers • It is the relative quantities of weak acid and conjugate base that are important; the actual concentrations do not matter • Diluting a buffer solution will not change its pH • An increase in the concentration of the buffer components increases the buffer capacity – more acid or base can be added without change in pH

  11. Henderson-Hasselbalch Equations(a.k.a. David Hasselhoff) • pH = pKa + log [A-] [HA] pOH = pKb + log [HB+] [B] The Henderson-Hasselbalch equation is generally valid when the ratio of [conj base]/[acid] is less than 10 and greater than 0.1

  12. Using Henderson-Hasselbalch • Ex. Suppose you dissolve 15.0g of NaHCO3 and 18.0g of Na2CO3 in enough water to make 1.00L of solution. Calculate the pH using Henderson-Hasselbalch. • 15.0g NaHCO3 = 0.179 mol NaHCO3/ 1 L • 18.0g Na2CO3 = 0.170 mol Na2CO3 / 1 L • Ka = 4.8x10-11 (from a table) • pH = -log(4.8x10-11) + log (.170/.179) • pH = 10.3

  13. Preparing a Buffer Solution • Ex. Prepare a 1.0L buffer solution with a pH of 4.30 • Select an acid whose pKa is close to the desired pH (from a table) • Use either the general equation for a buffer or Henderson-Hasselbalch to calculate the concentration ratio of acid/base needed • 2.8/1 acid/base concentration ratio

  14. How Does a Buffer Maintain pH? • If you add 1.0 mL of 1.0M HCl to 1.0L of water, how does the pH change? • Water has pH = 7 • 0.001 M H+ has pH = 3 • If you add 1.0mL of 1.0M HCl to 1.0L of 0.7M acetic acid/0.6M sodium acetate buffer, how does the pH change? • The pH of the buffer is 4.68 (how do you find this?)

  15. How Does a Buffer Maintain pH? • 0.001 mol of H+ from the HCl combines with 0.001 mol of acetate ion and produces 0.001 mol of acetic acid • Stoich to get new concentrations from the reaction (0.599M acetate ion, 0.701M acetic acid) • ICE these concentrations to get equilibrium concentrations and set up buffer equation to solve for [H+] • pH = 4.68

  16. 16.3 Acid-Base Titrations • The pH at the equivalence point of a strong acid/strong base titration is 7 (neutral) • If weak acid is titrated with strong base then pH > 7 at equivalence point due to conj base of weak acid • If weak base is titrated with strong acid then pH < 7 at equivalence point due to conj acid of weak base

  17. Titration of a Strong Acid with a Strong Base

  18. Titration of a Strong Acid with a Strong Base • pH of the initial solution is the pH of the acid • As NaOH is added to the acid solution, amount of HCl declines, volume of solution increases, so H+ concentration decreases and pH slowly increases • The equivalence point is the midpoint of the vertical portion of the curve; pH is 7 here • After all HCl has been used, pH rises slowly as more NaOH is added (and volume increases) • The pH at any other point is found using stoichiometry and relationship between pH and [H+]

  19. Titration of a Weak Acid with a Strong Base

  20. Titration of a Weak Acid with a Strong Base • The pH before any titration begins is found from the Ka of the weak acid and the acid concentration • Anywhere between the start and the equivalence point, the Henderson-Hasselbalch equation can be used • At the half-equivalence point the concentration of the weak acid is equal to the concentration of the conj base, so pH=pKa • At the equivalence point only the conj base remains; the pH is controlled by the conj base Kb and concentration • Beyond the equivalence the pH is found from the volume of the excess base added

  21. Titration of a Weak Polyprotic Acid

  22. Titration of a Weak Polyprotic Acid • The curve can be divided into three parts: • The portion of the curve up to the first equivalence point has a pH determined by the excess of the polyprotic acid • The portion of the curve between the first and second equivalence points has a pH determined by the excess of the amphiprotic substance • The portion of the curve after the second equivalence point is has a pH determined by the excess of the fully deprotonated conj base

  23. Titration of a Weak Base with a Strong Acid

  24. Titration of a Weak Base with a Strong Acid • At the half-equivalence point, [OH-] = Kb of the weak base • The pH at the equivalence point is weakly acidic due to the conj acid of the weak base

  25. 16.4 pH Indicators • Usually a weak acid or base (treat it as such mathematically) • Often a large organic molecule that has different shapes in acid and base solution • The different structures have different colors that allows for monitoring changing pH • Choose an indicator with a Ka near that of the acid being titrated so that the color change occurs at the right stage in the titration

  26. Acid-Base IndicatorsFigure 18.8

  27. Indicators for Acid-Base Titrations

  28. Natural IndicatorsRed rose extract at different pH’s and with Al3+ ions Rose extract In CH3OH Add Al3+ Add HCl Add NH3 Add NH3/NH4+ See pages 848–849

  29. 16.5 Solubility of Salts • Salts are considered to be insoluble if less than 0.01 moles can be dissolved per liter of water • The equilibrium constant for the solubility of a salt is called the solubility product (Ksp), and from this molar solubility can be calculated • Addition of a common ion depresses the solubility of a salt (Le Chatelier) • Direct comparisons of the solubility of two salts on the basis of their Ksp values can only be made for salts having the same ion ratio

  30. BaCl2 Ba2+ + 2Cl- x 2xKsp = [Ba2+][2Cl-]2Ksp = (x)(4x2)Ksp = 4x3 In solubility problems, s is often substituted for x when solving for molar solubility. • NaClKsp = s2 • BaCl2Ksp= 4s3 • AlCl3Ksp = 27s4 • Al2(SO4)3Ksp = 108s5

  31. Barium SulfateKsp = 1.1 x 10-10 (b) BaSO4 is opaque to x-rays. Drinking a BaSO4 cocktail enables a physician to exam the intestines. (a) BaSO4 is a common mineral, appearing a white powder or colorless crystals.

  32. Common Ion Effect PbCl2(s)  Pb2+(aq) + 2 Cl-(aq) Ksp = 1.9 x 10-5 How will the addition of lead(II) ion or chloride ion impact the solubility of lead(II) chloride? If a saturated solution of lead(II) chloride is prepared, what will happen when sodium chloride solution is added to the mixture?

  33. Common Ion Effect and Salt Solubility • Ex. If solid AgCl is placed in 1.00L of 0.55M NaCl, what mass of AgCl will dissolve? • AgCl  Ag+ + Cl- Ksp = 1.8x10-10 • s = 1.3x10-5 mol/L

  34. Common Ion Effect and Salt Solubility • Assume x is very small compared to 0.55 (because Ksp is so small) • Ksp = 1.8x10-10 = (x)(0.55) • X = 4.4x10-10 mol/L (which is less than 1.3x10-5 mol/L, as predicted by Le Chatelier’s principle

  35. Effect of Basic Anions on Salt Solubility • Any salt containing an anion that is the conjugate base of a weak acid will dissolve in water to a greater extent than given by Ksp PbS  Pb2+ + S2- • The conj base can hydrolyze water, which lowers the [conj base] causing more of the salt to dissolve to reestablish equilibrium S2- + H2O  HS- + OH- • Salts of phosphate, acetate, carbonate, cyanide, and sulfide can be affected

  36. Effect of Basic Anions on Salt Solubility • Insoluble salts in which the anion is the conjugate base of a weak acid will dissolve in strong acids • Anions such as acetate, carbonate, hydroxide, phosphate, and sulfide dissolve in strong acids. Ex. Mg(OH)2 + 2 H3O+ Mg2+ + 4 H2O • Salts are not soluble in strong acid if the anion is the conj base of a strong acid. Ex. AgCl is not soluble in strong acid because Cl- is a very weak base of a very strong acid

  37. 16.6 Precipitation Reactions • Precipitation is the reverse process of dissolving • If you write a dissolving reaction, its equilibrium constant expression, and its Ksp, you can write the reverse reaction and its equilbrium constant expression; notice that it has 1/Ksp!

  38. Ksp and the Reaction Quotient, Q • If Q = Ksp the solution is saturated • The ion concentrations are at equilibrium values • If Q<Ksp the solution is not saturated • If more of the solid is present it will continue to dissolve until equilibrium is reached; if there is no solid present, more can be added • If Q>Ksp the solution is supersaturated • The ion concentrations are too high and precipitation will occur until equilibrium is reached

  39. Solubility and the Reaction Quotient • Solid PbI2(Ksp = 9.8 x 10-9) is placed in a beaker of water. After a period of time, the lead(II) concentration is measured and found to be 1.1 x 10-3 M. Has the system yet reached equilibrium? • Q = [Pb2+][2x I-]2 • Q = 5.3 x 10-9 This is less than Ksp, more PbI2can dissolve. • Can you figure out how much more can be added?

  40. 16.7 Solubility and Complex Ions • Metal ions form complex ions with Lewis bases, such as ammonia and water • The formation of complex ions increases the solubility of metal ions as predicted by the Ksp

  41. 16.8 Solubility, Ion Separations, and Qualitative Analysis • Add CrO42- to solid PbCl2. The less soluble salt, PbCrO4, precipitates • PbCl2(s) + CrO42-  PbCrO4 + 2 Cl- • Salt Ksp PbCl2 1.7 x 10-5 • PbCrO4 1.8 x 10-14 Separating Salts by Differences in Ksp

  42. Separating Salts by Differences in Ksp • PbCl2(s) + CrO42- PbCrO4 + 2 Cl- • Salt Ksp • PbCl2 1.7 x 10-5 • PbCrO4 1.8 x 10-14 PbCl2(s)  Pb2+ + 2 Cl- K1 = Ksp Pb2+ + CrO42- PbCrO4 K2 = 1/Ksp Knet = K1 • K2 = 9.4 x 108 Net reaction is product-favored

  43. Separating Salts by Differences in Ksp A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first? Ksp for Ag2CrO4 = 9.0 x 10-12 Ksp for PbCrO4 = 1.8 x 10-14 Solution The substance whose Ksp is first exceeded precipitates first. The ion requiring the lesser amount of CrO42- ppts. first. You MUST use molar solubility to determine this!

  44. Separating Salts by Differences in Ksp A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first? Ksp for Ag2CrO4 = 9.0 x 10-12 Ksp for PbCrO4 = 1.8 x 10-14 Solution Calculate [CrO42-] required by each ion. [CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+] = 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M [CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2 = 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M PbCrO4 precipitates first

  45. Dissolving Precipitates by forming Complex Ions Examine the solubility of AgCl in ammonia. AgCl(s) Ag+ + Cl- Ksp = 1.8 x 10-10 Ag+ + 2 NH3 --> Ag(NH3)2+ Kform = 1.6 x 107 ------------------------------------- AgCl(s) + 2 NH3Ag(NH3)2+ + Cl- Knet = Ksp • Kform = 2.9 x 10-3 By adding excess NH3, the equilibrium shifts to the right.

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