1 / 46

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.

harding-raymond
Download Presentation

Chapter 16

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

  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. pH of a Buffer • General buffer equation Ka = [H+][A-] [HA] pH = -log [H+] Check 100Ka < [HAinitial] to be sure you can use the short cut

  10. Some Commonly Used Buffer Systems

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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?)

  16. 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 • ICE this 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

  17. 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

  18. Titration of a Strong Acid with a Strong Base

  19. 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+]

  20. Titration of a Weak Acid with a Strong Base

  21. 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

  22. Titration of a Weak Polyprotic Acid

  23. 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

  24. Titration of a Weak Base with a Strong Acid

  25. 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

  26. 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

  27. Acid-Base IndicatorsFigure 18.8

  28. Indicators for Acid-Base Titrations

  29. 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

  30. 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

  31. 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

  32. 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.

  33. 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?

  34. 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

  35. 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

  36. 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

  37. 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

  38. 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!

  39. 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

  40. 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?

  41. 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

  42. 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

  43. 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

  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 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!

  45. 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

  46. 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.

More Related