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Chapter 17 Additional Aspects of Acid-Base Equilibria

Chapter 17 Additional Aspects of Acid-Base Equilibria. Dr. Peter Warburton peterw@mun.ca http://www.chem.mun.ca/zcourses/1051.php. The common-ion effect. Say we make two acid solutions: 0.100 M HCl (a strong acid) and 0.100 M CH 3 COOH (a weak acid).

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Chapter 17 Additional Aspects of Acid-Base Equilibria

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  1. Chapter 17Additional Aspects of Acid-Base Equilibria Dr. Peter Warburton peterw@mun.ca http://www.chem.mun.ca/zcourses/1051.php

  2. The common-ion effect • Say we make two acid solutions: • 0.100 M HCl (a strong acid) and • 0.100 M CH3COOH (a weak acid). • A 0.100 M HCl solutionby itself would have a pH 1.0([H3O+] = 0.100M) since the reaction goes to completion.

  3. The common-ion effect • A 0.100 M CH3COOH solution • (Ka = 1.8 x 10-5)by itself would have a • pH 2.8 since an equilibrium is established where the equilibrium concentration of acetic acid  0.100 M • x = [H3O+] = [CH3COO-] = 1.3 x 10-3 M

  4. The common-ion effect

  5. The common-ion effect • Say we put together a solution that is BOTH • 0.100 M HCl (a strong acid) and • 0.100 M CH3COOH (a weak acid). • The two reactions of • the acids with water take place • in the same container • at the same time!

  6. The common-ion effect • Both reactions are a source of H3O+ and so we could expect that • [H3O+] = 0.100 M + 1.3 x 10-3 M • [H3O+] = 0.1013 M • and pH  0.99 • which appears to be true

  7. The common-ion effect • However, both reactions share the common ion H3O+, and so they cannot be treated as independent reactions. Something that affects one reaction must also affect the other reaction. • Le Chatalier’s Principle!

  8. The common-ion effect • Imagine we start with the 0.100 M CH3COOH and then we add 0.100 M H3O+by adding 0.100 M HCl. • Le Chatalier’ Principle tells us our reaction will shift back towards reactants!

  9. The common-ion effect • If we assume x is much smaller than 0.100 M we will quickly find that x = Ka • x = [CH3COO-] = 1.8 x 10-5 M • The value of x has decreased because of the added H3O+!

  10. H3O+ as a common-ion

  11. OH- as a common-ion

  12. Salts as source of a basic common-ion

  13. Salts as source of a basic common-ion On the left is a solution of 0.100 M CH3COOH while on theright is a solution that is both 0.100 M CH3COOH and 0.100 M CH3COONawhich is a source of the basic common ion CH3COO-. The reaction has shifted back towards reactants!

  14. Salts as source of an acidic common-ion

  15. Salts as source of an acidic common-ion On the left is a solution of 0.100 M NH3while on theright is a solution that is both 0.100 M NH3and 0.100 M NH4Clwhich is a source of the acidic common ion NH4+. The reaction has shifted back towards reactants!

  16. Buffer solutions • Solutions that contain both a weak acid and its conjugate base are buffer solutions. • These solutions are resistant to changes in pH.

  17. Buffer solutions • The system has enough of the • original acid and conjugate base molecules • in the solution to react with • added acid or addedbase, • so the new equilibrium mixture will be • very close in composition to the original equilibrium mixture.

  18. Buffer solutions • A 0.10 molL-1 acetic acid – 0.10 molL-1 acetate mixture has a pH of 4.74 and is a buffer solution! • CH3COOH (aq) + H2O (l)H3O+ (aq) + CH3COO- (aq)

  19. Buffer solutions • If we rearrange the Ka expression • and make the assumption that x is much less than 0.10, we see • If [CH3COOH] = [CH3COO-], • then [H3O+] = 1.8 x 10-5M = Ka • and pH = pKa = 4.74

  20. Buffer solutions

  21. Buffer solutions • What happens if we add 0.01 mol of NaOH(strong base) to 1.00 L of the acetic acid – acetate buffer solution? • CH3COOH (aq) + OH- (aq)→H2O (l) + CH3COO- (aq) • This goes to completion and keeps occurring until we run out of the limiting reagent OH- New[CH3COOH] = 0.09 Mandnew[CH3COO-] = 0.11 M

  22. Buffer solutions • With the assumption that x is much smaller than 0.09 mol (an assumption we always need to check after calculations are done!), we find Note we’ve made the assumption that x << 0.09! pH = - log [H3O+] pH = - log 1.5 x 10-5 pH = 4.82

  23. Buffer solutions • Adding 0.01 mol of OH- to 1.00 L of water would have given us a pH of 12.0! • There is no significant amount of acidin water for the base to react with.

  24. Buffer solutions • What happens if we add 0.01 mol of HCl(strong acid) to 1.00 L of the acetic acid – acetate buffer solution? • CH3COO- (aq) + H3O+ (aq)→H2O (l) + CH3COOH (aq) • This goes to completion and keeps occurring until we run out of the limiting reagentH3O+ New[CH3COOH] = 0.11 Mandnew[CH3COO-] = 0.09 M

  25. Buffer solutions • With the assumption that x is much smaller than 0.09 mol (an assumption we always need to check after calculations are done!), we find Note we’ve made the assumption that x << 0.09! pH = - log [H3O+] pH = - log 2.2 x 10-5 pH = 4.66

  26. Buffer solutions • Adding 0.01 mol of H3O+ to 1.00 L of water would have given us a pH of 2.0! • There is no significant amount of basein water for the acid to react with.

  27. Adding acid or base to a buffer

  28. Problem • Calculate the pH of a 0.100 L buffer solution that is 0.25 mol/L in HF and 0.50 mol/L in NaF. • With the assumption that x is much smaller than 0.25 mol (an assumption we always need to check after calculations are done!), we find

  29. Problem pH = - log [H3O+] pH = - log 1.75 x 10-4 pH = 3.76

  30. Problem • a) What is the change in pH on addition of 0.002 mol of HNO3? New[HF] = 0.27 Mandnew[F-] = 0.48 M

  31. Problem Notice we’ve made the assumption that x << 0.27. We should check this! pH = - log [H3O+] pH = - log 1.97 x 10-4 pH = 3.71

  32. Problem • b) What is the change in pH on addition of 0.004 mol of KOH? New[HF] = 0.21 Mandnew[F-] = 0.54 M

  33. Problem Notice we’ve made the assumption that x << 0.21. We should check this! pH = - log [H3O+] pH = - log 1.36 x 10-4 pH = 3.87

  34. Predicting whether a solution is a buffer • Any solution that becomes a mixture of a conjugate acid-base pair will be a buffer. • 1)Weakacid-baseconjugate pairs like CH3COOH and CH3COO- orNH4+ and NH3. • 2) Weakacid reacting withsmall amounts of strong base like CH3COOH and NaOH. • 3) Weakbasereacting withsmall amounts of strong acid like NH3 and HCl.

  35. Problem • Describe how a mixture of a strong acid such as HCl and a salt of a weak acid such as CH3COONa can be a buffer solution.

  36. Problem • What is the pH of a buffer solution prepared by dissolving 23.1 g of NaCHO2(molar mass is68.01 gmol-1)in a sufficient volume of 0.432 M HCHO2 to make 500.0 mLof the buffer? • Ka of formic acid is 1.8 x 10-4 Answer: pH = 3.94

  37. The Henderson-Hasselbalch equation • We’ve seen that, for solutions with both members of a conjugate acid-base pair, that • pH = pKa + log [base] / [acid] • This is called the Henderson-Hasselbalch Equation.

  38. The Henderson-Hasselbalch Equation • If we have a buffer solution of a conjugate acid-base pair, then the pH of the solutionwill be close to the pKa of the acid. • This pKa value is modified by the logarithm of ratio of the concentrations of the base and acid in the solution to give the actual pH.

  39. The Henderson-Hasselbalch equation • pH = pKa + log [base] / [acid] • ALWAYS remember when you use the H-H eqn that there is theassumptionthat the equilibrium concentrations ofacidandbaseare relatively unchangedfrom the initial concentrations. • That is we have assumedx is very small compared to the initial concentrations!

  40. The Henderson-Hasselbalch equation • pH = pKa + log [base] / [acid] • Generally this assumptionis valid as long as we know • 0.10 < [base] / [acid] < 10 • AND • [base] / Ka> 100AND [acid] / Ka> 100

  41. Alternate Henderson-Hasselbalch equation • We can always look at a buffer solution as a base combined with its conjugate acid • B (aq)+ H2O (l) OH- (aq)+ BH+ (aq) • pOH = pKb + log [acid] / [base]

  42. Alternate Henderson-Hasselbalch equation • If we have a buffer solution of a conjugate acid-base pair, then the pOH of the solutionwill be close tothe pKb of the base. • This pKb value is modified by the logarithm of ratio of the concentrations of the acidand base in the solution to give the actual pH.

  43. Problem • Use the Henderson-Hasselbalch equation to calculate the pH of a buffer solution prepared by mixing equal volumes of 0.20 mol/L NaHCO3 and 0.10 mol/L Na2CO3. • Ka of HCO3- = 4.7 x 10-11 • (see Table 16.4) • We should also check the validity of using H-H at the end to be sure!

  44. Problem answer • If we mix equal volumes, the total volume is TWICE the volume for the original acid or basesolutions. • Since the number of moles of acid or baseDON’T CHANGE on mixing, • the concentrations will be • half the given values. • pH = pKa + log [base] / [acid] • pH = (-log 4.7 x 10-11)+ log (0.05) / (0.10) • pH = 10.33 – 0.30 • pH = 10.03

  45. Preparing buffer solutions • If we want to create a buffer solution of a specific pH, the H-H equation tells us we need to pick a conjugateacid-base pair with a pKafor the acidclose to the pH we want, and then we adjustthe amounts of the conjugate acid and base.

  46. How do we adjust the concentrations?

  47. Problem • How many grams of (NH4)2SO4(molar mass is 132.141gmol-1)must be dissolved in 0.500 L of 0.35 M NH3 to produce a solution with pH = 9.00? Assume that the solution volume remains at 0.500 L. • Kb for ammonia is 1.8 x 10-5. Answer: 21 g

  48. Buffer capacity • Buffer capacity is the measure of the ability of a buffer to absorb acid or basewithout significant change in pH. • Larger volumes of buffer solutions have a higher buffer capacity, and buffer solutions of higher initial concentrations of the conjugate acid-base pair have a larger buffer capacity.

  49. Buffer range • We’ve seen that as long as • 0.10 < [base] / [acid] < 10 • then the assumption that the Henderson-Hasselbalch equation is based upon • (x << [base] and x << [acid]) • is likely to be valid.

  50. Buffer range pH = pKa + log [base] / [acid] pH = pKa + log 10 pH = pKa + 1.0 OR pH = pKa + log [base] / [acid] pH = pKa + log 0.10 pH = pKa - 1.0

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