1 / 86

CHAPTER 17

CHAPTER 17. ACID – BASE EQUILIBRIA. I. INTRODUCTION. A) Acid strength is measured by the extent of the overall reaction of the acid with water. 1) Strong acids go 100% to the right. HCl + H 2 O  H 3 O + + Cl -. 2) Weak acids go "slightly" to the right. HA + H 2 O 34 H 3 O + + A -.

lilli
Download Presentation

CHAPTER 17

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 17 ACID – BASE EQUILIBRIA

  2. I. INTRODUCTION A) Acid strength is measured by the extent of the overall reaction of the acid with water. 1) Strong acids go 100% to the right. HCl + H2O  H3O+ + Cl-

  3. 2) Weak acids go "slightly" to the right. HA + H2O 34 H3O+ + A- Why can we make [H2O] Kc = to Ka?

  4. Ka is known as the acid dissociation constant and the acid ionization constant. CH3COOH + H2O 34 H3O+ + CH3COO-

  5. 3) Some Ka's at 25oC: 4) Some questions I will want you to be able to answer from these data are: a) Which of the above acids is the strongest? b) Which is the weakest acid? c) What is the strongest conjugate base?

  6. II. HOW CAN WE FIND Ka's? A) The most common way to obtain a Ka is to measure the pH of a solution prepared by dissolving a known amount of weak acid to form a given volume of solution. B) An example: The pH of a 0.10 M solution of HOCl is 4.23. What is the value of Ka?

  7. How do we get the values of [H3O+] , [OCl-] and [HOCl]? From the pH we can obtain what value? 4.23 = - log [H3O+] - 4.23 = log [H3O+] 5.9 X 10-5 = [H3O+] Then we know the value of the [OCl-] = ? The value of [HOCl] = ?

  8. We set up a table like we did in Chapter 15.

  9. Then we put the values in the appropriate equation for Ka.

  10. III. DEGREE OF DISSOCIATION AND PERCENT DISSOCIATION (OR IONIZATION). A) The degree of dissociation is the fraction of molecules that react with water to give ions.

  11. If the [H+] in a 0.60 M solution of HF is 0.021 M, what is the degree of dissociation? % ionization is obtained by multiplying the degree of dissociation by 100 and adding the % sign.

  12. C) The larger the Ka the greater the degree of dissociation (the greater the % dissociation). The stronger the acid the greater the % ionization. D) As the original concentration decreases the percent dissociation increases. How do we explain this situation?

  13. IV. Calculation of the Concentration of Species in a Weak Acid Solution Using Ka - THE APPROXIMATE METHOD

  14. A) What are the concentrations of all species in a 0.10 M of acetic acid (CH3COOH)? What is the pH of the solution? Ka for acetic acid is 1.7 X 10-5. You will need a balanced equation and a chart. CH3COOH + H2O 34 H3O+ + CH3COO-

  15. Then we put the results of the chart into the equation for Ka.

  16. To avoid using the quadratic equation and focus more attention on what is happening chemically, we THINK AS FOLLOWS: Since Ka is so small, x can be ASSUMED to be much less than 0.10. That makes 0.10 - x = 0.10. Then we don't have to use the quadratic equation to obtain x.

  17. The above expression for Ka becomes:

  18. BUT YOU will have to check your assumption: To check if the assumption is ok, we will use the 5% rule as given in the text. How do we calculate the 5%? Another check to see if the assumption is ok is to check to see if the precision rule allows you drop the x. In the above case: 0.10 - 0.0013 = 0.10 (The answer can only be reported to hundredths.)

  19. The answer to the question then includes the following: [CH3COOH] = 0.10 M [H2O] = 55.5 M [H3O+] = 1.3 X 10-3 M [CH3COO-] = 1.3 X 10-3 M And there is one more species in the solution. What is it and what is its concentration?

  20. To answer the question, we still have to calculate the pH. pH = - log 1.3 x 10-3 = 2.89

  21. B) Calculate the [H3O+] in a 0.100 M solution of nitrous acid, HNO2, for which the Ka is 4.5 X 10-4. You will need an equation and a chart. HNO2+ +H2O 34 H3O+ + NO2-

  22. Now we have to see if the assumption is ok!! 0.100 - 0.0067 gives an answer to three digits (0.093) so that indicates the approximation is not appropriate for this problem.

  23. Also, IF YOU TAKE: You cannot use the approximate method, you must use the quadratic equation. What answer do we obtain with the quadratic equation?

  24. Work some extra credit problems. C) Calculating concentrations of species in a solution of a diprotic acid. (What is a diprotic acid? 1) We will examine the stepwise ionization of oxalic acid, a diprotic acid found in spinach and rhubarb.

  25. H2C2O4 + H2O 34 H3O+ + HC2O4- Ka1 = 5.90 x 10-2 HC2O4- + H2O 34 H3O+ + C2O42- Ka2 = 6.40 x 10-5 Now what do we do with this information? If the Ka's are far apart, we can make the problem easier. We first compare the Ka's to see which one is more important in delivering H3O+'s to the solution.

  26. In this problem Ka1 is much greater than Ka2. We will assume that all the H3O+'s came from the first equation, when calculating the [H3O+], [HC2O4-], and the [C2O42-] in the 0.10 M H2C2O4 solution. You will need an equation and a chart.

  27. H2C2O4+ H2O34 H3O+ + HC2O4- Ka1 = 5.90 x 10-2 We now put these values in the general equation:

  28. Ka is fairly large so the approximate method cannot be used. You must use the quadratic equation to obtain the value of x. The above Ka relation becomes: x2 + 5.90 X 10-2x - 5.9 X 10-3 = 0 The positive root gives x = _____ = [H3O+] = [HC2O4-] We then proceed to the second ionization, where we make a second chart corresponding to the second reaction.

  29. HC2O4- + H2O 34 H3O+ + C2O42- Ka2 = 6.40 x 10-5 We then place these values in the equation for Ka2.

  30. Since Ka2 is small we can assume that y << 0.053. When we do this the above equation becomes:

  31. We obtain the following values for the concentrations of the following species in solution: [H3O+] = 0.053 M [HC2O4-] = 0.053 M [C2O42-] = 6.4 X 10-4 M [H2O] = 55.5 M [H2C2O4] = 0.10 - 0.053 = 0.05 M [OH-] = 1.00 X 10-14 divided by 0.053 is 1.9 X 10-13 M

  32. V. Base Ionization Equilibrium A) Weak bases have a Kb. NH3 + H2O 34 NH4+ + OH- B) Some other weak bases are what can be called organic relatives of ammonia.

  33. CH3NH2 + H2O 34 CH3NH3+ + OH-Kb = 4.4 X 10-4 (CH3)2NH + H2O 34 (CH3)2NH2+ + OH- Kb = 5.1 X 10-4 (CH3)3N + H2O 34 (CH3)3NH+ + OH- Kb = 7.4 X 10-5 C6H5NH2 + H2O 34 C6H5NH3+ + OH-Kb = 4.2 X 10-10

  34. C) Calculate the pH of a 1.0 M solution of methylamine, CH3NH2. What will be your guess at the answer? _____ We need the equation, a chart and the set-up for Kb . CH3NH2 + H2O 34 CH3NH3+ + OH-

  35. We then assume that x << 1.0 and we obtain the following expression:

  36. We then obtain the [OH-] = [CH3NH3+] = 2.1 X 10-2 M. We have to check the assumption: [OH-] = 2.1 X 10-2 M pOH = - log [OH-] = - log 2.1 X 10-2 = 1.68 This is not the answer to the question. Remember the question is about pH, not pOH.

  37. So the pH = 14.00 - 1.68 = 12.32 Some extra credit problems for you to do in class. VI. Acid and Base Properties of Salts A) I want you to be able to predict whether the solution of a salt is acidic, basic or neutral. B) How will you be able to do this?

  38. C) You will have to recall the 6 strong acids and 6 strong bases, and call all others are WEAK. 1) Salts which have cations from strong bases and the anions of strong acids have no effect upon the [H+] when dissolved in water. EXAMPLE: NaCl

  39. Na+ is the cation of NaOH, a very strong base, therefore Na+ is a very weak conjugate acid. It does not react with water to take an OH- from it. Cl- is the anion of HCl, a very strong acid, therefore Cl- is a very weak conjugate base. It does not react with water to take an H+ from it. We expect that the water solution of NaCl to be neutral.

  40. Na+ + H2O ---> N.R. Cl- + H2O ---> N.R. Neutral solution results. We have not changed the concentration of H3O+ or the concentration of the OH- ions by inserting NaCl into the water.

  41. 2) Salts which have cations from strong bases and the anions of weak acids produce a basic solution when dissolved in water. EXAMPLES: NaC2H3O2 sodium acetate NaC2H3O2 in water dissociates 100% to Na+ ions and CH3COO- ions. Na+ is the cation of NaOH, a very strong base, therefore Na+ is a very ________ ________ _______________ . It does not react with water to take an OH from it.

  42. CH3COO- is the anion of the weak acid, acetic acid, therefore its conjugate base is stronger and will react with water to produce more OH- ions. Na+ + H2O ----> N.R. CH3COO- + H2O 34 CH3COOH + OH- Do you see how a basic solution results from the action of the conjugate base with water?

  43. Another example is NaF. What happens with it? It dissociates 100% into ________ions and ________ions. Which of these ions can react with water? What kind of solution will it make? Will the pH of the solution be greater or less than 7?

  44. 3) A salt which has the cation from a weak base and the anion of a strong acid produce an acidic solution.EXAMPLE: NH4Cl NH4Cl ----> NH4+ + Cl- Cl- + H2O ----> N.R. (Why?) NH4+ + H2O 34 NH3 + H3O+ We see that the concentration of the H3O+ increases so the pH of the solution will be __________________ 7.

  45. 4) Salts derived from a weak acid and a weak base have both ions reacting with water (undergoing hydrolysis). a) Whether the solution is acidic or basic depends on the relative acid - base strengths of the two ions. To determine this, you will need to compare the cation's Ka with the Kb of the anion.

  46. b) If the Kb of the anion > the Ka of the cation, the solution is basic. c) If the Kb of the anion < the Ka of the cation, the solution is acidic. d) If the Kb of the anion = the Ka of the cation, the solution is neutral. B) How do we obtain Ka's and Kb's for ions?

  47. What is the Kb for the CH3COO- ion? We start with CH3COONa. It dissociates 100% CH3COONa ----> CH3COO- + Na+ Na+ + H2O ----> N.R. CH3COO- + H2O 34 CH3COOH + OH-

  48. Generally the values for the Ka's and Kb's for ions are not given in books. How can we obtain them if this is the case? What will be given in the text and how can we use it? The Ka for the weak acid will be given. How is that related to the Kb for the conjugate base?

  49. Compare Ka with Kb.

  50. If we look closely, we see that part of Kb is upside down from the Ka. So let's invert it. BUT this still isn't equal to Kb. We have to eliminate H3O+ from the denominator and get OH- in the numerator. How can we do this?

More Related