1. Here, we’ll show you how to calculate the pH and % ionization of a weak acid with a given concentration and a known Ka value. K a to pH and Percent Ionization

2. For the (a) part of this question, we’re asked to find the pH of a 0.25 M solution of ethanoic acid, CH3COOH. a) Find the pH of 0.25 M CH 3 COOH

3. The “b” part of the question asks us to find the % ionization in 0.25 M ethanoic acid. a) Find the pH of 0.25 M CH 3 COOH b) Find the % ionization of 0.25 M CH 3 COOH

4. We’ll start with the “a” part. We’re asked to find the pH of a 0.25 M solution of ethanoic acid, CH3COOH. a) Find the pH of 0.25 M CH 3 COOH

5. Whenever we’re asked to do a calculation with an acid, the first thing we always have to do is identify the acid as strong or weak. a) Find the pH of 0.25 M CH 3 COOH Strong Acid or Weak Acid?

6. We see by its position on the acid table (click) that ethanoic acid is a weak acid a) Find the pH of 0.25 M CH 3 COOH Strong Acid or Weak Acid? Weak Acid

7. When we’re doing calculations involving weak acids, we must use an ICE table. a) Find the pH of 0.25 M CH 3 COOH Strong Acid or Weak Acid? Weak Acid ICE Table

8. We set up an ice table like this a) Find the pH of 0.25 M CH 3 COOH [I] [C] [E]

9. We start by writing the equilibrium equation for ionization of ethanoic acid here. a) Find the pH of 0.25 M CH 3 COOH [I] [C] [E] Equilibrium Equation

10. Next, we draw borders in so that columns line up nicely with the substances in the equation. a) Find the pH of 0.25 M CH 3 COOH [I] [C] [E] Equilibrium Equation

11. Water is a liquid in the equilibrium equation so we can ignore the column below water. We’ll colour it blue here. a) Find the pH of 0.25 M CH 3 COOH [I] [C] [E]

12. We’ll start out with the initial concentration row. a) Find the pH of 0.25 M CH 3 COOH [I]1 [C]1 [E]1

13. The initial concentration of the ethanoic acid is 0.25 M, so will add 0.25 to this cell. a) Find the pH of 0.25 M CH 3 COOH [I]0.251 [C]1 [E]1

14. No H3O+ or CH3COOminus was added, so we can consider their concentrations to be 0 before ionization. a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]1 [E]1

15. Now we’ll look at the (click) changes in concentration as the ionization occurs. a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]1 [E]1

16. Because there were no products initially, the equilibrium will (click) shift to the right during the ionization. a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]1 [E]1 Shift to the Right

17. As a shift to the right occurs, the concentrations of hydronium and ethanoate ions will both increase so we’ll write + signs here. a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]1+?+?+?+? [E]1 Shift to the Right

18. And the concentration of CH3COOH will decrease, so we’ll write a minus sign here. a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+?+?+?+? [E]1 Shift to the Right

19. We’re not given any equilibrium concentrations, so we don’t know how much these will increase a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+?+?+?+? [E]1 Shift to the Right

20. So we’ll write + x for both of them, as they both have a coefficient of 1 in the equilibrium equation. a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+x+x+x+x [E]1 Shift to the Right

21. Because the coefficient on CH3COOH is also 1, we can state that it will go down by x, so we write minus x here. a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+x+x+x+x [E]1 Shift to the Right

22. Now, we’ll look at the concentrations of everything (click) at equilibrium. a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+x+x+x+x [E]1

23. The hydronium and ethanoate ions will both be 0 a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+x+x+x+x [E] 0.25–x 1xx

24. Plus x a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+x+x+x+x [E] 0.25–x 1xx

25. Which is equal to a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+x+x+x+x [E] 0.25–x 1xx 0 + x

26. x. a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+x+x+x+x [E] 0.25–x 1xx 0 + x

27. The concentration of CH3COOH started out as 0.25 a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+x+x+x+x [E] 0.25–x 1xx

28. and went down by x, a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+x+x+x+x [E] 0.25–x 1xx

29. so its equilibrium concentration will be… a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+x+x+x+x [E] 0.25–x 1xx

30. 0.25 minus x a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+x+x+x+x [E] 0.25–x 1xx

31. The question now is, How do we find x? a) Find the pH of 0.25 M CH 3 COOH [I]0.25100 [C]–x–x1+x+x+x+x [E] 0.25–x 1xx How do we find x ?

32. We start solving for x by writing the Ka expression for ethanoic acid a) Find the pH of 0.25 M CH 3 COOH

33. Using the equilibrium equation, we can see that the Ka expression is equal to the concentration of hydronium times concentration of ethanoate over the concentration of ethanoic acid. a) Find the pH of 0.25 M CH 3 COOH

34. The equilibrium concentrations of hydronium and ethanoate are both equal to x, a) Find the pH of 0.25 M CH 3 COOH

35. so for their product in the Ka expression, we can substitute x times x or x squared. a) Find the pH of 0.25 M CH 3 COOH

36. The concentration of ethanoic acid at equilibrium is 0.25 - x a) Find the pH of 0.25 M CH 3 COOH

37. So we’ll substitute 0.25 –x in for the concentration of CH3COOH a) Find the pH of 0.25 M CH 3 COOH

38. The degree of ionization for ethanoic acid is very low. So we make the assumption that x is insignificant compared to 0.25. This can be written as 0.25 – x is almost equal to 0.25 a) Find the pH of 0.25 M CH 3 COOH Assume 0.25–x  0.25

39. We can check this assumption later when we determine the % ionization. In general the assumption is valid if the percent ionization is 5% or less. a) Find the pH of 0.25 M CH 3 COOH Assume 0.25–x  0.25 Valid if the Percent Ionization is 5% or Less

40. Using this assumption will help us avoid having to use a quadratic equation. In Chemistry 12, this assumption is generally used here. When you use it, you Must Always state it. a) Find the pH of 0.25 M CH 3 COOH Assume 0.25–x  0.25 Always STATE this assumption if you’re using it!

41. so taking out the x on the bottom, we can state that Ka is approximately equal to x 2 over 0.25 a) Find the pH of 0.25 M CH 3 COOH

42. Rearranging this equation to solve for x 2 gives us x 2 = 0.25 times the Ka a) Find the pH of 0.25 M CH 3 COOH

43. Taking the square root of both sides gives us x is equal to the square root of 0.25 times the ka. a) Find the pH of 0.25 M CH 3 COOH

44. At this point, we’ll remind ourselves that x is equal to the equilibrium concentration of hydronium. a) Find the pH of 0.25 M CH 3 COOH

45. Looking on the acid table, we see that the Ka for ethanoic acid is 1.8 × 10 -5, so we substitute that for Ka in the equation. a) Find the pH of 0.25 M CH 3 COOH

46. 0.25 times 1.8 × 10-5 is equal to 4.5 × 10 -6, so we’ll substitute that in here, under the square root sign. a) Find the pH of 0.25 M CH 3 COOH

47. The square root of 4.5 x 10-6 is 2.12 x 10-3. Because we now have a value for the concentration of hydronium, we’ll include the unit Molarity. a) Find the pH of 0.25 M CH 3 COOH

48. Both the value of Ka and the given concentration have 2 significant figures, which limits our final answer to 2 significant figures. a) Find the pH of 0.25 M CH 3 COOH

49. We’ve written the concentration of hydronium with 3 significant figures here. We’ll use this and round to 2 significant figures at the end. a) Find the pH of 0.25 M CH 3 COOH

50. We’re asked for the pH. The pH is the negative log of the hydronium ion concentration, which is the negative log of 2.12 × 10 -3. a) Find the pH of 0.25 M CH 3 COOH

51. And that comes out to 2.67366. a) Find the pH of 0.25 M CH 3 COOH

52. But we round this to 2.67 in order to give us 2 decimal places in this pH value, or 2 significant figures in our final answer. a) Find the pH of 0.25 M CH 3 COOH

53. So now we have the final answer for Part “a”. The pH of 0.25M ethanoic acid is equal to 2.67 a) Find the pH of 0.25 M CH 3 COOH

54. Part “b” of this question asks us to find the percent ionization of 0.25 M ethanoic acid. b) Find the % ionization of 0.25 M CH 3 COOH

55. At one point in our calculations, we had determined that the hydronium ion concentration in this solution is 2.12 x 10 -3 molar. b) Find the % ionization of 0.25 M CH 3 COOH

56. We can use this to help us find the percent ionization b) Find the % ionization of 0.25 M CH 3 COOH

57. We’ll make a note of it up here. b) Find the % ionization of 0.25 M CH 3 COOH

58. The formula for percent ionization is the hydronium ion concentration divided by the initial concentration of the acid times 100 percent. b) Find the % ionization of 0.25 M CH 3 COOH

59. We’ll substitute 2.12 × 10 -3 M in for the concentration of hydronium b) Find the % ionization of 0.25 M CH 3 COOH

60. And 0.25 M in for the initial concentration of ethanoic acid. b) Find the % ionization of 0.25 M CH 3 COOH

61. We can cancel the unit Molarity and we multiply by 100% b) Find the % ionization of 0.25 M CH 3 COOH

62. And we get 0.85% b) Find the % ionization of 0.25 M CH 3 COOH

63. So now we’ve answered question “b”. The percent ionization of 0.25 M ethanoic acid is 0.85%. Notice it’s quite small, less than 1% of the ethanoic acid molecules have ionized. b) Find the % ionization of 0.25 M CH 3 COOH

64. remember during the calculations for pH, we assumed that x was insignificant compared to 0.25 M. We said this was valid if the % ionization is 5% or less. a) Find the pH of 0.25 M CH 3 COOH Assume 0.25–x  0.25 Valid if the Percent Ionization is 5% or less

65. We have just determined that the % ionization is 0.85 %, which is much less than 5%. Assume 0.25–x  0.25 Valid if the Percent Ionization is 5% or less Percent Ionization = 0.85% a) Find the pH of 0.25 M CH 3 COOH

66. So this verifies that our assumption was definitely valid! Assume 0.25–x  0.25 Valid if the Percent Ionization is 5% or less Percent Ionization = 0.85% a) Find the pH of 0.25 M CH 3 COOH