1. Chapter 15 Acid-Base Titration and pH 1

2. Solution Concentrations* Molarity – one mole of solute dissolved in enough solvent (water) to make exactly one liter of solution. Molality – one mole of solute dissolved in exactly 1,000 grams of solvent. Normality – one gram equivalent weight (gew) of solute dissolved in enough solvent to make exactly one liter of solution. These are on your Ch 14/15 handout titled: “ph/Acid/Base Equations” 2

3. Molarity (M) Formulae M = grams of solute given GFW of solute liters of solvent Grams of solute needed= M(GFW of solute)(Liters Solvent) L of solvent needed = g solute/GFW solute / M 3

4. Some molarity problems are on pages 420 and 421 You have 3.50L of solution that contains 90.0g of NaCl. What is the molarity of the solution? g NaCl x 1mol NaCl = mol NaCl, the solute g NaCl mol of Solute = molarity of solution L of solution 90.0g NaCl x 1 mol NaCl = 1.54 mol NaCl 58.44 g NaCl 1.54 mol NaCl / 3.50 L = 0.440 M NaCl 4

5. Another molarity problem. P421 practice. 1.What is the molarity of a solution composed of 5.85g KI, dissolved in enough water to make 0.125 L of solution? 5.85g KI x 1 mol KI = 0.0352 mol KI 166g KI 0.0352 mol KI = 0.282 M KI 0.125 L Complete #2 and # on P421. Check your answers in the back of the book. 5

6. molality, m – one mole of solute dissolved in exactly 1,000 g of solvent. m = g of solute given X 1000 GFW of solute X g of solvent g of solute needed = m(GFW)(g solvent) 1000 g solvent needed = g solute(1000) GFW solute x m Molality = molarity if water is the solvent (aqueous solutions) 6

7. Normality, N – one gram equivalent weight of solute dissolved in enough solvent to make exactly one liter of solution. N = _________g solute_____ GEW solute x L solvent g solute = N X GEW solute X L of solvent L solvent = ____g of solute___ GEW of solute X N Normality to Molarity M = N(valence of cation)(subscript of cation) GEW = _________GFW of solute_______ charge X subscript of solute cation GEW – gram equivalent weight 7

8. pH – What is it? pH is an indication of the hydronium ion concentration present in a solution. [H 3 0 + ] is the symbol for concentration of hydronium ion in moles per liter or mol/L pOH is an indication of the hydroxide ion concentration present in a solution. [OH-] is the symbol for concentration of hydroxide ion in mol/L 8

9. Water self ionizes H 2 0(l) + H 2 0(l) H 3 0 + (aq) + OH - (aq) In the above reaction, two water molecules produce a hydronium ion and a hydroxide ion by transfer of a proton. Water is self Ionizing. At 25 o C, the concentrations of H 3 0 + and OH - are each only 1.0x10 -7 mol/L of water. 9

10. Math product of these ions is a constant k w, the ionization constant of water. K w = [H 3 0 + ] [OH - ] = 1.0x10 -7 (1.0x10 -7 ) =1.0x10 -14 This occurs at 25 o C. If the temperature changes, the ion product, K w changes. When both [H 3 0 + ] and[OH - ] are 1.0x10 -7, the solution is neutral. If [H 3 0 + ] is greater than 1.0x10 -7, the solution is Acidic. (10 -6 or 10 -4 would be greater) If [OH - ] is greater than 1.0x10 -7, the solution is Basic. 10

11. New info K w - the ionization constant of water = 1x10 -14. [H 3 0 + ] – concentration of hydronium ions in mol/L or M. Kw = [H 3 0 + ] [OH - ] 11

12. Calculating without a calculator K w = [H 3 0 + ] [OH - ] = 1.0x10 -7 (1.0x10 -7 ) =1.0x10 -14 Let’s say that the [H 3 0 + ] is 1.0x 10 -6 and you are asked to find the [OH - ]. K w = [H 3 0 + ] [OH - ] --> [OH - ] = K w = 1.0x10 -14 [H 3 0 + ] 1.0x10 -6 -14 – (-6) = -14 + 6 = -8 so: [OH - ] = 10 -8 mol/Liter More practice: 10 -14 /10 -2 = 10 -12 and 10 -14 /10 -9 = 10 -5 12

13. Calculating [H 3 0 + ] and [OH - ] Your own scientific calculator is a MUST here!!! Find these keys: 2 nd, either EE or EXP, and change sign (-) or (+/-) on your calculator. Let’s practice putting in numbers in sci. not. 1x10 -7 : Press keys in this sequence: 1 2 nd EE (-) 7 on your display you see something similar to this: 1E -7 2 x10-4: 2 2 nd EE (-) 4 display: 2 E -4 13

14. For concentration, M means moles/L The [H 3 0 + ] is 2.34 x 10 -5 M in a solution. Calculate the [OH - ] of the solution. [OH - ] = K w = 1.0x10 -14 [H 3 0 + ] 2.34 x 10 -5 Key sequence: 1 2 nd EE (-) 14 : 2.34 2 nd EE (-) 5 enter Display: 4.27 E -10 which means: 4.27 x 10 -10 M 14

15. Calculate hydronium and hydroxide ion concentrations in a solution that is 1x10 -4 M HCl. HCl is a strong acid that ionizes completely. So the concentration of H 3 0 + is 1x10 -4 M. Find [OH-]: [OH - ] = K w = 1.0x10 -14 [H 3 0 + ] 1x10 -4 Answer: [OH - ] = 10 -10 M Asgn: Page 502 in book: Practice 2,3,4 15

16. New info K w - the ionization constant of water = 1x10 -14. [H 3 0 + ] – concentration of hydronium ions in mol/L or M. Kw = [H 3 0 + ] [OH - ] 16

17. The pH Scale is used to show how acidic or basic (alkaline) a solution is. pH of a solution is the negative of the common logarithm of the hydronium ion concentration. pH = - log [H 3 0 + ] A common logarithm of a number is “the power to which 10 must be raised to equal the number.” 17

18. “pHinding” pH The logarithm of 1.0x10 -7 is - 7.0 The pH = - log [H 3 0 + ] = - log 1.0x10 -7 = 7.0 pOH is the negative log of [OH-]. pOH = -log [OH-]. In a neutral solution where [OH-] is 1.0x10 -7, the pOH = -log [OH-] = -log 1.0x10 -7 = 7.0 18

19. K w = [H 3 0 + ] [OH - ] = 1.0x10 -7 (1.0x10 -7 ) =1.0x10 -14 From above: pH + pOH = 14.0 pOH can also be found by: pOH = 14.0 – pH = 14.0 – 7.0 = 7.0 19

20. Determine the pH of these solutions: a.1x10 -3 M HCl. HCl is a strong acid that ionizes completely. So the concentration of H 3 0 + is 1x10 -3 M. pH = - log [H 3 0 + ] = - log (1.0x10 -3 ) = 3.0 b.1x10 -4 M NaOH. NaOH is a strong base that ionizes completely. The concentration of OH - is 1x10 -4 M [H 3 0 + ] = K w = 10 -14 = 10 -10 [OH - ] 10 -4 pH = -log(10 -10 ) = 10.0 20

21. More pH samples… Find the pH of a solution where [H 3 0 + ] is 2.8 x 10 -5 M? pH = -log [H 3 0 + ] = -log 2.8 x 10 -5 = 4.55 Key sequence: (-) log 2.8 2 nd EE (-) 5 = Find the pH of a 4.7 x 10 -2 M NaOH solution. [H 3 0 + ] = K w = 1 x 10 -14 [OH - ] 4.7x10 -2 [H 3 0 + ] = 2.1x10 -13 pH = -log [H 3 0 + ] = - log 2.1x10 -13 pH = 12.7 21

22. Calculating [H 3 0 + ] & [OH - ] from whole number pH pH = -log [H 3 0 + ] log [H 3 0 + ] = -pH [H 3 0 + ] = antilog (-pH) [H 3 0 + ] = 10 -pH If pH = 7, [H 3 0 + ] = 1 x 10 -7 If pH = 2, [H 3 0 + ] = 1x10 -2 22

23. Calculating [H 3 0 + ] & [OH - ] from whole number pH Concentration to pH: pH = -log [H 3 0 + ] pH to Concentration: [H 3 0 + ] = 10 -pH Key sequence: 10 x (-) pH 23

24. Find the [H 3 0 + ] of a solution with pH 7.52. (Not a whole number) pH = -log [H 3 0 + ] log [H 3 0 + ] = -pH [H 3 0 + ] =antilog(-pH) = antilog (-7.52) = 1x10 -7.52 = 3.0 x 10 -8 M H 3 0 + Flow chart: [2 nd ] [log] [+/- ] [7.52] = 3.01x10 -8 M 24

25. Practice on page 508 2. pH = 12.0 so… [H 3 0 + ] = 1x10 -12 M 3. The pH of an aqueous solution is measured as 1.50. Calculate the [H 3 0 + ] and [OH - ]. [H 3 0 + ] = [2 nd ] [Log] [+/-] [1.5] [H 3 0 + ] = 1x10 -1.5 = 3.16x10 -2 M [OH - ] = ___Kw___ = 1 x10 -14 = 3.16 x 10 -13 M 3.16x10 -2 3.16x10 -2

26. The pH of an aqueous solution is 3.67. Determine [H 3 0 + ]. [H 3 0 + ] = antilog (-pH) = antilog (-3.67) … OR [2 nd ] [log] [+/-] [3.67] = 2.14 x10 -4 M

27. Review Acid + Base --> Salt + Water K w = [H 3 0 + ][OH - ] = 1x10 -14 pH = -log [H 3 0 + ] [H 3 0 + ] = 10 -pH  “ taking the antilog “ pOH = -log [OH - ] pH + pOH = 14.0 27

28. Solve these now What is the pH of a 2.69 x 10 -3 M HCl solution? What is the [OH - ] of a solution with a pH of 11.2? 28

29. 29

30. “pHinding” pH Assignment 505/1. b and d 506/1-4 508/1-4 Ch15 Rev: 125/2-4d and 126/5,6a 523/6-12 a and b ONLY on each 523/13-15 a only on each and 16 a-e 30

31. In this section we are going to look at: indicators, pH meters, and titrations. You used 2 indictors in a lab recently to determine how acidic or basic several solutions were. You used litmus paper and pH paper (Hydrion) Acid-base indicators are sensitive to pH of acids and bases. They will change color as a result of the ions present. 31

32. In acidic solutions, an indicator will be one color (litmus turns red) and in basic solutions, an indicator will be another color (litmus burns blue). 32

33. Indicator Samples Methyl red, Bromthymol blue, Methyl orange, Phenolphthalein, Phenol red are indicator samples. These will ionize in solution and, depending upon their acid or base strength will change color over a range of pH values until the end point is reached. The range over which an indicator changes color is called its transition interval. 33

34. Reading Indicator Values Litmus gives a very broad reading – a solution is either acidic or basic. Indicators are more specific in reading the pH of an acid or base. But the most accurate method of measuring pH is with a pH meter. A pH meter determines the pH of a solution by electrically measuring the voltage between the two electrodes placed in a solution. 34

35. Titration Titration is the controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration. More simply: it is using a known concentration of a solution to determine the concentration of a solution of unknown concentration. 35

36. Titration Set up and Procedure Retrieved from web.ysu.edu. Your book has a more complete explanation. 1.Fill one buret with an acid. Record volume. 2.Fill other buret with standard solution base. Record volume. 3.Indicator (Phenolphthalein) will be in a flask. 4.Add a given amount of A to the flask. 5.Begin adding B to the flask until the pink color of the indicator begins to form. Swirl the contents constantly. 6.As the pink color begins to remain for longer periods of time, you are nearing the end point. 7.When the pink color remains after 30 seconds of swirling, the equivalence point is reached. 8.Record the exact volume of the base put in the flask. 36

37. Equivalence point The point at which two solutions used in a titration are present in chemically equivalent amounts. End point The point in a titration at which an indicator changes color. The figure below shows typical pH curves for various acid- base titrations. The equivalence points and end points are different for the various combinations of strong and weak acids and bases. Retrieved from: chemguide.co.uk 37

38. Determination of the acidity/alkalinity of salt solutions produced by neutralization reactions. The relative pH of a neutralized solution can be determined utilizing the following scale showing A/B strengths. pH 1-3 strong acid with a weak base pH 3-5 strong acid with a moderate base pH 5-7 moderate acid with a weak base pH 7(neutral)equal strength acid/base reacton pH 7-9moderate base with a weak acid pH 9-12strong base with a moderate acid pH 12-14strong base with a weak acid See your Ch 14/15 handout for more information. 38

39. Molarity and Titration Standard solution – the solution that contains the precisely known concentration of a solute. Primary standard – highly purified solid compound used to check the concentration of the known solution in a titration. Knowing the molarity and volume of a known solution used in a titration, the molarity of a given volume of a solution with unknown concentration can be found. 39

40. Formula to calculate molarity in a titration. Ma x Va = Mb x Vb Ma – molarity of acid Va – volume of acid Mb – molarity of base Vb – volume of base See Ch 14/15 handout for more info. 40

41. Sample Problem 524/24 Determine the number of moles of the first substance that would be chemically equivalent of the second substance in these acid-base titrations: a.NaOH with 1.0 mol HCl b.HNO 3 with 0.75 mol KOH c.Ba(OH) 2 with 0.20 mol HF Answers: a. NaOH + HCl  NaCl + H 2 0 1.0mol HCl 1 mol NaOH = 1.0 mol NaOH 1 mol HCl b. HNO 3 + KOH  KNO 3 + H 2 0 0.75 mol KOH 1 mol HNO 3 = 0.75 mol 1 mol KOH c. Ba(OH) 2 + 2HF  BaF 2 + 2H 2 O 0.20 mol HF 1 mol Ba(OH) 2 = 0.10 mol Ba(OH) 2 2 mol HF 41

42. Sample: 524/37 Suppose that 10.1 mL of HNO 3 is neutralized by 71.4 mL of a 4.2 x 10 -3 M solution of KOH in a titration. Calculate the concentration of the HNO 3 solution. KOH + HNO 3  KNO 3 + H 2 0 4.2 x 10 -3 mol KOH 71.4 mL L L 1mol HNO 3 1 L 1000 mL 1mol KOH = 3.0 x 10 -4 mol HNO 3 3.0 x 10 -4 mol HNO 3 1000 mL = 3.0 x 10 -2 M HNO 3 10.1 mL L 42

47. 14.3 Assignments: 489/1,2 491/19-25 492/26-30,36,37 (Overall practice problems) Sample problem on next page:

48. Titration links: https://www.youtube.com/watch?v=HwKFcjGH N_4 https://www.youtube.com/watch?v=8UiuE7Xx5l 8 48

49. Titration Consider the following neutralization reaction: OH – (aq) + HA(aq) H 2 O(l) + A – (aq) known concentration unknown concentration strong baseweak acid Can be determined using the mole ratio: 1 mole OH – ~ 1 mole HA equivalence point: in a titration, the point where just enough base (or acid) has been added to neutralize all the acid (or base). Determining an unknown concentration by performing a chemical reaction with a solution of known concentration

50. Titration OH – (aq) + HA(aq) H 2 O(l) + A – (aq) known concentration unknown concentration strong base Volume of OH – added (read on the burette) Moles of OH – added Moles of HA originally present in the beaker Original concentration of HA weak acid Experimental setup

51. How do you know when the equivalence point is reached? That is, how do you know when enough NaOH has been added? Titration You can use a pH meter, or a pH indicator

52. Titration The equivalence point (V NaOH = 5 mL) is located at the steepest part of the curve Phenolphthalein and methyl orange go through their entire color change in that pH range

53. Titration Why is the titration curve shaped that way? Why is it “flat” at the beginning and the end, and steep in the middle?

54. Chemical reactions with acids and bases: Corrosion Electrolysis Neutralization Salts of weak acids Salts of strong acids Titration experiments Buffers We need to understand the concept of buffers

55. Acid-base conjugates

56. HC 2 H 3 O 2 H + + C 2 H 3 O 2 – So for this weak acid: acid base acid-base conjugates HC 2 H 3 O 2 is the conjugate acid of C 2 H 3 O 2 – C 2 H 3 O 2 – is the conjugate base of HC 2 H 3 O 2

57. buffer: a solution that resists changes in pH by chemical action. A solution that contains a mixture of a weak acid and its conjugate base is an example of a buffer Add an acid The conjugate base will neutralize the added acid HA H + + A – acid base

58. buffer: a solution that resists changes in pH by chemical action. A solution that contains a mixture of a weak acid and its conjugate base is an example of a buffer Add a base The conjugate acid will neutralize the added base HA H + + A – acid base

59. Buffer capacity Add an acid The conjugate base will neutralize the added acid HA H + + A – acid base The conjugate acid will neutralize the added base Add a base Buffer: There is little or no change in pH

60. Buffer capacity Add an acid The conjugate base will neutralize the added acid HA H + + A – acid base The conjugate acid will neutralize the added base Add a base At some point you run out of conjugate base At some point you run out of conjugate acid The buffer capacity is reached! If more acid or base is added, the pH will change drastically Buffer:

61. Why is the titration curve shaped that way? (1)It’s a buffer! The pH is relatively constant HA H + + A – acid base (1) OH – (aq) + HA(aq) H 2 O(l) + A – (aq)

62. Why is the titration curve shaped that way? (2) OH – (aq) + HA(aq) H 2 O(l) + A – (aq) (1)It’s a buffer! The pH is relatively constant (2)The buffer capacity is reached (We ran out of HA to neutralize the added base) HA H + + A – acid base (1)

63. Why is the titration curve shaped that way? (3) OH – (aq) + HA(aq) H 2 O(l) + A – (aq) (1)It’s a buffer! The pH is relatively constant (2)The buffer capacity is reached (We ran out of HA to neutralize the added base) (3) There is an excess of NaOH. The pH is the pH of NaOH (not the pH of A – ) because NaOH is a strong base HA H + + A – acid base (2) (1)