Introduction to Acids & Bases II. Packet #26

Transcription

1 Introduction to Acids & Bases II Packet #26 1

2 Review I Svante Arrhenius was the first person to recognize the essential nature of acids and bases. 2

3 Review II Arrhenius postulated that: Acids produce hydrogen ions (H + ) in an aqueous solution Bases produce hydroxide ions (OH - ) in an aqueous solution. The limitation of Arrhenius concept was that OH - was the only base that could be produced. H 2 O HCl(g) H + (aq) + Cl - (aq) H 2 O NaOH(s) Na + (aq) + OH - (aq) Top reaction The product produced is hydrochloric acid. Both products are in the aqueous form Bottom reaction The bottom reaction occurs when sodium hydroxide, is placed in water. Produces a strong base as the solution. 3

4 Review III Bronsted-Lowry In the Bronsted-Lowry model: - An acid is a proton (H + ) donor A base is a proton (H + ) acceptor This results in the formation of a new acid, called the conjugate acid, and a new base called the conjugate base. 4

5 Review IV Bronsted-Lowry Acids & Bases HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) Acid + Base Conjugate acid + Conjugate Base Example HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl - (aq) HCl(aq) and Cl - become the acid-conjugate base pair H 2 O(aq) and H 3 O + (aq) become the base-acid conjugate base pair. A conjugate acid-base pair consists of two substances relate to each other by the donating and accepting of a Thursday, single April 16, proton Ryan Barrow

6 BEHAVIOR OF WATER 6

7 Introduction Water molecules are highly polar and are in continuous motion. Occasionally, water molecules collide and a self-ionization reaction occurs. 7

8 Introduction II In pure water, the equilibrium concentration of hydrogen ions (H + ) and hydroxide ions (OH - ) are both 1 * 10-7 M The concentrations are equal and the aqueous solution is described as a neutral solution. 8

9 Introduction III In an aqueous solution, when H + increases while OH - decreases OR when H + decreases while OH - increases, Le Chatelier s principle applies. 9

10 Le Chatelier s Principle Le Chatelier s principle The principle that when an external force is applied to a system at equilibrium, the system adjusts so as to minimize the effect of the applied force. 10

11 Introduction IV If additional ions (either H + or OH - ) are added to a solution, the equilibrium shifts. While one ion increases, the other decreases. More water molecules are produced in the process. 11

12 Introduction VI For aqueous solutions, the PRODUCT of the hydrogen ion concentration and the hydroxide ion concentration equals 1.0 * [H + ] * [OH - ] = 1.0 *

13 Introduction VII As substances are ADDED to a dilute aqueous solution, the concentration of the ions will change. Therefore, the product of the concentration of the ions, H + and OH -, equaling 1.0 * 10-14, is the ion-product constant for water. (K w ) K w = [H + ] * [OH - ] = 1.0 *

14 Overview Acids & Bases Aqueous Solution {Solute + Solvent(H2O)} Acid Base Acidic Solution Basic Solution pka pkb Acid + Water Base + Water ph poh 14

15 Equations Associated with Acids, Bases, Acidic Solutions & Basic Solutions K a = [H+][Conj. Base] / (Molarity of solution [H+]) pk a = -log(k a ) ph = pk a + log 10 ( [Conj. Base] / (Molarity of solution K b = [OH-][Conj. Acid] / (Molarity of solution [OH-]) pk b = -log(k b ) poh = pk b + log 10 ( [Conj. Acid] / (Molarity of [H+]) ph = -log[h + ] ph = 14 - poh [H + ] = 10 -ph solution [OH-]) poh = -log[oh - ] poh = 14 - ph [OH - ] = 10 -poh 15

16 Not to be confused with ACIDS! FORMATION OF ACIDIC SOLUTIONS 16

17 Formation of Acidic Solutions Remember, not all aqueous solutions are neutral. Having equal amounts of H + and OH - When some substances dissolve in water, they release H + HCl(g) H + + Cl - When this occurs within the solution, the hydrogen ion (H + ) concentration is greater than the hydroxide ion (OH - ) concentration Remember, there are already H + ions present from the selfionization of H 2 O. The solution is now considered an acidic solution as the concentration of hydrogen ions (H + ) is greater than hydroxide ions (OH - ) The concentration of hydrogen ions (H + ) is greater than 1.0 * 10-7 M 17

18 INTRODUCTION TO LOGARITHMS 18

19 Definition of Logarithm log b n = p b p = n b > 0 and b = 16 log 2 16 = = 1000 log = 3 19

20 Not to be confused with BASES! FORMATION OF BASIC SOLUTIONS 20

21 Formation of Basic(Alkaline) Solutions Remember, not all aqueous solutions are neutral. Having equal amounts of H + and OH - When some substances dissolve in water, they release OH - NaOH(s) Na + (aq) + OH - (aq) When this occurs within the solution, the hydroxide ion (OH - ) concentration is greater than the hydrogen ion (H + ) concentration Remember, there are already OH - ions present from the selfionization of H 2 O. The solution is now considered an acidic solution as the concentration of hydrogen ions (H + ) is greater than hydroxide ions (OH - ) The concentration of hydrogen ions (H + ) is less than 1.0 * 10-7 M while the hydroxide ion (OH - ) is greater than 1.0 * 10-7 M. 21

22 pk a An acid dissociation constant (K a ) is the ratio of the concentration of the dissociated (or ionized) form of an acid to the concentration of the undissociated (nonionized) form. The dissociated form includes both the H 3 O + and/or the anion H +. 22

23 PH 23

24 The Connection ph Expressing hydrogen-ion concentration in morality is long-winded. Therefore, the more widely used system for expressing [H + ] is the ph scale. The ph of a solution is the negative logarithm of the hydrogen-ion concentration. ph = -log[h + ] 24

25 The Connection ph Let us investigate a neutral solution ph = -log[h + ] [H + ] = 1.0 * 10-7 ph = -log(1 * 10-7 ) ph = -(log 1 + log 10-7 ) ph = -(0.0 + (-7.0) ph = 7 25

26 POH 26

27 The Connection poh The poh of a solution is the negative logarithm of the hydroxide-ion [OH - ] concentration. poh = -log[oh - ] 27

28 Relationship Among [H + ], [OH - ] and ph 28

29 Problem Set I What is the ph of a solution with a hydrogen-ion concentration of 4.2 * M? Find the ph of the following solutions [H + ] = 1 * 10-4 M [H + ] = M What is the ph values of the following solutions, based on their hydrogen-ion concentrations? [H + ] = 1.0 * M [H + ] = 0.045M 29

30 Problem Set II The ph of an unknown solution is What is its hydrogen-ion concentration [H + ]? Calculate [H + ] for each of the following: - ph = 5.00 ph = ph = 4.00 ph =

31 Problem Set III What is the ph of a solution if [OH - ] = 4.0 * M? Calculate the ph of each solution: - [OH - ] = 4.3 * 10-5 M [OH - ] = 4.5 * M [H + ] = 5.0 * 10-5 M [H + ] = 8.3 * M 31

32 STRENGTHS OF ACIDS AND BASES 32

33 Strengths of Acids and Bases I Acids are classified as strong or weak depending on the degree to which they ionize in water. Strong acids are completely ionized in aqueous solutions. Weak acids ionize only slightly in aqueous solutions. 33

34 Relative Strengths of Acids & Bases 34

35 Acid Dissociation Constant PKA 35

36 pk a An acid dissociation constant (K a ) is the ratio of the concentration of the dissociated (or ionized) form of an acid to the concentration of the undissociated (nonionized) form. The dissociated form includes both the H 3 O + and/or the anion H +. 36

37 pk a An acid dissociation constant (K a ) is the ratio of the concentration of the dissociated (or ionized) form of an acid to the concentration of the undissociated (nonionized) form. The dissociated form includes both the H 3 O + and/or the anion H +. 37

38 pka Weak acids have small K a values. The stronger an acid is, the larger its K a value. A stronger acid would have a higher [H 3 O + ] / [H + ]. 38

39 pka Weak acids have small K a values. The stronger the acid, the larger its K a value. 39

40 pka Table 40

41 Problem Set IV A M solution of ethanoic acid (CH 3 COOH) is only partially ionized. From measurements of ph of the solution, [H + ] is determined to be 1.34 * 10-3 M. What is the acid dissociation constant (K a ) of ethanoic acid? What is the ph of the solution? What is the poh of the solution? What is the [OH - ] of the solution? K a = [H+][Conj. Base] / (Molarity of solution [H+]) pk a = -log(k a ) ph = pk a + log 10 ( [Conj. Base] / (Molarity of solution [H+]) ph = -log[h + ] ph = 14 - poh [H + ] = 10 -ph K b = [OH-][Conj. Acid] / (Molarity of solution [OH-]) pk b = -log(k b ) poh = pk b + log 10 ( [Conj. Acid] / (Molarity of solution [OH-]) poh = -log[oh - ] poh = 14 - ph [OH - ] = 10 -poh [H + ][OH - ] = 1 *

42 Problem Set V In an exactly 0.1M solution of methanoic acid, [H + ] = 4.2 * 10-3 M. Calculate the K a, pk a, ph, POH, {OH - }, K b, and pk b of the methanoic acid. In an exactly 0.2M of a monoprotic weak acid, [H + ] = 9.86 * 10-4 M. Calculate the K a, pk a, ph, POH, {OH - }, K b, and pk b for this acid? K a = [H+][Conj. Base] / (Molarity of solution [H+]) pk a = -log(k a ) ph = pk a + log 10 ( [Conj. Base] / (Molarity of solution [H+]) ph = -log[h + ] ph = 14 - poh [H + ] = 10 -ph K b = [OH-][Conj. Acid] / (Molarity of solution [OH-]) pk b = -log(k b ) poh = pk b + log 10 ( [Conj. Acid] / (Molarity of solution [OH-]) poh = -log[oh - ] poh = 14 - ph [OH - ] = 10 -poh [H + ][OH - ] = 1 *

43 Problem Set VI Aspirin has a pk a of 3.4. What is the ratio of [A ] aka(conj. Base) to [HA] in: (a) the blood (ph = 7.4) (b) the stomach (ph = 1.4) K a = [H+][Conj. Base] / (Molarity of solution [H+]) pk a = -log(k a ) ph = pk a + log 10 ( [Conj. Base] / (Molarity of solution [H+]) ph = -log[h + ] ph = 14 - poh [H + ] = 10 -ph K b = [OH-][Conj. Acid] / (Molarity of solution [OH-]) pk b = -log(k b ) poh = pk b + log 10 ( [Conj. Acid] / (Molarity of solution [OH-]) poh = -log[oh - ] poh = 14 - ph [OH - ] = 10 -poh [H + ][OH - ] = 1 *

44 Problem Set VII You need to produce a buffer solution that has a ph of You already have a solution that contains 10.0 mmol (millimoles) of acetic acid. How many millimoles of sodium acetate will you need to add to this solution? The pk a of acetic acid is

45 Base Dissociation Constant PK B 45

46 pk b Just like there are strong acids and weak acids, there are also strong bases and weak bases. Strong bases dissociate completely into metal ions and hydroxide ions in an aqueous solution. Weak bases react with water to form the hydroxide ion and the conjugate acids of the base. Example: - NH 3 (aq) + H 2 O(l) NH 4 (aq) + OH - (aq) 46

47 pk b The base association constant (K b ) is the ratio of the concentration of the conjugate acid times the concentration of the hydroxide ion to the conjugate base. 47

48 Relative Strengths of Acids & Bases 48

49 pk b Table 49

50 Special Note The words concentrated and dilute indicates how much of an acid or base is dissolved in a solution. The terms also refer to the number of moles of the acid or base in a given volume of solution. The words strong or weak refer to the extent of ionization or dissociation of an acid or base. HCl(aq) is a strong acid. HCl completely dissociates into ions. NH 3 is a weak base. The amount of ionization is small. 50

51 Overview I Acids & Bases Aqueous Solution {Solute + Solvent(H2O)} Acid Base Acidic Solution Basic Solution pka pkb Acid + Water Base + Water ph poh 51

52 Overview II Keep in mind Both basic solutions and acidic solutions have a calculated ph and poh Both acids and bases have a pk a and pk b Acidic Solution Aqueous Solution {Solute + Solvent(H2O)} Acid + Water Basic Solution Base + Water Acids & Bases Acid pka Base pkb ph poh 52

53 Overview III Solve for number of moles of acid/base g sample / molar mass Solve the molarity of the solution. Moles of solute / Liters of solution Calculate the concentration of the acid/base Convert the molarity into exponential format. Calculate the ph. Calculate the poh ph = -(log[h + ]) ph + poh = 14 Calculate the concentration of the acid/base Thursday, April 16, 2015 [H + ] + [OH - ] Ryan = 1 * Barrow

54 Overview IV Equations Associated with Acids, Bases, Acidic Solutions & Basic Solutions K a = [H+][Conj. Base] / (Molarity of solution [H+]) pk a = -log(k a ) ph = pk a + log 10 ( [Conj. Base] / (Molarity of solution K b = [OH-][Conj. Acid] / (Molarity of solution [OH-]) pk b = -log(k b ) poh = pk b + log 10 ( [Conj. Acid] / (Molarity of [H+]) ph = -log[h + ] ph = 14 - poh [H + ] = 10 -ph solution [OH-]) poh = -log[oh - ] poh = 14 - ph [OH - ] = 10 -poh 54

55 Problem Set VIII Calculate the ph and poh for the following solution at 25 C. 1 * 10-3 M [OH - ] 2.85 * 10-4 M [H + ] The ph of a soil sample was measure to be What is the [H + ] and [OH - ] of the soil? Calculate the [H + ], ph, poh and [OH - ] for a 4L solution that contains 28g of HCl. Calculate the [H + ], ph, poh and [OH - ] for a 6L solution that contains 460g of sulfuric acid (H 2 SO 4 ). 55

56 BUFFERS 56

57 Buffers A buffer solution is one that retains a constant ph despite the addition of small quantities of acids or bases. Buffers contain both hydrogen ion donors and acceptors. Hydrogen carbonate ions may act as an acceptor or a donor Hydrogen carbonate salts and phosphate salts are responsible for the buffering of blood. ph of

58 ph Scale 58

59 REVIEW 59

60 OTHER SLIDES THAT MAY BE HELPFUL 60

61 Dissociation Water has a slight tendency to dissociate into ions according to the equation 2H 2 O H 2 O + + OH - H 2 O H + + OH - Hydronium ions and hydroxide ions 61

62 Dissociation II Apart from dissociating itself, water readily causes the dissociation of other substances placed in it. Making it an excellent solvent. 62

63 Neutral ph In 1 liter of water, this dissociation produces 1/10,000,000 (10-7 ) mole of hydrogen ions. Equivalent to a ph of 7 Neutral 63

64 Acidic ph If the concentration of hydrogen ions is higher than what is seen at neutral ph, then the solution would be acidic An acid, therefore, is a substance that donates hydrogen ions Increases hydrogen ion concentration Example If the concentration of hydrogen ions is 1 / 1000 (10-3 ), then the ph would be 3 64

65 Basic ph If the concentration of hydrogen ions is lower than what is seen at neutral ph, the solution would be basic A base is a substance that is hydrogen ion acceptor Decreases hydrogen ion concentration Example If the concentration of hydrogen ions is 1 / 1,000,000,000 (10-9 ) the ph would be 9. 65