Chapter 15 Acids and Bases

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1 Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Chapter 15 Acids and Bases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall

2 Stomach Acid & Heartburn the cells that line your stomach produce hydrochloric acid to kill unwanted bacteria to help break down food to activate enzymes that break down food if the stomach acid backs up into your esophagus, it irritates those tissues, resulting in heartburn acid reflux GERD gastroesophageal reflux disease chronic leaking of stomach acid into the esophagus Tro, Chemistry: A Molecular Approach 2

3 Curing Heartburn mild cases of heartburn can be cured by neutralizing the acid in the esophagus swallowing saliva which contains bicarbonate ion taking antacids that contain hydroxide ions and/or carbonate ions Tro, Chemistry: A Molecular Approach 3

4 Properties of Acids sour taste react with active metals i.e., Al, Zn, Fe, but not Cu, Ag, or Au 2 Al + 6 HCl 2 AlCl H 2 corrosive react with carbonates, producing CO 2 marble, baking soda, chalk, limestone CaCO HCl CaCl 2 + CO 2 + H 2 O change color of vegetable dyes blue litmus turns red react with bases to form ionic salts Tro, Chemistry: A Molecular Approach 4

5 Common Acids Chemical Name Formula Uses Strength Nitric Acid HNO 3 explosive, fertilizer, dye, glue Strong Sulfuric Acid H 2 SO 4 explosive, fertilizer, dye, glue, batteries Strong Hydrochloric Acid HCl metal cleaning, food prep, ore refining, stomach acid Strong Phosphoric Acid H 3 PO 4 fertilizer, plastics & rubber, food preservation Moderate plastics & rubber, food Acetic Acid HC 2 H 3 O 2 preservation, Vinegar Weak Hydrofluoric Acid HF metal cleaning, glass etching Weak Carbonic Acid H 2 CO 3 soda water Weak Boric Acid H 3 BO 3 eye wash Weak Tro, Chemistry: A Molecular Approach 5

6 Structures of Acids binary acids have acid hydrogens attached to a nonmetal atom HCl, HF Tro, Chemistry: A Molecular Approach 6

7 Structure of Acids oxy acids have acid hydrogens attached to an oxygen atom H 2 SO 4, HNO 3 Tro, Chemistry: A Molecular Approach 7

8 carboxylic acids have COOH group HC 2 H 3 O 2, H 3 C 6 H 5 O 7 only the first H in the formula is acidic the H is on the COOH Structure of Acids Tro, Chemistry: A Molecular Approach 8

9 Properties of Bases also known as alkalis taste bitter alkaloids plant product that is alkaline often poisonous solutions feel slippery change color of vegetable dyes different color than acid red litmus turns blue react with acids to form ionic salts neutralization Tro, Chemistry: A Molecular Approach 9

10 Chemical Name sodium hydroxide potassium hydroxide calcium hydroxide sodium bicarbonate magnesium hydroxide ammonium hydroxide Common Bases Formula NaOH KOH Common Name lye, caustic soda caustic potash Uses soap, plastic, petrol refining soap, cotton, electroplating Strength Strong Strong Ca(OH) 2 slaked lime cement Strong NaHCO 3 baking soda cooking, antacid Weak Mg(OH) 2 NH 4 OH, {NH 3 (aq)} milk of magnesia ammonia water antacid detergent, fertilizer, explosives, fibers Weak Weak Tro, Chemistry: A Molecular Approach 10

11 Structure of Bases most ionic bases contain OH ions NaOH, Ca(OH) 2 some contain CO 3 2- ions CaCO 3 NaHCO 3 molecular bases contain structures that react with H + mostly amine groups Tro, Chemistry: A Molecular Approach 11

12 Indicators chemicals which change color depending on the acidity/basicity many vegetable dyes are indicators anthocyanins litmus from Spanish moss red in acid, blue in base phenolphthalein found in laxatives red in base, colorless in acid Tro, Chemistry: A Molecular Approach 12

13 Arrhenius Theory bases dissociate in water to produce OH - ions and cations ionic substances dissociate in water NaOH(aq) Na + (aq) + OH (aq) acids ionize in water to produce H + ions and anions because molecular acids are not made of ions, they cannot dissociate they must be pulled apart, or ionized, by the water HCl(aq) H + (aq) + Cl (aq) in formula, ionizable H written in front HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 (aq) Tro, Chemistry: A Molecular Approach 13

14 Arrhenius Theory HCl ionizes in water, producing H + and Cl ions NaOH dissociates in water, producing Na + and OH ions Tro, Chemistry: A Molecular Approach 14

15 Hydronium Ion the H + ions produced by the acid are so reactive they cannot exist in water H + ions are protons!! instead, they react with a water molecule(s) to produce complex ions, mainly hydronium ion, H 3 O + H + + H 2 O H 3 O + there are also minor amounts of H + with multiple water molecules, H(H 2 O) + n Tro, Chemistry: A Molecular Approach 15

16 Arrhenius Acid-Base Reactions the H + from the acid combines with the OH - from the base to make a molecule of H 2 O it is often helpful to think of H 2 O as H-OH the cation from the base combines with the anion from the acid to make a salt acid + base salt + water HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) Tro, Chemistry: A Molecular Approach 16

17 Problems with Arrhenius Theory does not explain why molecular substances, like NH 3, dissolve in water to form basic solutions even though they do not contain OH ions does not explain how some ionic compounds, like Na 2 CO 3 or Na 2 O, dissolve in water to form basic solutions even though they do not contain OH ions does not explain why molecular substances, like CO 2, dissolve in water to form acidic solutions even though they do not contain H + ions does not explain acid-base reactions that take place outside aqueous solution Tro, Chemistry: A Molecular Approach 17

18 Brønsted-Lowry Theory in a Brønsted-Lowry Acid-Base reaction, an H + is transferred does not have to take place in aqueous solution broader definition than Arrhenius acid is H donor, base is H acceptor base structure must contain an atom with an unshared pair of electrons in an acid-base reaction, the acid molecule gives an H + to the base molecule H A + :B :A + H B + Tro, Chemistry: A Molecular Approach 18

19 Brønsted-Lowry Acids Brønsted-Lowry acids are H + donors any material that has H can potentially be a Brønsted-Lowry acid because of the molecular structure, often one H in the molecule is easier to transfer than others HCl(aq) is acidic because HCl transfers an H + to H 2 O, forming H 3 O + ions water acts as base, accepting H + HCl(aq) + H 2 O(l) Cl (aq) + H 3 O + (aq) acid base Tro, Chemistry: A Molecular Approach 19

20 Brønsted-Lowry Bases Brønsted-Lowry bases are H + acceptors any material that has atoms with lone pairs can potentially be a Brønsted-Lowry base because of the molecular structure, often one atom in the molecule is more willing to accept H + transfer than others NH 3 (aq) is basic because NH 3 accepts an H + from H 2 O, forming OH (aq) water acts as acid, donating H + NH 3 (aq) + H 2 O(l) NH 4+ (aq) + OH (aq) base acid Tro, Chemistry: A Molecular Approach 20

21 Amphoteric Substances amphoteric substances can act as either an acid or a base have both transferable H and atom with lone pair water acts as base, accepting H + from HCl HCl(aq) + H 2 O(l) Cl (aq) + H 3 O + (aq) water acts as acid, donating H + to NH 3 NH 3 (aq) + H 2 O(l) NH 4+ (aq) + OH (aq) Tro, Chemistry: A Molecular Approach 21

22 Brønsted-Lowry Acid-Base Reactions one of the advantages of Brønsted-Lowry theory is that it allows reactions to be reversible H A + :B :A + H B + the original base has an extra H + after the reaction so it will act as an acid in the reverse process and the original acid has a lone pair of electrons after the reaction so it will act as a base in the reverse process :A + H B + H A + :B Tro, Chemistry: A Molecular Approach 22

23 Conjugate Pairs In a Brønsted-Lowry Acid-Base reaction, the original base becomes an acid in the reverse reaction, and the original acid becomes a base in the reverse process each reactant and the product it becomes is called a conjugate pair the original base becomes the conjugate acid; and the original acid becomes the conjugate base Tro, Chemistry: A Molecular Approach 23

24 Brønsted-Lowry Acid-Base Reactions H A + :B :A + H B + acid base conjugate conjugate base acid HCHO 2 + H 2 O CHO 2 + H 3 O + acid base conjugate conjugate base acid H 2 O + NH 3 HO + NH + 4 acid base conjugate conjugate base acid Tro, Chemistry: A Molecular Approach 24

25 Conjugate Pairs In the reaction H 2 O + NH 3 HO + NH 4 + H 2 O and HO constitute an Acid/Conjugate Base pair NH 3 and NH 4+ constitute a Base/Conjugate Acid pair Tro, Chemistry: A Molecular Approach 25

26 Ex 15.1a Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reaction H 2 SO 4 + H 2 O HSO 4 + H 3 O + When the H 2 SO 4 becomes HSO 4, it lost an H + so H 2 SO 4 must be the acid and HSO 4 its conjugate base When the H 2 O becomes H 3 O +, it accepted an H + so H 2 O must be the base and H 3 O + its conjugate acid H 2 SO 4 + H 2 O HSO 4 + H 3 O + acid base conjugate conjugate base acid Tro, Chemistry: A Molecular Approach 26

27 Ex 15.1b Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reaction HCO 3 + H 2 O H 2 CO 3 + HO When the HCO 3 becomes H 2 CO 3, it accepted an H + so HCO 3 must be the base and H 2 CO 3 its conjugate acid When the H 2 O becomes OH, it donated an H + so H 2 O must be the acid and OH its conjugate base HCO 3 + H 2 O H 2 CO 3 + HO base acid conjugate conjugate acid base Tro, Chemistry: A Molecular Approach 27

28 H 2 O Practice Write the formula for the conjugate acid of the following NH 3 CO 3 2 H 2 PO 4 1 Tro, Chemistry: A Molecular Approach 28

29 Practice Write the formula for the conjugate acid of the following H 2 O H 3 O + NH 3 NH 4 + CO 3 2 HCO 3 H 2 PO 4 1 H 3 PO 4 Tro, Chemistry: A Molecular Approach 29

30 H 2 O Practice Write the formula for the conjugate base of the following NH 3 CO 3 2 H 2 PO 4 1 Tro, Chemistry: A Molecular Approach 30

31 Practice Write the formula for the conjugate base of the following H 2 O HO NH 3 NH 2 CO 3 2 since CO 3 2 does not have an H, it cannot be an acid H 2 PO 4 1 HPO 4 2 Tro, Chemistry: A Molecular Approach 31

32 Arrow Conventions chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions a single arrow indicates all the reactant molecules are converted to product molecules at the end a double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products in these notes Tro, Chemistry: A Molecular Approach 32

33 Strong or Weak a strong acid is a strong electrolyte practically all the acid molecules ionize, a strong base is a strong electrolyte practically all the base molecules form OH ions, either through dissociation or reaction with water, a weak acid is a weak electrolyte only a small percentage of the molecules ionize, a weak base is a weak electrolyte only a small percentage of the base molecules form OH ions, either through dissociation or reaction with water, Tro, Chemistry: A Molecular Approach 33

34 Strong Acids The stronger the acid, the more willing it is to donate H use water as the standard base strong acids donate practically all their H s 100% ionized in water strong electrolyte [H 3 O + ] [strong acid] HCl H + + Cl - HCl + H 2 O H 3 O + + Cl - Tro, Chemistry: A Molecular Approach 34

35 Weak Acids weak acids donate a small fraction of their H s most of the weak acid molecules do not donate H to water much less than 1% ionized in water [H 3 O + ] << [weak acid] HF H + + F - HF + H 2 O H 3 O + + F - Tro, Chemistry: A Molecular Approach 35

36 Polyprotic Acids often acid molecules have more than one ionizable H these are called polyprotic acids the ionizable H s may have different acid strengths or be equal 1 H monoprotic, 2 H diprotic, 3 H triprotic HCl monoprotic, H 2 SO 4 diprotic, H 3 PO 4 triprotic polyprotic acids ionize in steps each ionizable H removed sequentially removing of the first H automatically makes removal of the second H harder H 2 SO 4 is a stronger acid than HSO 4 Tro, Chemistry: A Molecular Approach 36

37 Increasing Acidity Acids Conjugate Bases HClO 4-1 ClO 4 H 2 SO 4 HSO 4-1 HI I -1 HBr Br -1 HCl Cl -1 HNO 3 NO 3-1 H 3 O +1 HSO CH 4 CH 3 Tro, Chemistry: A Molecular Approach 37 OH -1 O -2 H 2 O SO 4-2 H 2 SO 3 HSO 3-1 H 3 PO 4 H 2 PO 4-1 HNO 2 NO 2-1 HF F -1 HC 2 H 3 O 2 C 2 H 3 O 2-1 H 2 CO 3 HCO 3-1 H 2 S HS -1 NH 4 +1 NH 3 HCN CN -1 HCO 3-1 CO 3-2 HS -1 S -2 H 2 O OH -1 CH 3 -C(O)-CH 3 CH 3 -C(O)-CH 2-1 NH 3 NH 2-1 Increasing Basicity

38 Strengths of Acids & Bases commonly, acid or base strength is measured by determining the equilibrium constant of a substance s reaction with water HAcid + H 2 O Acid -1 + H 3 O +1 Base: + H 2 O HBase +1 + OH -1 the farther the equilibrium position lies to the products, the stronger the acid or base the position of equilibrium depends on the strength of attraction between the base form and the H + stronger attraction means stronger base or weaker acid Tro, Chemistry: A Molecular Approach 38

39 General Trends in Acidity the stronger an acid is at donating H, the weaker the conjugate base is at accepting H higher oxidation number stronger oxyacid H 2 SO 4 > H 2 SO 3 ; HNO 3 > HNO 2 cation stronger acid than neutral molecule; neutral stronger acid than anion H 3 O +1 > H 2 O > OH -1 ; NH 4 +1 > NH 3 > NH 2-1 base trend opposite Tro, Chemistry: A Molecular Approach 39

40 Acid Ionization Constant, K a acid strength measured by the size of the equilibrium constant when react with H 2 O HAcid + H 2 O Acid -1 + H 3 O +1 the equilibrium constant is called the acid ionization constant, K a larger K a stronger acid K a [Acid 1 ] [H O [HAcid] ] Tro, Chemistry: A Molecular Approach 40

41 41

42 Name Formula K a1 K a2 K a3 Benzoic C 6 H 5 COOH 6.14 x 10-5 Propanoic CH 3 CH 2 COOH 1.34 x 10-5 Formic HCOOH 1.77 x 10-5 Acetic CH 3 COOH 1.75 x 10-5 Chloroacetic ClCH 2 COOH 1.36 x 10-5 Trichloroacetic Cl 3 C-COOH 1.29 x 10-4 Oxalic HOOC-COOH 5.90 x x 10-5 Nitric HNO 3 strong Nitrous HNO x 10-4 Phosphoric H 3 PO x x x Phosphorous H 3 PO x x 10-7 Arsenic H 3 AsO x x x Arsenious H 3 AsO x x very small Perchloric HClO 4 > 10 8 Chloric HClO 3 5 x 10 2 Chlorous HClO x 10-2 Hypochlorous HClO 3.0 x 10-8 Boric H 3 BO x Carbonic H 2 CO x x 10-11

43 Autoionization of Water Water is actually an extremely weak electrolyte therefore there must be a few ions present about 1 out of every 10 million water molecules form ions through a process called autoionization H 2 O H + + OH H 2 O + H 2 O H 3 O + + OH all aqueous solutions contain both H 3 O + and OH the concentration of H 3 O + and OH are equal in water [H 3 O + ] [OH ] C Tro, Chemistry: A Molecular Approach 43

44 Ion Product of Water the product of the H 3 O + and OH concentrations is always the same number the number is called the ion product of water and has the symbol K w [H 3 O + ] x [OH ] K w 1 x 25 C if you measure one of the concentrations, you can calculate the other as [H 3 O + ] increases the [OH ] must decrease so the product stays constant inversely proportional Tro, Chemistry: A Molecular Approach 44

45 Acidic and Basic Solutions all aqueous solutions contain both H 3 O + and OH ions neutral solutions have equal [H 3 O + ] and [OH ] [H 3 O + ] [OH ] 1 x 10-7 acidic solutions have a larger [H 3 O + ] than [OH ] [H 3 O + ] > 1 x 10-7 ; [OH ] < 1 x 10-7 basic solutions have a larger [OH ] than [H 3 O + ] [H 3 O + ] < 1 x 10-7 ; [OH ] > 1 x 10-7 Tro, Chemistry: A Molecular Approach 45

46 Example 15.2b Calculate the [OH ] at 25 C when the [H 3 O + ] 1.5 x 10-9 M, and determine if the solution is acidic, basic, or neutral Given: Find: [H 3 O + ] 1.5 x 10-9 M [OH ] Concept Plan: Relationships: [H 3 O + ] [OH ] K w H O [ ][OH ] Solution: K w [OH - ] [H O 3 + Kw [H O ][OH - ] 3 + ] [OH - ] M Check: The units are correct. The fact that the [H 3 O + ] < [OH ] means the solution is basic

47 Complete the Table [H + ] vs. [OH - ] [H + ] H + H + H + H + H+ OH - [OH - ] OH - OH - OH - OH - Tro, Chemistry: A Molecular Approach 47

48 Complete the Table [H + ] vs. [OH - ] Acid Base [H + ] H + H + H + H + H+ OH - OH - OH - OH - OH - [OH - ] even though it may look like it, neither H + nor OH - will ever be 0 the sizes of the H + and OH - are not to scale because the divisions are powers of 10 rather than units Tro, Chemistry: A Molecular Approach 48

49 ph the acidity/basicity of a solution is often expressed as ph ph -log[h 3 O + ], [H 3 O + ] 10 -ph exponent on 10 with a positive sign ph water -log[10-7 ] 7 need to know the [H + ] concentration to find ph ph < 7 is acidic; ph > 7 is basic, ph 7 is neutral Tro, Chemistry: A Molecular Approach 49

50 Sig. Figs. & Logs when you take the log of a number written in scientific notation, the digit(s) before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number log(2.0 x 10 6 ) log(10 6 ) + log(2.0) since the part of the scientific notation number that determines the significant figures is the decimal part, the sig figs are the digits after the decimal point in the log log(2.0 x 10 6 ) 6.30 Tro, Chemistry: A Molecular Approach 50

51 ph the lower the ph, the more acidic the solution; the higher the ph, the more basic the solution 1 ph unit corresponds to a factor of 10 difference in acidity normal range 0 to 14 ph 0 is [H + ] 1 M, ph 14 is [OH ] 1 M ph can be negative (very acidic) or larger than 14 (very alkaline) 51

52 ph of Common Substances Substance 1.0 M HCl 0.1 M HCl stomach acid lemons soft drinks plums apples cherries unpolluted rainwater human blood egg whites milk of magnesia (sat d Mg(OH) 2 ) household ammonia 1.0 M NaOH ph to to to to to to to to to

53 Tro, Chemistry: A Molecular Approach 53

54 Example 15.3b Calculate the ph at 25 C when the [OH ] 1.3 x 10-2 M, and determine if the solution is acidic, basic, or neutral Given: Find: [OH ] 1.3 x 10-2 M ph Concept Plan: Relationships: Solution: Check: K w 3 K [H O [OH ] w ] [H O + 3 H O [ ][OH - ] ][OH - ] 14 2 [H 3 O + ] [H ph ] ph is unitless. The fact that the ph > 7 means the solution is basic 3 O + ph + ph - log[h3o ] - log ph M ( 13)

55 poh another way of expressing the acidity/basicity of a solution is poh poh -log[oh ], [OH ] 10 -poh poh water -log[10-7 ] 7 need to know the [OH ] concentration to find poh poh < 7 is basic; poh > 7 is acidic, poh 7 is neutral Tro, Chemistry: A Molecular Approach 55

56 ph and poh Complete the Table ph [H + ] H + H + H + H + H+ OH OH - - OH - OH - [OH - ] poh OH - Tro, Chemistry: A Molecular Approach 56

57 ph and poh Complete the Table ph [H + ] H + H + H + H + H+ OH OH - - OH - OH - [OH - ] OH - poh Tro, Chemistry: A Molecular Approach 57

58 Relationship between ph and poh the sum of the ph and poh of a solution at 25 C can use poh to find ph of a solution [H O ][OH ( ( + ) ( 14 log [H ) 3O ][OH ] log ( log ( + [H O ] ) + ( log ([OH - ] ) ph + ] K - poh w Tro, Chemistry: A Molecular Approach 58

59 pk a way of expressing the strength of an acid or base is pk pk a -log(k a ), K a 10 -pka pk b -log(k b ), K b 10 -pkb the stronger the acid, the smaller the pk a larger K a smaller pk a because it is the log Tro, Chemistry: A Molecular Approach 59

60 Finding the ph of a Strong Acid there are two sources of H 3 O + in an aqueous solution of a strong acid the acid and the water for the strong acid, the contribution of the water to the total [H 3 O + ] is negligible shifts the K w equilibrium to the left so far that [H 3 O + ] water is too small to be significant except in very dilute solutions, generally < 1 x 10-4 M for a monoprotic strong acid [H 3 O + ] [HAcid] for polyprotic acids, the other ionizations can generally be ignored 0.10 M HCl has [H 3 O + ] 0.10 M and ph 1.00 Tro, Chemistry: A Molecular Approach 60

61 Finding the ph of a Weak Acid there are also two sources of H 3 O + in and aqueous solution of a weak acid the acid and the water however, finding the [H 3 O + ] is complicated by the fact that the acid only undergoes partial ionization calculating the [H 3 O + ] requires solving an equilibrium problem for the reaction that defines the acidity of the acid HAcid + H 2 O Acid + H 3 O + Tro, Chemistry: A Molecular Approach 61

62 Ex 15.6 Find the ph of M HNO 2 (aq) 25 C Write the reaction for the acid with water HNO 2 + H 2 O NO 2 + H 3 O + Construct an ICE table for the reaction Enter the initial concentrations assuming the [H 3 O + ] from water is 0 initial change equilibrium [HNO 2 ] [NO 2- ] 0 [H 3 O + ] 0 since no products initially, Q c 0, and the reaction is proceeding forward Tro, Chemistry: A Molecular Approach 62

63 Ex 15.6 Find the ph of M HNO 2 (aq) 25 C represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HNO 2 ] [NO 2- ] [H 3 O + ] initial change x equilibrium x x x K a [ HNO ] ( x)( x) - + [NO2][H3O ] 2 ( x) Tro, Chemistry: A Molecular Approach 63

64 K a Ex 15.6 Find the ph of M HNO 2 (aq) 25 C determine the value of K a from Table 15.5 since K a is very small, approximate the [HNO 2 ] eq [HNO 2 ] init and solve for x [ - ][ + NO H O ] [ HNO ] ( x( )( xx)( ) x) ( ) x) x initial change 1 K a for HNO x 10-4 equilibrium ( 4 )( 1 ) Tro, Chemistry: A Molecular Approach 64 x [HNO 2 ] x x [NO 2- ] 0 x x [H 3 O + ] 0 x

65 Ex 15.6 Find the ph of M HNO 2 (aq) check if the approximation is valid by seeing if x < 5% of [HNO 2 ] init C initial change K a for HNO x 10-4 equilibrium x 9.6 x % [HNO 2 ] [NO 2- ] 0 x Tro, Chemistry: A Molecular Approach 65 -x 4.8% the approximation is valid < 5% [H 3 O + ] 0 x

66 Ex 15.6 Find the ph of M HNO 2 (aq) substitute x into the equilibrium concentration definitions and solve 25 C initial change K a for HNO x 10-4 equilibrium x 9.6 x 10-3 [HNO 2 ] x x [NO 2- ] x [ ] ( ) HNO x M [ ] [ ] NO H O x M 2 3 [H 3 O + ] 0 x Tro, Chemistry: A Molecular Approach 66

67 Ex 15.6 Find the ph of M HNO 2 (aq) substitute [H 3 O + ] into the formula for ph and solve 25 C initial change K a for HNO x 10-4 equilibrium [HNO 2 ] x [NO 2- ] [H 3 O + ] ph ( + ) 3O ( 3) log H log 9.6 Tro, Chemistry: A Molecular Approach 67

68 Ex 15.6 Find the ph of M HNO 2 (aq) check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated 25 C initial change K a for HNO x 10-4 equilibrium [HNO 2 ] K a to the given K a [ - ][ + NO H O ] though not exact, the answer is reasonably close K a [NO 2- ] Tro, Chemistry: A Molecular Approach x [ HNO ] ( ) ( 0.190) [H 3 O + ] 0

69 Practice - What is the ph of a M solution of nicotinic acid, HC 6 H 4 NO 2? (K a 1.4 x 25 C) Tro, Chemistry: A Molecular Approach 69

70 Practice - What is the ph of a M solution of nicotinic acid, HC 6 H 4 NO 2? Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations assuming the [H 3 O + ] from water is 0 HC 6 H 4 NO 2 + H 2 O C 6 H 4 NO 2 + H 3 O + initial change equilibrium [HA] [A - ] 0 [H 3 O + ] 0 Tro, Chemistry: A Molecular Approach 70

71 Practice - What is the ph of a M solution of nicotinic acid, HC 6 H 4 NO 2? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression HC 6 H 4 NO 2 + H 2 O C 6 H 4 NO 2 + H 3 O + initial change equilibrium K a [HA] [ HC H NO ] [A - ] [H 3 O + ] x x x x ( x)( x) - + [C6H4NO2][H3O ] ( x) Tro, Chemistry: A Molecular Approach

72 Practice - What is the ph of a M solution of nicotinic acid, HC 6 H 4 NO 2? K a 1.4 x 25 C determine the value of K a since K a is very small, approximate the [HA] eq [HA] init and solve for x K a [ - ][ + A H O ] [ HA ] ( x( )( xx)( ) x) ( ) x) 2 x HC 6 H 4 NO 2 + H 2 O C 6 H 4 NO 2 + H 3 O + initial change equilibrium 2 ( 5 )( 2 ) Tro, Chemistry: A Molecular Approach 72 x [HA] x x [A 2- ] 0 x x [H 3 O + ] 0 x

73 Practice - What is the ph of a M solution of nicotinic acid, HC 6 H 4 NO 2? K a 1.4 x 25 C check if the approximation is valid by seeing if x < 5% of [HC 6 H 4 NO 2 ] init K a for HC 6 H 4 NO x 10-5 initial change equilibrium x 4.1 x 10-4 [HA] x [A 2- ] 0 x [H 3 O + ] 0 x % 3.4% < 5% the approximation is valid Tro, Chemistry: A Molecular Approach 73

74 Practice - What is the ph of a M solution of nicotinic acid, HC 6 H 4 NO 2? K a 1.4 x 25 C substitute x into the equilibrium concentration definitions and solve initial change equilibrium [HA] x x [A 2- ] 0 x [H 3 O + ] 0 x x 4.1 x [ ] ( ) HC H NO x M [ ] [ + ] 4 C H NO H O x M Tro, Chemistry: A Molecular Approach 74

75 Practice - What is the ph of a M solution of nicotinic acid, HC 6 H 4 NO 2? K a 1.4 x 25 C substitute [H 3 O + ] into the formula for ph and solve initial change [HA] x [A 2- ] 0 [H 3 O + ] 0 equilibrium ph log ( + ) 3O ( 4 ) log H Tro, Chemistry: A Molecular Approach 75

76 Practice - What is the ph of a M solution of nicotinic acid, HC 6 H 4 NO 2? K a 1.4 x 25 C check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values match initial change equilibrium K a [HA] [A 2- ] Tro, Chemistry: A Molecular Approach 76 [C 6 H -x 4 NO ( 4 ) ( ) 0 [ HC H NO ] ][H O ] 5 [H 3 O + ]

77 Ex 15.7 Find the ph of M HClO 2 (aq) 25 C Write the reaction for the acid with water HClO 2 + H 2 O ClO 2 + H 3 O + Construct an ICE table for the reaction Enter the initial concentrations assuming the [H 3 O + ] from water is 0 initial change equilibrium [HClO 2 ] [ClO 2- ] 0 [H 3 O + ] 0 Tro, Chemistry: A Molecular Approach 77

78 Ex 15.7 Find the ph of M HClO 2 (aq) 25 C represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression initial change equilibrium K a [HClO 2 ] [ HClO ] x x x ( x)( x) - + [ClO2][H3O ] 2 -x [ClO 2- ] [H 3 O + ] 0 0 ( x) Tro, Chemistry: A Molecular Approach 78

79 Ex 15.7 Find the ph of M HClO 2 (aq) 25 C determine the value of K a from Table 15.5 since K a is very small, approximate the [HClO 2 ] eq [HClO 2 ] init and solve for x K a [ - ][ + ClO H O ] 2 3 [ HClO ] initial change ( x)( x) ( ) 2 x K a for HClO x 10-2 equilibrium [HClO 2 ] ( 2 )( 1 ) Tro, Chemistry: A Molecular Approach 79 x x x [ClO 2- ] 0 x x [H 3 O + ] 0 x

80 Ex 15.7 Find the ph of M HClO 2 (aq) 25 C check if the approximation is valid by seeing if x < 5% of [HNO 2 ] init initial change 2 1 K a for HClO x 10-2 equilibrium x 3.3 x % [HClO 2 ] 33% 5% Tro, Chemistry: A Molecular Approach x x > the approximation is invalid [ClO 2- ] 0 x [H 3 O + ] 0 x

81 Ex 15.7 Find the ph of M HClO 2 (aq) 25 C if the approximation is invalid, solve for x using the quadratic formula K a K a for HClO x 10-2 [ - ][ + ClO H O ] [ HClO ] ( x)( x) ( x) x 2 ( 1 ) x x 0 x 0.011± x (1)( ) ( 0.011) 2(1) x 0.028or Tro, Chemistry: A Molecular Approach 81

82 Ex 15.7 Find the ph of M HClO 2 (aq) 25 C substitute x into the equilibrium concentration definitions and solve initial change x K a for HClO x 10-2 [ HClO ] ( ) x M [ - ] [ + ] H O x M 2 ClO 3 equilibrium [HClO 2 ] x x [ClO 2- ] x [H 3 O + ] x Tro, Chemistry: A Molecular Approach 82

83 Ex 15.7 Find the ph of M HClO 2 (aq) 25 C substitute [H 3 O + ] into the formula for ph and solve initial change ph -log H3 log K a for HClO x 10-2 equilibrium ( + ) O [HClO 2 ] x ( 0.028) [ClO 2- ] [H 3 O + ] Tro, Chemistry: A Molecular Approach 83

84 Ex 15.7 Find the ph of M HClO 2 (aq) 25 C check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a for HClO x 10-2 K a to the given K a [ - ][ + ClO H O ] the answer matches initial change equilibrium K ( ) ( 0.072) [ HClO ] Tro, Chemistry: A Molecular Approach 84 a [HClO 2 ] x [ClO 2- ] [H 3 O + ]

85 Ex What is the K a of a weak acid if a M solution has a ph of 4.25? Use the ph to find the equilibrium [H 3 O + ] [H O 3 + ] 10 -ph M Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations and [H 3 O + ] equil initial change equilibrium HA + H 2 O A + H 3 O + [HA] [A - ] 0 [H 3 O + ] 0 5.6E-05 Tro, Chemistry: A Molecular Approach 85

86 Ex What is the K a of a weak acid if a M solution has a ph of 4.25? fill in the rest of the table using the [H 3 O + ] as a guide if the difference is insignificant, [HA] equil [HA] initial substitute into the K a expression and compute K a initial change equilibrium K K a a HA + H 2 O A + H 3 O [A ][H3O [ HA] [HA] [A - ] Tro, Chemistry: A Molecular Approach E E E-05 ] [H 3 O + ] E E E-05 ( 5)( ) ( 0.100)

87 Percent Ionization another way to measure the strength of an acid is to determine the percentage of acid molecules that ionize when dissolved in water this is called the percent ionization the higher the percent ionization, the stronger the acid molarity of ionized acid Percent Ionization 100% initial molarity of acid since [ionized acid] equil [H 3 O + ] equil Percent Ionization [HA] 100% Tro, Chemistry: A Molecular Approach 87 [H 3 O + ] equil init

88 Ex What is the percent ionization of a 2.5 M HNO 2 solution? Write the reaction for the acid with water Construct an ICE table for the reaction Enter the Initial Concentrations Define the Change in Concentration in terms of x Sum the columns to define the Equilibrium Concentrations initial change HNO 2 + H 2 O NO 2 + H 3 O + equilibrium [HNO 2 ] [NO 2- ] [H 3 O + ] x 2.5 x x x 0 Tro, Chemistry: A Molecular Approach 88

89 Ex What is the percent ionization of a 2.5 M HNO 2 solution? determine the value of K a from Table 15.5 since K a is very small, approximate the [HNO 2 ] eq [HNO 2 ] init and solve for x [ - ][ + NO H O ] 2 3 K a [ HNO2] initial change ( x)( x) ( 2.5) K a for HNO x 10-4 equilibrium x 2.5 Tro, Chemistry: A Molecular Approach 89 x [HNO 2 ] 2.5 -x 2.5-x ( )( 2.5) x [NO 2- ] 2 0 x [H 3 O + ] 0 x

90 Ex What is the percent ionization of a 2.5 M HNO 2 solution? substitute x into the Equilibrium Concentration definitions and solve x 3.4 x 10-2 initial change HNO 2 + H 2 O NO 2 + H 3 O + equilibrium [HNO 2 ] 2.5 -x [NO 2- ] 0 [H 3 O + ] x x x [ HNO ] ( ) x M [ - ] [ + ] H O x M 2 NO 3 Tro, Chemistry: A Molecular Approach 90

91 Ex What is the percent ionization of a 2.5 M HNO 2 solution? Apply the Definition and Compute the Percent Ionization since the percent ionization is < 5%, the x is small approximation is valid initial change HNO 2 + H 2 O NO 2 + H 3 O + equilibrium Percent Ionization [HNO 2 ] [NO 2- ] Tro, Chemistry: A Molecular Approach x 2.5 [H 3 O [HNO 100% 1.4% 0 + ] 2 equil ] init [H 3 O + ] %

92 Relationship Between [H 3 O + ] equilibrium & [HA] initial increasing the initial concentration of acid results in increased H 3 O + concentration at equilibrium increasing the initial concentration of acid results in decreased percent ionization this means that the increase in H 3 O + concentration is slower than the increase in acid concentration Percent Ionization [H 3 O + [HA] ] equil init 100% Tro, Chemistry: A Molecular Approach 92

93 Why doesn t the increase in H 3 O + keep up with the increase in HA? the reaction for ionization of a weak acid is: HA (aq) + H 2 O (l) A (aq) + H 3 O+ (aq) according to Le Châtelier s Principle, if we reduce the concentrations of all the (aq) components, the equilibrium should shift to the right to increase the total number of dissolved particles we can reduce the (aq) concentrations by using a more dilute initial acid concentration the result will be a larger [H 3 O + ] in the dilute solution compared to the initial acid concentration this will result in a larger percent ionization 93

94 Finding the ph of Mixtures of Acids generally, you can ignore the contribution of the weaker acid to the [H 3 O + ] equil for a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H 3 O + ] to shift the weak acid equilibrium to the left so far that the weak acid s added [H 3 O + ] is negligible for mixtures of weak acids, generally only need to consider the stronger for the same reasons as long as one is significantly stronger than the other, and their concentrations are similar Tro, Chemistry: A Molecular Approach 94

95 Ex Find the ph of a mixture of M HF(aq) solution and M HClO 2 (aq) Write the reactions for the acids with water and determine their K a s If the K a sare sufficiently different, use the strongest acid to construct an ICE table for the reaction Enter the initial concentrations assuming the [H 3 O + ] from water is 0 HF + H 2 O F + H 3 O + K a 3.5 x 10-4 HClO + H 2 O ClO + H 3 O + K a 2.9 x 10-8 H 2 O + H 2 O OH + H 3 O + K w 1.0 x initial change equilibrium [HF] [F - ] 0 [H 3 O + ] 0 Tro, Chemistry: A Molecular Approach 95

96 Ex Find the ph of a mixture of M HF(aq) solution and M HClO 2 (aq) represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HF] [F - ] [H 3 O + ] initial change x equilibrium x x x K a [ HF] ( x)( x) - + [F ][H3O ] ( x) Tro, Chemistry: A Molecular Approach 96

97 Ex Find the ph of a mixture of M HF(aq) solution and M HClO 2 (aq) determine the value of K a for HF since K a is very small, approximate the [HF] eq [HF] init and solve for x [ ][ - + H O ] F 3 K a [ HF ] 4 ( x( )( xx)( ) x) ( 0.150) x) 2 x initial change 1 K a for HF 3.5 x 10-4 equilibrium ( 4 )( 1 ) Tro, Chemistry: A Molecular Approach 97 x [HF] x x [F - ] 0 x x [H 3 O + ] 0 x

98 Ex Find the ph of a mixture of M HF(aq) solution and M HClO 2 (aq) check if the approximation is valid by seeing if x < 5% of [HF] init initial change K a for HF 3.5 x 10-4 equilibrium x 7.2 x % [HF] [F - ] Tro, Chemistry: A Molecular Approach 98 -x 4.8% the approximation is valid < 0 x 5% [H 3 O + ] 0 x

99 Ex Find the ph of a mixture of M HF(aq) solution and M HClO 2 (aq) substitute x into the equilibrium concentration definitions and solve initial change K a for HF 3.5 x 10-4 equilibrium x 7.2 x 10-3 [HF] x x [F - ] 0 [ ] ( ) HF x M [ ] [ ] F H O x M 3 x [H 3 O + ] 0 x Tro, Chemistry: A Molecular Approach 99

100 Ex Find the ph of a mixture of M HF(aq) solution and M HClO 2 (aq) substitute [H 3 O + ] into the formula for ph and solve initial change K a for HF 3.5 x 10-4 equilibrium [HF] x [F - ] [H 3 O + ] ph ( + ) 3O ( 3) log H log 7.2 Tro, Chemistry: A Molecular Approach 100

101 Ex Find the ph of a mixture of M HF(aq) solution and M HClO 2 (aq) check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a for HF 3.5 x 10-4 K a to the given K a [ -][ + F H O ] though not exact, the answer is reasonably close initial change equilibrium K a [ HF] ( ) ( 0.143) Tro, Chemistry: A Molecular Approach 101 [HF] x [F - ] [H 3 O + ]

102 Strong Bases the stronger the base, the more willing it is to accept H use water as the standard acid for strong bases, practically all molecules are dissociated into OH or accept H s strong electrolyte multi-oh strong bases completely dissociated [HO ] [strong base] x (# OH) NaOH Na + + OH - Tro, Chemistry: A Molecular Approach 102

103 Example 15.11b Calculate the ph at 25 C of a M Sr(OH) 2 solution and determine if the solution is acidic, basic, or neutral Given: Find: [Sr(OH) 2 ] 1.5 x 10-3 M ph Concept Plan: Relationships: [Sr(OH) 2 ] [OH ]2[Sr(OH) 2 ] [OH ] K w H O [ ][OH [H 3 O + ] ] ph + ph - log[h3o ] Solution: [OH ] 2(0.0015) M K w [H O 3 [H O + ] 3 + ][OH - ] [H 3 O ph + ] ph log M ( 12 ) 10 Check: ph is unitless. The fact that the ph > 7 means the solution is basic

104 Example 15.11b (alternative) Calculate the ph at 25 C of a M Sr(OH) 2 solution and determine if the solution is acidic, basic, or neutral Given: Find: [Sr(OH) 2 ] 1.5 x 10-3 M ph Concept Plan: Relationships: [Sr(OH) 2 ] [OH ]2[Sr(OH) 2 ] [OH ] poh ph poh -log[oh - ] ph + poh Solution: [OH ] 2(0.0015) M poh poh - log ( 3) 10 ph poh ph Check: ph is unitless. The fact that the ph > 7 means the solution is basic

105 Practice - Calculate the ph of a M Ba(OH) 2 solution and determine if it is acidic, basic, or neutral Tro, Chemistry: A Molecular Approach 105

106 Practice - Calculate the ph of a M Ba(OH) 2 solution and determine if it is acidic, basic, or neutral Ba(OH) 2 Ba OH - therefore [OH - ] 2 x x 10-3 M [H 3 O + ] K w [H 3 O + ][OH ] 1.00 x x x M ph -log [H 3 O + ] -log (5.0 x ) ph ph > 7 therefore basic Tro, Chemistry: A Molecular Approach 106

107 Weak Bases in weak bases, only a small fraction of molecules accept H s weak electrolyte most of the weak base molecules do not take H from water much less than 1% ionization in water [HO ] << [weak base] finding the ph of a weak base solution is similar to finding the ph of a weak acid NH 3 + H 2 O NH 4+ + OH - Tro, Chemistry: A Molecular Approach 107

108 Tro, Chemistry: A Molecular Approach 108

109 Structure of Amines Tro, Chemistry: A Molecular Approach 109

110 Ex Find the ph of M NH 3 (aq) solution Write the reaction for the base with water NH 3 + H 2 O NH 4+ + OH Construct an ICE table for the reaction Enter the initial concentrations assuming the [OH ] from water is 0 initial change equilibrium [NH 3 ] [NH 4+ ] 0 [OH ] 0 since no products initially, Q c 0, and the reaction is proceeding forward Tro, Chemistry: A Molecular Approach 110

111 Ex Find the ph of M NH 3 (aq) solution represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [NH 3 ] [NH 4+ ] [OH ] initial change x equilibrium x x x K b [ NH ] ( )( ) + [NH4 ][OH ] x x 3 ( x) Tro, Chemistry: A Molecular Approach 111

112 Ex Find the ph of M NH 3 (aq) solution determine the value of K b from Table 15.8 since K b is very small, approximate the [NH 3 ] eq [NH 3 ] init and solve for x K b for NH x 10-5 initial change equilibrium [NH 3 ] x x [NH 4+ ] 0 x [OH ] 0 x K b [ + ][ NH OH ] [ NH ] ( x( )( x)( ) x) ( ) x) 2 x x ( 5)( 1 ) x Tro, Chemistry: A Molecular Approach 112 3

113 Ex Find the ph of M NH 3 (aq) solution check if the approximation is valid by seeing if x < 5% of [NH 3 ] init K b for NH x 10-5 initial change equilibrium x 1.33 x 10-3 [NH 3 ] x [NH 4+ ] 0 x [OH ] 0 x % 1.33% < 5% the approximation is valid Tro, Chemistry: A Molecular Approach 113

114 Ex Find the ph of M NH 3 (aq) solution substitute x into the equilibrium concentration definitions and solve initial change K b for NH x 10-5 equilibrium x 1.33 x 10-3 [NH 3 ] x 3 3 [NH 4+ ] [ ] ( ) NH x M -x 0 x 1.33E-3 [OH ] 0 x 1.33E-3 [NH + 4 ] - [OH ] x M Tro, Chemistry: A Molecular Approach 114

115 Ex Find the ph of M NH 3 (aq) solution use the [OH - ] to find the [H 3 O + ] using K w substitute [H 3 O + ] into the formula for ph and solve initial change K b for NH x 10-5 equilibrium [NH 3 ] x [NH 4+ ] E-3 [OH ] E-3 K w [H O 3 [H O 3 [H O + + ] ] 3 + ][OH - ] ph ( + ) 3O ( 12 ) log H log 7.52 Tro, Chemistry: A Molecular Approach 115

116 Ex Find the ph of M NH 3 (aq) solution use the [OH - ] to find the poh use poh to find ph poh log initial change ( ) -log OH ( 3) 10 K b for NH x 10-5 equilibrium [NH 3 ] x ph [NH 4+ ] E [OH ] E-3 poh Tro, Chemistry: A Molecular Approach 116

117 Ex Find the ph of M NH 3 (aq) solution check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated initial change K b for NH x 10-5 equilibrium [NH 3 ] x K b to the given K b [ + ][ NH OH ] though not exact, the answer is reasonably close K b 4 [ NH ] ( ) ( 0.099) 3 2 [NH 4+ ] E [OH ] E-3 5 Tro, Chemistry: A Molecular Approach 117

118 Practice Find the ph of a M morphine solution, K b 1.6 x 10-6 Tro, Chemistry: A Molecular Approach 118

119 Practice Find the ph of a M morphine solution Write the reaction for the base with water Construct an ICE table for the reaction Enter the initial concentrations assuming the [OH ] from water is 0 B + H 2 O BH + + OH initial change equilibrium [B] [BH + ] 0 [OH ] 0 since no products initially, Q c 0, and the reaction is proceeding forward Tro, Chemistry: A Molecular Approach 119

120 Practice Find the ph of a M morphine solution represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [B] [BH + ] [OH ] initial change x equilibrium x x x K b [ B] ( x)( x) + [BH ][OH ] ( x) Tro, Chemistry: A Molecular Approach 120

121 Practice Find the ph of a M morphine solution determine the value of K b since K b is very small, approximate the [B] eq [B] init and solve for x K b for Morphine 1.6 x 10-6 initial change equilibrium [B] x x [BH + ] 0 x [OH ] 0 x K b [ + ][ BH OH ] [ B ] ( x( )( x)( ) x) ( ) x) 2 x x ( 6 )( 3) x Tro, Chemistry: A Molecular Approach 121 5

122 Practice Find the ph of a M morphine solution check if the approximation is valid by seeing if x < 5% of [B] init K b for Morphine 1.6 x 10-6 initial change equilibrium x 4.9 x 10-5 [B] x [BH + ] 0 x [OH ] 0 x % 3.3% < 5% the approximation is valid Tro, Chemistry: A Molecular Approach 122

123 Practice Find the ph of a M morphine solution substitute x into the equilibrium concentration definitions and solve initial change K b for Morphine 1.6 x 10-6 equilibrium x 4.9 x 10-5 [B] x [BH + ] 5 [ ] ( ) Morphine x M -x 0 x 4.9E-5 [OH ] 0 x 4.9E-5 [BH + ] [OH ] x M Tro, Chemistry: A Molecular Approach 123

124 Practice Find the ph of a M morphine solution use the [OH - ] to find the [H 3 O + ] using K w substitute [H 3 O + ] into the formula for ph and solve K b for Morphine 1.6 x 10-6 initial change equilibrium [B] x [BH + ] 0 [OH ] E-5 4.9E-5 K w [H O 3 [H O 3 [H O + + ] ] 3 + ][OH ] ph log ( + ) 3O ( 10 ) log H Tro, Chemistry: A Molecular Approach 124

125 Practice Find the ph of a M morphine solution use the [OH - ] to find the poh use poh to find ph poh log 4.31 initial change ( ) -log OH ( 5 ) K b for Morphine 1.6 x 10-6 equilibrium ph [B] x [BH + ] 0 4.9E [OH ] 0 4.9E-5 poh Tro, Chemistry: A Molecular Approach 125

126 Practice Find the ph of a M morphine solution check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K b to the given K b [ + ][ BH OH ] the answer matches the given K b initial change K b for Morphine 1.6 x 10-6 equilibrium K b [B] x [ B] ( ) ( ) 2 [BH + ] 0 4.9E [OH ] 0 4.9E-5 Tro, Chemistry: A Molecular Approach 126

127 Acid-Base Properties of Salts salts are water soluble ionic compounds salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid are basic NaHCO 3 solutions are basic Na + is the cation of the strong base NaOH HCO 3 is the conjugate base of the weak acid H 2 CO 3 salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic NH 4 Cl solutions are acidic NH 4+ is the conjugate acid of the weak base NH 3 Cl is the anion of the strong acid HCl Tro, Chemistry: A Molecular Approach 127

128 Anions as Weak Bases every anion can be thought of as the conjugate base of an acid therefore, every anion can potentially be a base A (aq) + H 2 O(l) HA(aq) + OH (aq) the stronger the acid is, the weaker the conjugate base is an anion that is the conjugate base of a strong acid is ph neutral Cl (aq) + H 2 O(l) HCl(aq) + OH (aq) since HCl is a strong acid, this equilibrium lies practically completely to the left an anion that is the conjugate base of a weak acid is basic F (aq) + H 2 O(l) HF(aq) + OH (aq) since HF is a weak acid, the position of this equilibrium favors the right Tro, Chemistry: A Molecular Approach 128

129 Ex Use the Table to Determine if the Given Anion Is Basic or Neutral a) NO 3 the conjugate base of a strong acid, therefore neutral b) NO 2 the conjugate base of a weak acid, therefore basic Tro, Chemistry: A Molecular Approach 129

130 Relationship between K a of an Acid and K b of Its Conjugate Base many reference books only give tables of K a values because K b values can be found from them when you add equations, you multiply the K s HA( aq) + H O( l) A A ( aq) + H O ( aq) + H O( l) HA( aq) + OH 2 3 K a ( aq) ( aq) H2O( l) H O ( aq) + OH ( aq) K K a b [A ][H3O [HA] K b [H O ][OH ] [HA][OH [A ] ] K w ] K K a b + [A ][H3O ] [HA] [HA][H 3O [A ] + ] 130

131 Na + is the cation of a strong base ph neutral. The CHO 2 is the anion of a weak acid ph basic Write the reaction for the anion with water Construct an ICE table for the reaction Ex Find the ph of M NaCHO 2 (aq) solution CHO 2 + H 2 O HCHO 2 + OH initial change equilibrium [CHO 2 ] [HCHO 2 ] 0 [OH ] 0 Enter the initial concentrations assuming the [OH ] from water is 0 Tro, Chemistry: A Molecular Approach 131

132 Ex Find the ph of M NaCHO 2 (aq) solution represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x Calculate the value of K b from the value of K a of the weak acid from Table 15.5 substitute into the equilibrium constant expression initial change equilibrium K b K K a b K [CHO 2 ] x x b [ ] CHO x ( x)( x) [HCHO2][OH ] 2 K w [HCHO 2 ] ( x) [OH ] 0 x 132

133 Ex Find the ph of M NaCHO 2 (aq) solution since K b is very small, approximate the [CHO 2 ] eq [CHO 2 ] init and solve for x initial change K b for CHO x equilibrium [CHO 2 ] x x [HCHO 2 ] 0 x [OH ] 0 x K b [ CHO ] ( x( )( x)( ) x) [HCHO 2 ][OH ] 2 1 ( ) x) x x ( 11)( ) x Tro, Chemistry: A Molecular Approach 133 6

134 Ex Find the ph of M NaCHO 2 (aq) solution check if the approximation is valid by seeing if x < 5% of [CHO 2 ] init K b for CHO x initial change equilibrium x 2.4 x 10-6 [CHO 2 ] [HCHO 2 ] x x [OH ] 0 x % % < 5% the approximation is valid Tro, Chemistry: A Molecular Approach 134

135 Ex Find the ph of M NaCHO 2 (aq) solution substitute x into the equilibrium concentration definitions and solve initial K b for CHO x change -x equilibrium [CHO 2 ] [HCHO 2 ] x x 2.4E-6 [OH ] 0 x 2.4E-6 x 2.4 x 10-6 [ ] ( 6 CHO x ) M 2 [HCHO 2 ] [OH ] x Tro, Chemistry: A Molecular Approach M

136 Ex Find the ph of M NaCHO 2 (aq) solution use the [OH - ] to find the [H 3 O + ] using K w substitute [H 3 O + ] into the formula for ph and solve initial K b for CHO x change -x equilibrium [CHO 2 ] [HCHO 2 ] E-6 [OH ] 0 2.4E-6 K w [H O 3 [H O 3 [H O + + ] ] 3 + ][OH ] ph log ( + ) 3O ( ) log H Tro, Chemistry: A Molecular Approach 136

137 Ex Find the ph of M NaCHO 2 (aq) solution use the [OH - ] to find the poh use poh to find ph initial K b for CHO x change -x equilibrium [CHO 2 ] [HCHO 2 ] E-6 [OH ] 0 2.4E-6 poh log 5.62 ( ) -log OH ( 6 ) ph poh Tro, Chemistry: A Molecular Approach 137

138 Ex Find the ph of M NaCHO 2 (aq) solution check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K b to the given K b [ ][ HCHO OH ] though not exact, the answer is reasonably close initial change K b for CHO x equilibrium K b [CHO 2 ] x ( ) ( 0.100) [ CHO ] [HCHO 2 ] 0 2.4E [OH ] 0 2.4E-6 Tro, Chemistry: A Molecular Approach 138

139 Polyatomic Cations as Weak Acids some cations can be thought of as the conjugate acid of a base others are the counterions of a strong base therefore, some cation can potentially be an acid MH + (aq) + H 2 O(l) MOH(aq) + H 3 O + (aq) the stronger the base is, the weaker the conjugate acid is a cation that is the counterion of a strong base is ph neutral a cation that is the conjugate acid of a weak base is acidic NH 4+ (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) since NH 3 is a weak base, the position of this equilibrium favors the right Tro, Chemistry: A Molecular Approach 139

140 Metal Cations as Weak Acids cations of small, highly charged metals are weakly acidic alkali metal cations and alkali earth metal cations ph neutral cations are hydrated Al(H 2 O) 6 3+ (aq) + H 2 O(l) Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) Tro, Chemistry: A Molecular Approach 140

141 Ex Determine if the Given a) C 5 N 5 NH 2 + Cation Is Acidic or Neutral the conjugate acid of a weak base, therefore acidic b) Ca 2+ the counterion of a strong base, therefore neutral c) Cr 3+ a highly charged metal ion, therefore acidic Tro, Chemistry: A Molecular Approach 141

142 Classifying Salt Solutions as Acidic, Basic, or Neutral if the salt cation is the counterion of a strong base and the anion is the conjugate base of a strong acid, it will form a neutral solution NaCl Ca(NO 3 ) 2 KBr if the salt cation is the counterion of a strong base and the anion is the conjugate base of a weak acid, it will form a basic solution NaF Ca(C 2 H 3 O 2 ) 2 KNO 2 Tro, Chemistry: A Molecular Approach 142

143 Classifying Salt Solutions as Acidic, Basic, or Neutral if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution NH 4 Cl if the salt cation is a highly charged metal ion and the anion is the conjugate base of a strong acid, it will form an acidic solution Al(NO 3 ) 3 Tro, Chemistry: A Molecular Approach 143

144 Classifying Salt Solutions as Acidic, Basic, or Neutral if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the ph of the solution depends on the relative strengths of the acid and base NH 4 F since HF is a stronger acid than NH 4+, K a of NH 4+ is larger than K b of the F ; therefore the solution will be acidic Tro, Chemistry: A Molecular Approach 144

145 Ex Determine whether a solution of the following salts is acidic, basic, or neutral a) SrCl 2 Sr 2+ is the counterion of a strong base, ph neutral Cl is the conjugate base of a strong acid, ph neutral solution will be ph neutral b) AlBr 3 Al 3+ is a small, highly charged metal ion, weak acid Cl is the conjugate base of a strong acid, ph neutral solution will be acidic c) CH 3 NH 3 NO 3 CH 3 NH 3+ is the conjugate acid of a weak base, acidic NO 3 is the conjugate base of a strong acid, ph neutral solution will be acidic Tro, Chemistry: A Molecular Approach 145

146 Ex Determine whether a solution of the following salts is acidic, basic, or neutral d) NaCHO 2 Na + is the counterion of a strong base, ph neutral CHO 2 is the conjugate base of a weak acid, basic solution will be basic e) NH 4 F NH 4+ is the conjugate acid of a weak base, acidic F is the conjugate base of a weak acid, basic K a (NH 4+ ) > K b (F ); solution will be acidic Tro, Chemistry: A Molecular Approach 146

147 Polyprotic Acids since polyprotic acids ionize in steps, each H has a separate K a K a1 > K a2 > K a3 generally, the difference in K a values is great enough so that the second ionization does not happen to a large enough extent to affect the ph most ph problems just do first ionization except H 2 SO 4 use [H 2 SO 4 ] as the [H 3 O + ] for the second ionization [A 2- ] K a2 as long as the second ionization is negligible Tro, Chemistry: A Molecular Approach 147

148 148

149 Ex Find the ph of M H 2 SO 4 (aq) 25 C Write the reactions H 2 SO 4 + H 2 O HSO 4 + H 3 O + for the acid with water HSO 4 + H 2 O SO H 3 O + Construct an ICE table for the reaction Enter the initial concentrations assuming the [HSO 4 ] and [H 3 O + ] is [H 2 SO 4 ] initial change equilibrium [HSO 4 ] [SO 4 2 ] 0 [H 3 O + ] Tro, Chemistry: A Molecular Approach 149

150 Ex Find the ph of M H 2 SO 4 (aq) 25 C represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression initial change equilibrium K a [HSO 4 ] x x 2- [SO4 ][H3O - [HSO ] 4 + ] [SO 2 4 ] [H 3 O + ] x x ( x)( x) ( x) Tro, Chemistry: A Molecular Approach 150

151 Ex Find the ph of M H 2 SO 4 (aq) 25 C expand and solve for x using the quadratic formula x K a K a for HSO [SO4 ][H3O - [HSO ] ( x)( x) ] ( x) ( x) x ( x) Tro, Chemistry: A Molecular Approach 151 x x ( 0.022) 2 x ± 4(1)( ) 2(1) x or x + x 2

152 Ex Find the ph of M H 2 SO 4 (aq) 25 C substitute x into the equilibrium concentration definitions and solve x ( ) M [ HSO4 ] x initial change equilibrium ( x) M [ H3 O ] 2- [SO4 ] x K a for HSO [HSO 4 ] x [SO 2 4 ] x Tro, Chemistry: A Molecular Approach 152 x M 0 [H 3 O + ] x

153 Ex Find the ph of M H 2 SO 4 (aq) 25 C substitute [H 3 O + ] into the formula for ph and solve ph 3 initial change equilibrium K a for HSO [HSO 4 ] x ( + ) -log H O log( ) [SO 4 2 ] [H 3 O + ] Tro, Chemistry: A Molecular Approach 153

154 Ex 15.7 Find the ph of M HClO 2 (aq) 25 C check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a [ 2-][ + SO ] 4 H3O Ka [ - HSO ] the answer matches initial change equilibrium K a for HSO [HSO 4 ] ( )( ) ( ) Tro, Chemistry: A Molecular Approach 154 x [SO 4 2 ] [H 3 O + ]

155 Strengths of Binary Acids the more δ+ H-X δ- polarized the bond, the more acidic the bond the stronger the H-X bond, the weaker the acid binary acid strength increases to the right across a period H-C < H-N < H-O < H-F binary acid strength increases down the column H-F < H-Cl < H-Br < H-I Tro, Chemistry: A Molecular Approach 155

156 Strengths of Oxyacids, H-O-Y the more electronegative the Y atom, the stronger the acid helps weakens the H-O bond the more oxygens attached to Y, the stronger the acid further weakens and polarizes the H-O bond Tro, Chemistry: A Molecular Approach 156

157 Lewis Acid - Base Theory electron sharing electron donor Lewis Base nucleophile must have a lone pair of electrons electron acceptor Lewis Acid electrophile electron deficient when Lewis Base gives electrons from lone pair to Lewis Acid, a covalent bond forms between the molecules Nucleophile: + Electrophile Nucleophile:Electrophile product called an adduct other acid-base reactions also Lewis Tro, Chemistry: A Molecular Approach 157

158 Example - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile OH H C H + OH -1 OH H C H Electrophile + OH -1 Nucleophile OH H C H OH Tro, Chemistry: A Molecular Approach 158

159 Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile BF 3 + HF CaO + SO 3 KI + I 2 Tro, Chemistry: A Molecular Approach 159

160 Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile BF 3 + HF H +1 BF 4-1 Elec CaO + SO 3 Ca +2 SO 4-2 Nuc KI + I 2 KI 3 Nuc Elec Nuc Elec H F Ca+2 O -2 K +1 I -1 Tro, Chemistry: A Molecular Approach F B F F O S O O I I H +1 Ca +2 F O F -1 B F F O S O O -2 K +1 I I I -1

161 What Is Acid Rain? natural rain water has a ph of 5.6 naturally slightly acidic due mainly to CO 2 rain water with a ph lower than 5.6 is called acid rain acid rain is linked to damage in ecosystems and structures Tro, Chemistry: A Molecular Approach 161

162 What Causes Acid Rain? many natural and pollutant gases dissolved in the air are nonmetal oxides CO 2, SO 2, NO 2 nonmetal oxides are acidic CO 2 + H 2 O H 2 CO 3 2 SO 2 + O H 2 O 2 H 2 SO 4 processes that produce nonmetal oxide gases as waste increase the acidity of the rain natural volcanoes and some bacterial action man-made combustion of fuel weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced Tro, Chemistry: A Molecular Approach 162

163 ph of Rain in Different Regions

164 Sources of SO 2 from Utilities

165 Damage from Acid Rain acids react with metals, and materials that contain carbonates acid rain damages bridges, cars, and other metallic structures acid rain damages buildings and other structures made of limestone or cement acidifying lakes affecting aquatic life dissolving and leaching more minerals from soil making it difficult for trees Tro, Chemistry: A Molecular Approach 165

166 Acid Rain Legislation 1990 Clean Air Act attacks acid rain force utilities to reduce SO 2 result is acid rain in northeast stabilized and beginning to be reduced Tro, Chemistry: A Molecular Approach 166

167 Damage from Acid Rain 167

Chapter 16 - Acids and Bases

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