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Copyright  2011 Pearson Education, Inc. Chapter 15 Acids and Bases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular.

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Presentation on theme: "Copyright  2011 Pearson Education, Inc. Chapter 15 Acids and Bases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular."— Presentation transcript:

1 Copyright  2011 Pearson Education, Inc. Chapter 15 Acids and Bases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro

2 Copyright  2011 Pearson Education, Inc. 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 2 Tro, Chemistry: A Molecular Approach, 2/e

3 Copyright  2011 Pearson Education, Inc. 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 3 Tro, Chemistry: A Molecular Approach, 2/e

4 Copyright  2011 Pearson Education, Inc. GERD Chronic heartburn is a problem for some people GERD (gastroesophageal reflux disease) is chronic leaking of stomach acid into the esophagus In people with GERD, the muscles separating the stomach from the esophagus do not close tightly, allowing stomach acid to leak into the esophagus Physicians diagnose GERD by attaching a pH sensor to the esophagus to measure the acidity levels of the fluids over time 4 Tro, Chemistry: A Molecular Approach, 2/e

5 Copyright  2011 Pearson Education, Inc. Properties of Acids Sour taste React with “active” metals i.e., Al, Zn, Fe, but not Cu, Ag, or Au 2 Al + 6 HCl  AlCl 3 + 3 H 2 corrosive React with carbonates, producing CO 2 marble, baking soda, chalk, limestone CaCO 3 + 2 HCl  CaCl 2 + CO 2 + H 2 O Change color of vegetable dyes blue litmus turns red React with bases to form ionic salts 5 Tro, Chemistry: A Molecular Approach, 2/e

6 Copyright  2011 Pearson Education, Inc. Common Acids 6 Tro, Chemistry: A Molecular Approach, 2/e

7 Copyright  2011 Pearson Education, Inc. Structures of Acids 7 Binary acids have acid hydrogens attached to a nonmetal atom HCl, HF Tro, Chemistry: A Molecular Approach, 2/e

8 Copyright  2011 Pearson Education, Inc. Structure of Acids 8 Oxy acids have acid hydrogens attached to an oxygen atom H 2 SO 4, HNO 3 Tro, Chemistry: A Molecular Approach, 2/e

9 Copyright  2011 Pearson Education, Inc. Structure of Acids 9 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 Tro, Chemistry: A Molecular Approach, 2/e

10 Copyright  2011 Pearson Education, Inc. Properties of Bases 10 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, 2/e

11 Copyright  2011 Pearson Education, Inc. Common Bases 11 Tro, Chemistry: A Molecular Approach, 2/e

12 Copyright  2011 Pearson Education, Inc. 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 12 Tro, Chemistry: A Molecular Approach, 2/e

13 Copyright  2011 Pearson Education, Inc. Indicators Chemicals that change color depending on the solution’s acidity or 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 13 Tro, Chemistry: A Molecular Approach, 2/e

14 Copyright  2011 Pearson Education, Inc. 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) 14 Tro, Chemistry: A Molecular Approach, 2/e

15 Copyright  2011 Pearson Education, Inc. Arrhenius Theory HCl ionizes in water, producing H + and Cl – ions NaOH dissociates in water, producing Na + and OH – ions 15 Tro, Chemistry: A Molecular Approach, 2/e

16 Copyright  2011 Pearson Education, Inc. 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 water molecules 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 + 16 Tro, Chemistry: A Molecular Approach, 2/e

17 Copyright  2011 Pearson Education, Inc. 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) 17 Tro, Chemistry: A Molecular Approach, 2/e

18 Copyright  2011 Pearson Education, Inc. Problems with Arrhenius Theory Does not explain why molecular substances, such as NH 3, dissolve in water to form basic solutions – even though they do not contain OH – ions Does not explain how some ionic compounds, such as 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, such as 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 18 Tro, Chemistry: A Molecular Approach, 2/e

19 Copyright  2011 Pearson Education, Inc. Another Definition: Brønsted-Lowry Acid-Base Theory Brønsted and Lowry redefined acids and bases based on what happens in a reaction. Any reaction that involves H + being transferred from one molecule to another is an acid–base reaction regardless of whether it occurs in aqueous solution, or if there is OH − present All reactions that fit the Arrhenius definition also fit the Brønsted-Lowry definition, but many more do as well 19 Tro, Chemistry: A Molecular Approach, 2/e

20 Copyright  2011 Pearson Education, Inc. Brønsted-Lowry Theory In a Brønsted-Lowry acid–base reaction, an H + is transferred The acid is an H donor The base is an H acceptor base structure must contain an atom with an unshared pair of electrons In a Brønsted-Lowry acid-base reaction, the acid molecule gives an H + to the base molecule H–A + :B  :A – + H–B + 20 Tro, Chemistry: A Molecular Approach, 2/e

21 Copyright  2011 Pearson Education, Inc. 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 When HCl dissolves in water, the HCl is the acid 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) acidbase 21 Tro, Chemistry: A Molecular Approach, 2/e

22 Copyright  2011 Pearson Education, Inc. 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 When NH 3 dissolves in water, the NH 3 (aq) is the base 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 22 Tro, Chemistry: A Molecular Approach, 2/e

23 Copyright  2011 Pearson Education, Inc. A Warning! Because chemists know common bonding patterns, we often do not draw lone pair electrons on our structures. You need to be able to recognize when an atom in a molecule has lone pair electrons and when it doesn’t! 23 Tro, Chemistry: A Molecular Approach, 2/e

24 Copyright  2011 Pearson Education, Inc. Practice – Draw structures of the following that include lone pairs of electrons 24 HClO HCO 3 − Tro, Chemistry: A Molecular Approach, 2/e

25 Copyright  2011 Pearson Education, Inc. Amphoteric Substances Amphoteric substances can act as either an acid or a base because they have both a transferable H and an atom with lone pair electrons 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) 25 Tro, Chemistry: A Molecular Approach, 2/e

26 Copyright  2011 Pearson Education, Inc. 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 26 Tro, Chemistry: A Molecular Approach, 2/e

27 Copyright  2011 Pearson Education, Inc. 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 its conjugate acid; and the original acid becomes its conjugate base 27 Tro, Chemistry: A Molecular Approach, 2/e

28 Copyright  2011 Pearson Education, Inc. Brønsted-Lowry Acid–Base Reactions H–A + :B  :A – +H–B + acidbaseconjugate conjugate base acid HCHO 2 + H 2 O  CHO 2 – +H 3 O + acid baseconjugate conjugate base acid H 2 O + NH 3  HO – +NH 4 + acid baseconjugate conjugate base acid 28 Tro, Chemistry: A Molecular Approach, 2/e

29 Copyright  2011 Pearson Education, Inc. 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 29 Tro, Chemistry: A Molecular Approach, 2/e

30 Copyright  2011 Pearson Education, Inc. Practice – Write the formula for the conjugate acid of the following H 2 O NH 3 CO 3 2− H 2 PO 4 1− H 2 OH 3 O + NH 3 NH 4 + CO 3 2− HCO 3 − H 2 PO 4 1− H 3 PO 4 30 Tro, Chemistry: A Molecular Approach, 2/e

31 Copyright  2011 Pearson Education, Inc. Practice – Write the formula for the conjugate base of the following H 2 O NH 3 CO 3 2− H 2 PO 4 1− H 2 OHO − NH 3 NH 2 − CO 3 2− because CO 3 2− does not have an H, it cannot be an acid H 2 PO 4 1− HPO 4 2− 31 Tro, Chemistry: A Molecular Approach, 2/e

32 Copyright  2011 Pearson Education, Inc. Example 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 + acid baseconjugate conjugate base acid H 2 SO 4 + H 2 O  HSO 4 – +H 3 O + When the H 2 SO 4 becomes HSO 4 , it loses 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 accepts an H + so H 2 O must be the base and H 3 O + its conjugate acid 32 Tro, Chemistry: A Molecular Approach, 2/e

33 Copyright  2011 Pearson Education, Inc. Example 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 – base acidconjugate conjugate acid base HCO 3 – + H 2 O  H 2 CO 3 +HO – When the HCO 3  becomes H 2 CO 3, it accepts 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 donats an H + so H 2 O must be the acid and OH  its conjugate base 33 Tro, Chemistry: A Molecular Approach, 2/e

34 Copyright  2011 Pearson Education, Inc. Practice – Identify the Brønsted-Lowry acid, base, conjugate acid, and conjugate base in the following reaction HSO 4 − (aq) + HCO 3 − (aq)  SO 4 2− (aq) + H 2 CO 3(aq) Base Conjugate Acid Conjugate Base 34 Tro, Chemistry: A Molecular Approach, 2/e

35 Copyright  2011 Pearson Education, Inc. Practice—Write the equations for the following reacting with water and acting as a monoprotic acid & label the conjugate acid and base HBr + H 2 O  Br − + H 3 O + Acid Base Conj. Conj. base acid HSO 4 − + H 2 O  SO 4 2− + H 3 O + Acid Base Conj. Conj. base acid HBr HSO 4 − 35 Tro, Chemistry: A Molecular Approach, 2/e

36 Copyright  2011 Pearson Education, Inc. Practice—Write the equations for the following reacting with water and acting as a monoprotic-accepting base and label the conjugate acid and base I − + H 2 O  HI + OH − Base Acid Conj. Conj. acid base I − CO 3 2− CO 3 2− + H 2 O  HCO 3 − + OH − Base Acid Conj. Conj. acid base 36 Tro, Chemistry: A Molecular Approach, 2/e

37 Copyright  2011 Pearson Education, Inc. Comparing Arrhenius Theory and Brønsted-Lowry Theory 37 Brønsted – Lowry theory HCl(aq) + H 2 O(l)  Cl − (aq) + H 3 O + (aq) HF(aq) + H 2 O(l)  F − (aq) + H 3 O + (aq) NaOH(aq)  Na + (aq) + OH − (aq) NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH − (aq) Arrhenius theory HCl(aq)  H + (aq) + Cl − (aq) HF(aq)  H + (aq) + F − (aq) NaOH(aq)  Na + (aq) + OH − (aq) NH 4 OH(aq)  NH 4 + (aq) + OH − (aq) Tro, Chemistry: A Molecular Approach, 2/e

38 Copyright  2011 Pearson Education, Inc. 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 38 Tro, Chemistry: A Molecular Approach, 2/e

39 Copyright  2011 Pearson Education, Inc. 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,  39 Tro, Chemistry: A Molecular Approach, 2/e

40 Copyright  2011 Pearson Education, Inc. Strong Acids 40 HCl  H + + Cl − HCl + H 2 O  H 3 O + + Cl − 0.10 M HCl = 0.10 M H 3 O + The stronger the acid, the more willing it is to donate H we use water as the standard base to donate H to Strong acids donate practically all their H’s 100% ionized in water strong electrolyte [H 3 O + ] = [strong acid] [X] means the molarity of X Tro, Chemistry: A Molecular Approach, 2/e

41 Copyright  2011 Pearson Education, Inc. HF  H + + F − HF + H 2 O  H 3 O + + F − 0.10 M HF ≠ 0.10 M H 3 O + Weak Acids 41 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] Tro, Chemistry: A Molecular Approach, 2/e

42 Copyright  2011 Pearson Education, Inc. Strong & Weak Acids 42 Tro, Chemistry: A Molecular Approach, 2/e

43 Copyright  2011 Pearson Education, Inc. Tro, Chemistry: A Molecular Approach, 2/e

44 Copyright  2011 Pearson Education, Inc. 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 − + H 3 O + Base: + H 2 O  HBase + + OH − The farther the equilibrium position lies toward 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 44 Tro, Chemistry: A Molecular Approach, 2/e

45 Copyright  2011 Pearson Education, Inc. 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 + > H 2 O > OH − ; NH 4 + > NH 3 > NH 2 − trend in base strength opposite 45 Tro, Chemistry: A Molecular Approach, 2/e

46 Copyright  2011 Pearson Education, Inc. Acid Ionization Constant, K a Acid strength measured by the size of the equilibrium constant when reacts with H 2 O HAcid + H 2 O  Acid − + H 3 O + The equilibrium constant for this reaction is called the acid ionization constant, K a larger K a = stronger acid 46 Tro, Chemistry: A Molecular Approach, 2/e

47 Copyright  2011 Pearson Education, Inc. 47 Tro, Chemistry: A Molecular Approach, 2/e

48 Copyright  2011 Pearson Education, Inc. Autoionization of Water Water is actually an extremely weak electrolyte therefore there must be a few ions present About 2 out of every 1 billion 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 – ] = 10 −7 M @ 25 °C 48 Tro, Chemistry: A Molecular Approach, 2/e

49 Copyright  2011 Pearson Education, Inc. 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 aka the Dissociation Constant of Water [H 3 O + ] x [OH – ] = K w = 1.00 x 10 −14 @ 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 49 Tro, Chemistry: A Molecular Approach, 2/e

50 Copyright  2011 Pearson Education, Inc. 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 – ] = 1.00 x 10 −7 Acidic solutions have a larger [H 3 O + ] than [OH – ] [H 3 O + ] > 1.00 x 10 −7 ; [OH – ] < 1.00 x 10 −7 Basic solutions have a larger [OH – ] than [H 3 O + ] [H 3 O + ] 1.00 x 10 −7 50 Tro, Chemistry: A Molecular Approach, 2/e

51 Copyright  2011 Pearson Education, Inc. Practice – Complete the table [H + ] vs. [OH − ] OH − H+H+ H+H+ H+H+ H+H+ H+H+ [OH − ] [H + ] 10 0 10 −1 10 − 3 10 −5 10 −7 10 −9 10 −11 10 −13 10 −14 51 Tro, Chemistry: A Molecular Approach, 2/e

52 Copyright  2011 Pearson Education, Inc. Practice – Complete the table [H + ] vs. [OH − ] OH − H+H+ H+H+ H+H+ H+H+ H+H+ [OH − ]10 −14 10 −13 10 −11 10 −9 10 −7 10 −5 10 −3 10 −1 10 0 [H + ] 10 0 10 −1 10 −3 10 −5 10 −7 10 −9 10 −11 10 −13 10 −14 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 Acid Base 52 Tro, Chemistry: A Molecular Approach, 2/e

53 Copyright  2011 Pearson Education, Inc. 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 53 the units are correct; the fact that the [H 3 O + ] < [OH  ] means the solution is basic [H 3 O + ] = 1.5 x 10 −9 M [OH  ] Check: Solution: Conceptual Plan: Relationships: Given: Find: [H 3 O + ][OH  ] Tro, Chemistry: A Molecular Approach, 2/e

54 Copyright  2011 Pearson Education, Inc. Practice – Determine the [H 3 O + ] when the [OH − ] = 2.5 x 10 −9 M 54 Tro, Chemistry: A Molecular Approach, 2/e

55 Copyright  2011 Pearson Education, Inc. Practice – Determine the [H 3 O + ] when the [OH − ] = 2.5 x 10 −9 M 55 the units are correct; the fact that the [H 3 O + ] > [OH  ] means the solution is acidic [OH  ] = 2.5 x 10 −9 M [H 3 O + ] Check: Solution: Conceptual Plan: Relationships: Given: Find: [OH  ][H 3 O + ] Tro, Chemistry: A Molecular Approach, 2/e

56 Copyright  2011 Pearson Education, Inc. Measuring Acidity: pH 56 The acidity or basicity of a solution is often expressed as pH pH = −log[H 3 O + ] exponent on 10 with a positive sign pH water = −log[10 −7 ] = 7 need to know the [H 3 O + ] concentration to find pH pH 7 is basic, pH = 7 is neutral [H 3 O + ] = 10 −pH Tro, Chemistry: A Molecular Approach, 2/e

57 Copyright  2011 Pearson Education, Inc. 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) = 6 + 0.30303… = 6.30303... Because 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 57 Tro, Chemistry: A Molecular Approach, 2/e

58 Copyright  2011 Pearson Education, Inc. What Does the pH Number Imply? 58 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 of pH is 0 to 14 pH 0 is [H 3 O + ] = 1 M, pH 14 is [OH – ] = 1 M pH can be negative (very acidic) or larger than 14 (very alkaline) Tro, Chemistry: A Molecular Approach, 2/e

59 Copyright  2011 Pearson Education, Inc. 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 59 pH is unitless; the fact that the pH > 7 means the solution is basic [OH  ] = 1.3 x 10 −2 M pH Check: Solution: Conceptual Plan: Relationships: Given: Find: [H 3 O + ][OH  ]pH Tro, Chemistry: A Molecular Approach, 2/e

60 Copyright  2011 Pearson Education, Inc. Practice – Determine the pH @ 25 ºC of a solution that has [OH − ] = 2.5 x 10 −9 M 60 Tro, Chemistry: A Molecular Approach, 2/e

61 Copyright  2011 Pearson Education, Inc. Practice – Determine the pH @ 25 ºC of a solution that has [OH − ] = 2.5 x 10 −9 M 61 pH is unitless; the fact that the pH < 7 means the solution is acidic [OH  ] = 2.5 x 10 −9 M pH Check: Solution: Conceptual Plan: Relationships: Given: Find: [H 3 O + ][OH  ]pH Tro, Chemistry: A Molecular Approach, 2/e

62 Copyright  2011 Pearson Education, Inc. Practice – Determine the [OH − ] of a solution with a pH of 5.40 62 Tro, Chemistry: A Molecular Approach, 2/e

63 Copyright  2011 Pearson Education, Inc. Practice – Determine the [OH − ] of a solution with a pH of 5.40 63 because the pH < 7, [OH − ] should be less than 1 x 10 −7 ; and it is pH = 5.40 [OH − ], M Check: Solution: Conceptual Plan: Relationships: Given: Find: [H 3 O + ]pH[OH  ] Tro, Chemistry: A Molecular Approach, 2/e

64 Copyright  2011 Pearson Education, Inc. pOH 64 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 acidic, pOH = 7 is neutral pH + pOH = 14.0 Tro, Chemistry: A Molecular Approach, 2/e

65 Copyright  2011 Pearson Education, Inc. pH and pOH Practice – Complete the table pH pOH OH − H+H+ H+H+ H+H+ H+H+ H+H+ [OH − ]10 −14 10 −13 10 −11 10 −9 10 −7 10 −5 10 −3 10 −1 10 0 [H + ] 10 0 10 −1 10 −3 10 −5 10 −7 10 −9 10 −11 10 −13 10 −14 65 Tro, Chemistry: A Molecular Approach, 2/e

66 Copyright  2011 Pearson Education, Inc. pH and pOH Practice – Complete the table pH 0 1 3 5 7 9 11 13 14 pOH 14 13 11 9 7 5 3 1 0 OH − H+H+ H+H+ H+H+ H+H+ H+H+ [OH − ]10 −14 10 −13 10 −11 10 −9 10 −7 10 −5 10 −3 10 −1 10 0 [H + ] 10 0 10 −1 10 −3 10 −5 10 −7 10 −9 10 −11 10 −13 10 −14 Acid Base 66 Tro, Chemistry: A Molecular Approach, 2/e

67 Copyright  2011 Pearson Education, Inc. Relationship between pH and pOH pH + pOH = 14.00 at 25 °C you can use pOH to find pH of a solution 67 Tro, Chemistry: A Molecular Approach, 2/e

68 Copyright  2011 Pearson Education, Inc. Example: 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 68 pH is unitless; the fact that the pH > 7 means the solution is basic [OH  ] = 1.3 x 10 −2 M pH Check: Solution: Conceptual Plan: Relationships: Given: Find: pOH[OH  ]pH Tro, Chemistry: A Molecular Approach, 2/e

69 Copyright  2011 Pearson Education, Inc. Practice – Determine the pOH @ 25 ºC of a solution that has [H 3 O + ] = 2.5 x 10 −9 M 69 Tro, Chemistry: A Molecular Approach, 2/e

70 Copyright  2011 Pearson Education, Inc. Practice – Determine the pOH @ 25 ºC of a solution that has [H 3 O + ] = 2.5 x 10 −9 M 70 pH is unitless; the fact that the pH < 7 means the solution is acidic [H 3 O + ] = 2.5 x 10 −9 M pOH Check: Solution: Conceptual Plan: Relationships: Given: Find: pH[H 3 O + ]pOH Tro, Chemistry: A Molecular Approach, 2/e

71 Copyright  2011 Pearson Education, Inc. pKpK 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 The stronger the base, the smaller the pK b larger K b = smaller pK b 71 Tro, Chemistry: A Molecular Approach, 2/e

72 Copyright  2011 Pearson Education, Inc. [H 3 O + ] and [OH − ] in a Strong Acid or Strong Base Solution There are two sources of H 3 O + in an aqueous solution of a strong acid – the acid and the water There are two sources of OH − in an aqueous solution of a strong acid – the base and the water For a strong acid or base, the contribution of the water to the total [H 3 O + ] or [OH − ] is negligible the [H 3 O + ] acid shifts the K w equilibrium so far that [H 3 O + ] water is too small to be significant  except in very dilute solutions, generally < 1 x 10 −4 M 72 Tro, Chemistry: A Molecular Approach, 2/e

73 Copyright  2011 Pearson Education, Inc. Finding pH of a Strong Acid or Strong Base Solution For a monoprotic strong acid [H 3 O + ] = [HAcid] for polyprotic acids, the other ionizations can generally be ignored  for H 2 SO 4, the second ionization cannot be ignored 0.10 M HCl has [H 3 O + ] = 0.10 M and pH = 1.00 For a strong ionic base, [OH − ] = (number OH − )x[Base] for molecular bases with multiple lone pairs available, only one lone pair accepts an H, the other reactions can generally be ignored 0.10 M Ca(OH) 2 has [OH − ] = 0.20 M and pH = 13.30 73 Tro, Chemistry: A Molecular Approach, 2/e

74 Copyright  2011 Pearson Education, Inc. Finding the pH of a Weak Acid There are also two sources of H 3 O + in an 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 + 74 Tro, Chemistry: A Molecular Approach, 2/e

75 Copyright  2011 Pearson Education, Inc. [HNO 2 ][NO 2 − ][H 3 O + ] initial change equilibrium Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 75 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 because no products initially, Q c = 0, and the reaction is proceeding forward HNO 2 + H 2 O  NO 2  + H 3 O + [HNO 2 ][NO 2 − ][H 3 O + ] initial 0.200 0 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

76 Copyright  2011 Pearson Education, Inc. [HNO 2 ][NO 2 − ][H 3 O + ] initial 0.200 00 change equilibrium Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 76 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 +x+x +x+x xx 0.200  x xx HNO 2 + H 2 O (l)  NO 2 − (aq) + H 3 O + (aq) Tro, Chemistry: A Molecular Approach, 2/e

77 Copyright  2011 Pearson Education, Inc. Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 77 determine the value of K a from Table 15.5 because K a is very small, approximate the [HNO 2 ] eq = [HNO 2 ] init and solve for x K a for HNO 2 = 4.6 x 10 −4 [HNO 2 ][NO 2 − ][H 3 O + ] initial 0.200 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.200xx 0.200  x Tro, Chemistry: A Molecular Approach, 2/e

78 Copyright  2011 Pearson Education, Inc. Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 78 check if the approximation is valid by seeing if x < 5% of [HNO 2 ] init K a for HNO 2 = 4.6 x 10 −4 [HNO 2 ][NO 2 − ][H 3 O + ] initial 0.200 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.200 xx the approximation is valid x = 9.6 x 10 −3 Tro, Chemistry: A Molecular Approach, 2/e

79 Copyright  2011 Pearson Education, Inc. Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 79 substitute x into the equilibrium concentration definitions and solve K a for HNO 2 = 4.6 x 10 −4 [HNO 2 ][NO 2 − ][H 3 O + ] initial 0.200 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.200−x xx x = 9.6 x 10 −3 [HNO 2 ][NO 2 − ][H 3 O + ] initial 0.200 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.1900.0096 Tro, Chemistry: A Molecular Approach, 2/e

80 Copyright  2011 Pearson Education, Inc. Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 80 substitute [H 3 O + ] into the formula for pH and solve K a for HNO 2 = 4.6 x 10 −4 [HNO 2 ][NO 2 − ][H 3 O + ] initial 0.200 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.1900.0096 Tro, Chemistry: A Molecular Approach, 2/e

81 Copyright  2011 Pearson Education, Inc. Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 81 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a K a for HNO 2 = 4.6 x 10 −4 [HNO 2 ][NO 2 − ][H 3 O + ] initial 0.200 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.1900.0096 though not exact, the answer is reasonably close Tro, Chemistry: A Molecular Approach, 2/e

82 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? (K a = 1.4 x 10 −5 @ 25 °C) 82 Tro, Chemistry: A Molecular Approach, 2/e

83 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? 83 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 + [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

84 Copyright  2011 Pearson Education, Inc. [HA][A − ][H 3 O + ] initial 0.012 00 change equilibrium Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? 84 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 +x+x +x+x xx 0.012  x xx HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

85 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? K a = 1.4 x 10 −5 @ 25 °C determine the value of K a because K a is very small, approximate the [HA] eq = [HA] init and solve for x [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012xx 0.012  x HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

86 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? K a = 1.4 x 10 −5 @ 25 °C 86 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 2 = 1.4 x 10 −5 the approximation is valid x = 4.1 x 10 −4 [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012 xx Tro, Chemistry: A Molecular Approach, 2/e

87 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? K a = 1.4 x 10 −5 @ 25 °C 87 substitute x into the equilibrium concentration definitions and solve x = 4.1 x 10 −4 [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012−x xx Tro, Chemistry: A Molecular Approach, 2/e

88 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? K a = 1.4 x 10 −5 @ 25 °C 88 substitute [H 3 O + ] into the formula for pH and solve [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0120.00041 Tro, Chemistry: A Molecular Approach, 2/e

89 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? K a = 1.4 x 10 −5 @ 25 °C 89 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 [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0120.00041 Tro, Chemistry: A Molecular Approach, 2/e

90 Copyright  2011 Pearson Education, Inc. [HClO 2 ][ClO 2 − ][H 3 O + ] initial 0.100 0≈ 0 change equilibrium Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 90 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 HClO 2 + H 2 O  ClO 2  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

91 Copyright  2011 Pearson Education, Inc. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 91 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 [HClO 2 ][ClO 2 − ][H 3 O + ] initial 0.100 0≈ 0 change −x−x+x+x+x+x equilibrium 0.100−xxx Tro, Chemistry: A Molecular Approach, 2/e

92 Copyright  2011 Pearson Education, Inc. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 92 determine the value of K a from Table 15.5 because K a is very small, approximate the [HClO 2 ] eq = [HClO 2 ] init and solve for x K a for HClO 2 = 1.1 x 10 −2 [HClO 2 ][ClO 2 − ][H 3 O + ] initial 0.100 0≈ 0 change −x−x+x+x+x+x equilibrium 0.100−xxx Tro, Chemistry: A Molecular Approach, 2/e

93 Copyright  2011 Pearson Education, Inc. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 93 check if the approximation is valid by seeing if x < 5% of [HNO 2 ] init K a for HClO 2 = 1.1 x 10 −2 the approximation is invalid x = 3.3 x 10 −2 [HClO 2 ][ClO 2 − ][H 3 O + ] initial 0.100 0≈ 0 change −x−x+x+x+x+x equilibrium 0.100−xxx Tro, Chemistry: A Molecular Approach, 2/e

94 Copyright  2011 Pearson Education, Inc. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 94 if the approximation is invalid, solve for x using the quadratic formula K a for HClO 2 = 1.1 x 10 −2 Tro, Chemistry: A Molecular Approach, 2/e

95 Copyright  2011 Pearson Education, Inc. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 95 substitute x into the equilibrium concentration definitions and solve K a for HClO 2 = 1.1 x 10 −2 x = 0.028 [HClO 2 ][ClO 2 − ][H 3 O + ] initial 0.100 0≈ 0 change −x−x+x+x+x+x equilibrium 0.100−xxx [HClO 2 ][ClO 2 − ][H 3 O + ] initial 0.100 0≈ 0 change −x−x+x+x+x+x equilibrium 0.0720.028 Tro, Chemistry: A Molecular Approach, 2/e

96 Copyright  2011 Pearson Education, Inc. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 96 substitute [H 3 O + ] into the formula for pH and solve K a for HClO 2 = 1.1 x 10 −2 [HClO 2 ][ClO 2 − ][H 3 O + ] initial 0.100 0≈ 0 change −x−x+x+x+x+x equilibrium 0.0720.028 Tro, Chemistry: A Molecular Approach, 2/e

97 Copyright  2011 Pearson Education, Inc. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25°C 97 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a K a for HClO 2 = 1.1 x 10 −2 the answer matches [HClO 2 ][ClO 2 - ][H 3 O + ] initial 0.100 0≈ 0 change −x−x+x+x+x+x equilibrium 0.0720.028 Tro, Chemistry: A Molecular Approach, 2/e

98 Copyright  2011 Pearson Education, Inc. Example 15.8: What is the K a of a weak acid if a 0.100 M solution has a pH of 4.25? 98 use the pH to find the equilibrium [H 3 O + ] write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations and [H 3 O + ] equil HA + H 2 O  A  + H 3 O + [HA][A − ][H 3 O + ] initial change equilibrium [HA][A − ][H 3 O + ] initial 0.100 0 ≈ 0 change equilibrium 5.6E-05 Tro, Chemistry: A Molecular Approach, 2/e

99 Copyright  2011 Pearson Education, Inc. [HA][A − ][H 3 O + ] initial 0.100 00 change equilibrium Example 15.8: What is the K a of a weak acid if a 0.100 M solution has a pH of 4.25? 99 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 +5.6E-05 −5.6E-05 0.100  5.6E-05 HA + H 2 O  A  + H 3 O + 0.100 Tro, Chemistry: A Molecular Approach, 2/e

100 Copyright  2011 Pearson Education, Inc. Practice – What is the K a of nicotinic acid, HC 6 H 4 NO 2, if a 0.012 M solution of nicotinic acid has a pH of 3.40? 100 Tro, Chemistry: A Molecular Approach, 2/e

101 Copyright  2011 Pearson Education, Inc. Practice – What is the K a of nicotinic acid, HC 6 H 4 NO 2, if a 0.012 M solution of nicotinic acid has a pH of 3.40? 101 use the pH to find the equilibrium [H 3 O + ] write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations and [H 3 O + ] equil HA + H 2 O  A  + H 3 O + [HA][A − ][H 3 O + ] initial change equilibrium [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change equilibrium 4.0E-04 Tro, Chemistry: A Molecular Approach, 2/e

102 Copyright  2011 Pearson Education, Inc. [HA][A − ][H 3 O + ] initial 0.012 00 change equilibrium 0.012 Practice – What is the K a of nicotinic acid, HC 6 H 4 NO 2, if a 0.012 M solution of nicotinic acid has a pH of 3.40? 102 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 +4.0E-04 −4.0E-04 0.012  4.0E-04 HA + H 2 O  A  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

103 Copyright  2011 Pearson Education, Inc. Polyprotic Acids Acid molecules often 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 is 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  103 Tro, Chemistry: A Molecular Approach, 2/e

104 Copyright  2011 Pearson Education, Inc. 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 Because [ionized acid] equil = [H 3 O + ] equil 104 Tro, Chemistry: A Molecular Approach, 2/e

105 Copyright  2011 Pearson Education, Inc. Example 15.9: What is the percent ionization of a 2.5 M HNO 2 solution? 105 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 HNO 2 + H 2 O  NO 2  + H 3 O + [HNO 2 ][NO 2 − ][H 3 O + ] initial 2.50≈0 change equilibrium [HNO 2 ][NO 2 − ][H 3 O + ] initial 2.50≈ 0 change equilibrium +x+x +x+x xx 2.5  x xx Tro, Chemistry: A Molecular Approach, 2/e

106 Copyright  2011 Pearson Education, Inc. Example 15.9: What is the percent ionization of a 2.5 M HNO 2 solution? 106 determine the value of K a from Table 15.5 because K a is very small, approximate the [HNO 2 ] eq = [HNO 2 ] init and solve for x K a for HNO 2 = 4.6 x 10 −4 [HNO 2 ][NO 2 − ][H 3 O + ] initial 2.5 0≈ 0 change −x−x+x+x+x+x equilibrium 2.5−x ≈2.5xx Tro, Chemistry: A Molecular Approach, 2/e

107 Copyright  2011 Pearson Education, Inc. Example 15.9: What is the percent ionization of a 2.5 M HNO 2 solution? 107 substitute x into the equilibrium concentration definitions and solve HNO 2 + H 2 O  NO 2  + H 3 O + [HNO 2 ][NO 2 − ][H 3 O + ] initial 2.5 0 ≈ 0 change −x−x+x+x+x+x equilibrium 2.50.034 2.5  x xx x = 3.4 x 10 −2 Tro, Chemistry: A Molecular Approach, 2/e

108 Copyright  2011 Pearson Education, Inc. Example 15.9: What is the percent ionization of a 2.5 M HNO 2 solution? 108 apply the definition and compute the percent ionization HNO 2 + H 2 O  NO 2  + H 3 O + [HNO 2 ][NO 2 − ][H 3 O + ] initial 2.5 0 ≈ 0 change −x−x+x+x+x+x equilibrium 2.50.034 because the percent ionization is < 5%, the “x is small” approximation is valid Tro, Chemistry: A Molecular Approach, 2/e

109 Copyright  2011 Pearson Education, Inc. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? (K a = 1.4 x 10 −5 @ 25 °C) 109 Tro, Chemistry: A Molecular Approach, 2/e

110 Copyright  2011 Pearson Education, Inc. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? 110 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 + [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

111 Copyright  2011 Pearson Education, Inc. [HA][A − ][H 3 O + ] initial 0.012 00 change equilibrium Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? 111 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 +x+x +x+x xx 0.012  x xx HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

112 Copyright  2011 Pearson Education, Inc. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? K a = 1.4 x 10 −5 112 determine the value of K a because K a is very small, approximate the [HA] eq = [HA] init and solve for x [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012xx 0.012  x HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

113 Copyright  2011 Pearson Education, Inc. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? 113 substitute x into the equilibrium concentration definitions and solve x = 4.1 x 10 −4 [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012−x xx Tro, Chemistry: A Molecular Approach, 2/e

114 Copyright  2011 Pearson Education, Inc. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? 114 apply the definition and compute the percent ionization because the percent ionization is < 5%, the “x is small” approximation is valid [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0124.1E-04 Tro, Chemistry: A Molecular Approach, 2/e

115 Copyright  2011 Pearson Education, Inc. Relationship Between [H 3 O + ] equilibrium & [HA] initial Increasing the initial concentration of acid results in increased [H 3 O + ] 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 115 Tro, Chemistry: A Molecular Approach, 2/e

116 Copyright  2011 Pearson Education, Inc. 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 116 Tro, Chemistry: A Molecular Approach, 2/e

117 Copyright  2011 Pearson Education, Inc. 117 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, 2/e

118 Copyright  2011 Pearson Education, Inc. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 118 write the reactions for the acids with water and determine their K a s if the K a s are 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 [HF][F − ][H 3 O + ] initial 0.150 0 ≈ 0 change equilibrium 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 10 −14 Tro, Chemistry: A Molecular Approach, 2/e

119 Copyright  2011 Pearson Education, Inc. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 119 represent the change in the concentrations in term 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 0.150 00 change equilibrium +x+x+x+x xx 0.150  x xx Tro, Chemistry: A Molecular Approach, 2/e

120 Copyright  2011 Pearson Education, Inc. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 120 because K a is very small, approximate the [HF] eq = [HF] init and solve for x K a for HF = 3.5 x 10 −4 [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.150xx 0.150  x Tro, Chemistry: A Molecular Approach, 2/e

121 Copyright  2011 Pearson Education, Inc. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 121 check if the approximation is valid by seeing if x < 5% of [HF] init K a for HF = 3.5 x 10 −4 [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.150 xx the approximation is valid x = 7.2 x 10 −3 Tro, Chemistry: A Molecular Approach, 2/e

122 Copyright  2011 Pearson Education, Inc. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 122 substitute x into the equilibrium concentration definitions and solve K a for HF = 3.5 x 10 −4 [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.150-x xx x = 7.2 x 10 −3 [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.1430.0072 Tro, Chemistry: A Molecular Approach, 2/e

123 Copyright  2011 Pearson Education, Inc. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 123 substitute [H 3 O + ] into the formula for pH and solve K a for HF = 3.5 x 10 −4 [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.1430.0072 Tro, Chemistry: A Molecular Approach, 2/e

124 Copyright  2011 Pearson Education, Inc. 124 Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) K a for HF = 3.5 x 10 −4 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a though not exact, the answer is reasonably close [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.1430.0072 Tro, Chemistry: A Molecular Approach, 2/e

125 Copyright  2011 Pearson Education, Inc. Practice – Determine the pH @ 25 ºC of a solution that is a mixture of 0.045 M HCl and 0.15 M HF 125 Tro, Chemistry: A Molecular Approach, 2/e

126 Copyright  2011 Pearson Education, Inc. Practice – Determine the pH @ 25 ºC of a solution that is a mixture of 0.045 M HCl and 0.15 M HF 126 pH is unitless; the fact that the pH < 7 means the solution is acidic [HCl] = 4.5 x 10 −2 M, [HF] = 0.15 M pH Check: Solution: Conceptual Plan: Relationships: Given: Find: [H 3 O + ][HCl]pH Because HCl is a strong acid and HF is a weak acid, [H 3 O + ] = [HCl] Tro, Chemistry: A Molecular Approach, 2/e

127 Copyright  2011 Pearson Education, Inc. Strong Bases 127 NaOH  Na + + OH − The stronger the base, the more willing it is to accept H use water as the standard acid For ionic bases, practically all units are dissociated into OH – or accept H’s strong electrolyte multi-OH strong bases completely dissociated [HO – ] = [strong base] x (# OH) Tro, Chemistry: A Molecular Approach, 2/e

128 Copyright  2011 Pearson Education, Inc. Example 15.11: Calculate the pH at 25 °C of a 0.0015 M Sr(OH) 2 solution and determine if the solution is acidic, basic, or neutral pH is unitless; the fact that the pH > 7 means the solution is basic [Sr(OH) 2 ] = 1.5 x 10 −3 M pH Check: Solution: Conceptual Plan: Relationships: Given: Find: [H 3 O + ][OH  ]pH[Sr(OH) 2 ] [OH  ]=2[Sr(OH) 2 ] [OH  ] = 2(0.0015) = 0.0030 M 128 Tro, Chemistry: A Molecular Approach, 2/e

129 Copyright  2011 Pearson Education, Inc. Example 15.11: Calculate the pH at 25 °C of a 0.0015 M Sr(OH) 2 solution and determine if the solution is acidic, basic, or neutral pH is unitless; the fact that the pH > 7 means the solution is basic [Sr(OH) 2 ] = 1.5 x 10 −3 M pH Check: Solution: Conceptual Plan: Relationships: Given: Find: pOH[OH  ]pH[Sr(OH) 2 ] [OH  ]=2[Sr(OH) 2 ] [OH  ] = 2(0.0015) = 0.0030 M pH + pOH = 14.00 pH = 14.00 – 2.52 = 11.48 129 Tro, Chemistry: A Molecular Approach, 2/e

130 Copyright  2011 Pearson Education, Inc. Practice – Calculate the pH of the following strong acid or base solutions 0.0020 M HCl 0.0015 M Ca(OH) 2 [H 3 O + ] = [HCl] = 2.0 x 10 −3 M pH = −log(2.0 x 10 −3 ) = 2.70 [OH − ] = 2 x [Ca(OH) 2 ] = 3.0 x 10 −3 M pOH = −log(3.0 x 10 −3 ) = 2.52 pH = 14.00 − pOH = 14.00 − 2.52 pH = 11.48 130 Tro, Chemistry: A Molecular Approach, 2/e

131 Copyright  2011 Pearson Education, Inc. Weak Bases 131 NH 3 + H 2 O  NH 4 + + OH − 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 Tro, Chemistry: A Molecular Approach, 2/e

132 Copyright  2011 Pearson Education, Inc. Base Ionization Constant, K b Base strength measured by the size of the equilibrium constant when react with H 2 O :Base + H 2 O  OH − + H:Base + The equilibrium constant is called the base ionization constant, K b larger K b = stronger base 132 Tro, Chemistry: A Molecular Approach, 2/e

133 Copyright  2011 Pearson Education, Inc. 133 Tro, Chemistry: A Molecular Approach, 2/e

134 Copyright  2011 Pearson Education, Inc. Structure of Amines 134 Tro, Chemistry: A Molecular Approach, 2/e

135 Copyright  2011 Pearson Education, Inc. [NH 3 ][NH 4 + ][OH  ] initial change equilibrium Example 15.12:Find the pH of 0.100 M NH 3 (aq) 135 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 because no products initially, Q c = 0, and the reaction is proceeding forward NH 3 + H 2 O  NH 4 + + OH  [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

136 Copyright  2011 Pearson Education, Inc. [NH 3 ][NH 4 + ][OH  ] initial 0.100 00 change equilibrium Example 15.12: Find the pH of 0.100 M NH 3 (aq) 136 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 +x+x +x+x xx 0.100  x xx Tro, Chemistry: A Molecular Approach, 2/e

137 Copyright  2011 Pearson Education, Inc. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 137 determine the value of K b from Table 15.8 because K b is very small, approximate the [NH 3 ] eq = [NH 3 ] init and solve for x K b for NH 3 = 1.76 x 10 −5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.100xx 0.100  x Tro, Chemistry: A Molecular Approach, 2/e

138 Copyright  2011 Pearson Education, Inc. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 138 check if the approximation is valid by seeing if x < 5% of [NH 3 ] init K b for NH 3 = 1.76 x 10 −5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.100 xx the approximation is valid x = 1.33 x 10 −3 Tro, Chemistry: A Molecular Approach, 2/e

139 Copyright  2011 Pearson Education, Inc. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 139 substitute x into the equilibrium concentration definitions and solve K b for NH 3 = 1.76 x 10 −5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.100  x xx x = 1.33 x 10 −3 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.0991.33E-3 Tro, Chemistry: A Molecular Approach, 2/e

140 Copyright  2011 Pearson Education, Inc. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 140 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 NH 3 = 1.76 x 10 −5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0991.33E−3 Tro, Chemistry: A Molecular Approach, 2/e

141 Copyright  2011 Pearson Education, Inc. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 141 use the [OH − ] to find the pOH use pOH to find pH K b for NH 3 = 1.76 x 10 −5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0991.33E−3 Tro, Chemistry: A Molecular Approach, 2/e

142 Copyright  2011 Pearson Education, Inc. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 142 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K b to the given K b K b for NH 3 = 1.76 x 10 −5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0991.33E−3 though not exact, the answer is reasonably close Tro, Chemistry: A Molecular Approach, 2/e

143 Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution, K b = 1.6 x 10 −6 143 Tro, Chemistry: A Molecular Approach, 2/e

144 Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 144 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 because no products initially, Q c = 0, and the reaction is proceeding forward B + H 2 O  BH + + OH  [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

145 Copyright  2011 Pearson Education, Inc. [B][BH + ][OH  ] initial 0.0015 00 change equilibrium Practice – Find the pH of a 0.0015 M morphine solution 145 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 +x+x +x+x xx 0.0015  x xx B + H 2 O  BH + + OH  Tro, Chemistry: A Molecular Approach, 2/e

146 Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 146 determine the value of K b because K b is very small, approximate the [B] eq = [B] init and solve for x K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0015xx 0.0015  x Tro, Chemistry: A Molecular Approach, 2/e

147 Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 147 check if the approximation is valid by seeing if x < 5% of [B] init the approximation is valid x = 4.9 x 10 −5 K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0015 xx Tro, Chemistry: A Molecular Approach, 2/e

148 Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 148 substitute x into the equilibrium concentration definitions and solve x = 4.9 x 10 −5 K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0015  x xx [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5 Tro, Chemistry: A Molecular Approach, 2/e

149 Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 149 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 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5 Tro, Chemistry: A Molecular Approach, 2/e

150 Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 150 use the [OH − ] to find the pOH use pOH to find pH K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5 Tro, Chemistry: A Molecular Approach, 2/e

151 Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 151 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K b to the given K b the answer matches the given K b K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5 Tro, Chemistry: A Molecular Approach, 2/e

152 Copyright  2011 Pearson Education, Inc. 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 152 Tro, Chemistry: A Molecular Approach, 2/e

153 Copyright  2011 Pearson Education, Inc. 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) An anion that is the conjugate base of a weak acid is basic F − (aq) + H 2 O(l)  HF(aq) + OH − (aq) 153 Tro, Chemistry: A Molecular Approach, 2/e

154 Copyright  2011 Pearson Education, Inc. Example 15.13: 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 154 Tro, Chemistry: A Molecular Approach, 2/e

155 Copyright  2011 Pearson Education, Inc. 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 155 Tro, Chemistry: A Molecular Approach, 2/e

156 Copyright  2011 Pearson Education, Inc. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) solution 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 enter the initial concentrations – assuming the [OH  ] from water is ≈ 0 CHO 2 − + H 2 O  HCHO 2 + OH  [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change equilibrium

157 Copyright  2011 Pearson Education, Inc. [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change equilibrium 0.100  x Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) solution 157 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 +x+x+x+x xx xx Tro, Chemistry: A Molecular Approach, 2/e

158 Copyright  2011 Pearson Education, Inc. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 158 because K b is very small, approximate the [CHO 2 − ] eq = [CHO 2 − ] init and solve for x K b for CHO 2 − = 5.6 x 10 −11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change −x−x+x+x+x+x equilibrium 0.100xx 0.100  x Tro, Chemistry: A Molecular Approach, 2/e

159 Copyright  2011 Pearson Education, Inc. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 159 check if the approximation is valid by seeing if x < 5% of [CHO 2 − ] init the approximation is valid x = 2.4 x 10 −6 K b for CHO 2 − = 5.6 x 10 −11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change −x−x+x+x+x+x equilibrium 0.100 xx Tro, Chemistry: A Molecular Approach, 2/e

160 Copyright  2011 Pearson Education, Inc. [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change −x−x+x+x+x+x equilibrium 0.1002.4E-6 Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 160 substitute x into the equilibrium concentration definitions and solve x = 2.4 x 10 −6 K b for CHO 2 − = 5.6 x 10 −11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change −x−x+x+x+x+x equilibrium 0.100 −x xx Tro, Chemistry: A Molecular Approach, 2/e

161 Copyright  2011 Pearson Education, Inc. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 161 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 CHO 2 − = 5.6 x 10 −11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change −x−x+x+x+x+x equilibrium 0.1002.4E-6 Tro, Chemistry: A Molecular Approach, 2/e

162 Copyright  2011 Pearson Education, Inc. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 162 use the [OH − ] to find the pOH use pOH to find pH K b for CHO 2 − = 5.6 x 10 −11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change −x−x+x+x+x+x equilibrium 0.1002.4E−6 Tro, Chemistry: A Molecular Approach, 2/e

163 Copyright  2011 Pearson Education, Inc. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 163 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K b to the given K b though not exact, the answer is reasonably close K b for CHO 2 − = 5.6 x 10 −11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change −x−x+x+x+x+x equilibrium 0.1002.4E−6 Tro, Chemistry: A Molecular Approach, 2/e

164 Copyright  2011 Pearson Education, Inc. Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the K a of HA? 164 Tro, Chemistry: A Molecular Approach, 2/e

165 Copyright  2011 Pearson Education, Inc. If a 0.0015 M NaA solution has a pOH of 5.45, what is the K a of HA? Na + is the cation of a strong base – pOH neutral. Because pOH is < 7, the solutinon is basic. A − is basic. write the reaction for the anion with water construct an ICE table for the reaction enter the initial concentrations – assuming the [OH  ] from water is ≈ 0 A − + H 2 O  HA + OH  [A − ][HA][OH  ] initial 0.1000≈ 0 change equilibrium 165

166 Copyright  2011 Pearson Education, Inc. Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the K a of HA? use the pOH to find the [OH − ] use [OH − ] to fill in other items [A − ][HA][OH  ] initial 0.150≈ 0 change −3.6E−6 + 3.6E−6 equilibrium 0.153.6E-6 [A − ][HA][OH  ] initial 0.150≈ 0 change equilibrium 166 Tro, Chemistry: A Molecular Approach, 2/e

167 Copyright  2011 Pearson Education, Inc. calculate the value of K b of A − [A − ][HA][OH  ] initial 0.150≈ 0 change − 3.6E−6 + 3.6E−6 equilibrium 0.153.6E-6 167 Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the K a of HA? Tro, Chemistry: A Molecular Approach, 2/e

168 Copyright  2011 Pearson Education, Inc. use K b of A − to find K a of HA 168 Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the K a of HA? [A − ][HA][OH  ] initial 0.150≈ 0 change − 3.6E−6 + 3.6E−6 equilibrium 0.153.6E-6 Tro, Chemistry: A Molecular Approach, 2/e

169 Copyright  2011 Pearson Education, Inc. Polyatomic Cations as Weak Acids Some cations can be thought of as the conjugate acid of a weak base others are the counter-ions of a strong base Therefore, some cations can potentially be acidic 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 counter-ion 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)  because NH 3 is a weak base, the position of this equilibrium favors the right 169 Tro, Chemistry: A Molecular Approach, 2/e

170 Copyright  2011 Pearson Education, Inc. Metal Cations as Weak Acids Cations of small, highly charged metals are weakly acidic alkali metal cations and alkali earth metal cations are 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) 170 Tro, Chemistry: A Molecular Approach, 2/e

171 Copyright  2011 Pearson Education, Inc. Example 15.15: Determine if the given cation Is acidic or neutral a) C 5 N 5 NH 2 + the conjugate acid of the weak base pyridine, therefore acidic b) Ca 2+ the counter-ion of the strong base Ca(OH) 2, therefore neutral c) Cr 3+ a highly charged metal ion, therefore acidic 171 Tro, Chemistry: A Molecular Approach, 2/e

172 Copyright  2011 Pearson Education, Inc. Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the counter-ion 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 counter-ion 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 2 172 Tro, Chemistry: A Molecular Approach, 2/e

173 Copyright  2011 Pearson Education, Inc. 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 ) 3 173 Tro, Chemistry: A Molecular Approach, 2/e

174 Copyright  2011 Pearson Education, Inc. 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 because 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 174 Tro, Chemistry: A Molecular Approach, 2/e

175 Copyright  2011 Pearson Education, Inc. Example 15.16: Determine whether a solution of the following salts is acidic, basic, or neutral a) SrCl 2 Sr 2+ is the counter-ion 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 175 Tro, Chemistry: A Molecular Approach, 2/e

176 Copyright  2011 Pearson Education, Inc. Example 15.16: Determine whether a solution of the following salts is acidic, basic, or neutral d) NaCHO 2 Na + is the counter-ion 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 176 Tro, Chemistry: A Molecular Approach, 2/e

177 Copyright  2011 Pearson Education, Inc. the solution is pH neutral Co 3+ is a highly charged cation, pH acidic Practice – Determine whether a solution of the following salts is acidic, basic, or neutral 177 KNO 3 CoCl 3 Ba(HCO 3 ) 2 CH 3 NH 3 NO 3 K + is the counter-ion of a strong base, pH neutral NO 3 − is the counter-ion of a strong acid, pH neutral Cl − is the counter-ion of a strong acid, pH neutral the solution is pH acidic Ba 2+ is the counter-ion of a strong base, pH neutral HCO 3 − is the conjugate of a weak acid, pH basic the solution is pH basic CH 3 NH 3 + is the conjugate of a weak base, pH acidic NO 3 − is the counter-ion of a strong acid, pH neutral the solution is pH acidic Tro, Chemistry: A Molecular Approach, 2/e

178 Copyright  2011 Pearson Education, Inc. Ionization in Polyprotic Acids Because 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 178 Tro, Chemistry: A Molecular Approach, 2/e

179 Copyright  2011 Pearson Education, Inc. 179 Tro, Chemistry: A Molecular Approach, 2/e

180 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? (K a1 = 4.3 x 10 −7, K a2 = 5.6 x 10 −11 ) 180 Tro, Chemistry: A Molecular Approach, 2/e

181 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 181 write the reactions for the acid with water one H at a time construct an ICE table for the reaction enter the initial concentrations – assuming the second ionization is negligible H 2 CO 3 + H 2 O  HCO 3  + H 3 O + [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change equilibrium HCO 3 − + H 2 O  CO 3 2− + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

182 Copyright  2011 Pearson Education, Inc. [HA][A − ][H 3 O + ] initial 0.12 00 change equilibrium Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 182 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 +x+x +x+x xx 0.12  x xx H 2 CO 3 + H 2 O  HCO 3  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

183 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? determine the value of K a1 because K a1 is very small, approximate the [HA] eq = [HA] init and solve for x [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012xx 0.12  x K a1 = 4.3 x 10 −7, K a2 = 5.6 x 10 −11 183 Tro, Chemistry: A Molecular Approach, 2/e

184 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 184 check if the approximation is valid by seeing if x < 5% of [H 2 CO 3 ] init K a1 for H 2 CO 3 = 4.3 x 10 −7 the approximation is valid x = 2.27 x 10 −4 [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.12 xx Tro, Chemistry: A Molecular Approach, 2/e

185 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 185 substitute x into the equilibrium concentration definitions and solve x = 2.3 x 10 −4 [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.12−x xx Tro, Chemistry: A Molecular Approach, 2/e

186 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 186 substitute [H 3 O + ] into the formula for pH and solve [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.120.00023 Tro, Chemistry: A Molecular Approach, 2/e

187 Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 187 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 within sig figs [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.120.00023 K a1 for H 2 CO 3 = 4.3 x 10 −7 Tro, Chemistry: A Molecular Approach, 2/e

188 Copyright  2011 Pearson Education, Inc. Practice – What is the [CO 3 2− ] in a 0.12 M solution of carbonic acid, H 2 CO 3 ? (K a1 = 4.3 x 10 −7, K a2 = 5.6 x 10 −11 ) 188 Tro, Chemistry: A Molecular Approach, 2/e

189 Copyright  2011 Pearson Education, Inc. Practice – What is the [CO 3 2− ] in a 0.12 M solution of carbonic acid, H 2 CO 3 ? 189 write the reactions for the acid with water one H at a time construct an ICE table for the reaction enter the initial concentrations for the second ionization using the equilibrium concentrations from first ionization H 2 CO 3 + H 2 O  HCO 3  + H 3 O + [HCO 3 − ][CO 3 2− ][H 3 O + ] initial 0.00023 0 change equilibrium HCO 3 − + H 2 O  CO 3 2− + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

190 Copyright  2011 Pearson Education, Inc. [HCO 3 − ][CO 3 2− ][H 3 O + ] initial 0.00023 0 change equilibrium Practice – What is the [CO 3 2− ] in a 0.12 M solution of carbonic acid, H 2 CO 3 ? 190 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 +x+x +x+x xx 2.3E−4  x x HCO 3 − + H 2 O  CO 3 2− + H 3 O + 2.3E−4 +x Tro, Chemistry: A Molecular Approach, 2/e

191 Copyright  2011 Pearson Education, Inc. Practice – What is the [CO 3 2− ] in a 0.12 M solution of carbonic acid, H 2 CO 3 ? determine the value of K a2 because K a2 is very small, approximate the [HA] eq = [HA] init, [H 3 O + ] eq = [H 3 O + ] init, and solve for x using this approximation, it is seen that x = K a2. Therefore [CO 3 2− ] = K a2 K a1 = 4.3 x 10 −7, K a2 = 5.6 x 10 −11 [HCO 3 − ][CO 3 2− ][H 3 O + ] initial 0.00023 0 change −x−x+x+x+x+x equilibrium 2.3E−4 − x x 2.3E−4 + x [HCO 3 − ][CO 3 2− ][H 3 O + ] initial 0.00023 0 change −x−x+x+x+x+x equilibrium 2.3E−4x 191 Tro, Chemistry: A Molecular Approach, 2/e

192 Copyright  2011 Pearson Education, Inc. Ionization in H 2 SO 4 The ionization constants for H 2 SO 4 are H 2 SO 4 + H 2 O  HSO 4  + H 3 O + strong HSO 4  + H 2 O  SO 4 2  + H 3 O + K a2 = 1.2 x 10 −2 For most sulfuric acid solutions, the second ionization is significant and must be accounted for Because the first ionization is complete, use the given [H 2 SO 4 ] = [HSO 4 − ] initial = [H 3 O + ] initial 192 Tro, Chemistry: A Molecular Approach, 2/e

193 Copyright  2011 Pearson Education, Inc. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C 193 write the reactions for the acid with water construct an ICE table for the second ionization reaction enter the initial concentrations – assuming the [HSO 4 − ] and [H 3 O + ] is ≈ [H 2 SO 4 ] [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change equilibrium HSO 4  + H 2 O  SO 4 2  + H 3 O + H 2 SO 4 + H 2 O  HSO 4  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

194 Copyright  2011 Pearson Education, Inc. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 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 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.0100 −x x Tro, Chemistry: A Molecular Approach, 2/e194

195 Copyright  2011 Pearson Education, Inc. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C expand and solve for x using the quadratic formula K a for HSO 4 − = 0.012 Tro, Chemistry: A Molecular Approach, 2/e195

196 Copyright  2011 Pearson Education, Inc. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C substitute x into the equilibrium concentration definitions and solve x = 0.0045 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.0100 −xx K a for HSO 4 − = 0.012 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.00550.00450.0145 Tro, Chemistry: A Molecular Approach, 2/e196

197 Copyright  2011 Pearson Education, Inc. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C substitute [H 3 O + ] into the formula for pH and solve K a for HSO 4 − = 0.012 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.00550.00450.0145 Tro, Chemistry: A Molecular Approach, 2/e197

198 Copyright  2011 Pearson Education, Inc. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C 198 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the answer matches K a for HSO 4 − = 0.012 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.00550.00450.0145 Tro, Chemistry: A Molecular Approach, 2/e

199 Copyright  2011 Pearson Education, Inc. Strengths of Binary Acids The more  + H─X  − 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 acidity: H─C < H─N < H─O < H─F Binary acid strength increases down the column acidity: H─F < H─Cl < H─Br < H─I 199 Tro, Chemistry: A Molecular Approach, 2/e

200 Copyright  2011 Pearson Education, Inc. Relationship between Bond Strength and Acidity Acid Bond Energy kJ/mol Type of Acid HF565weak HCl431strong HBr364strong 200 Tro, Chemistry: A Molecular Approach, 2/e

201 Copyright  2011 Pearson Education, Inc. Strengths of Oxyacids, H–O–Y The more electronegative the Y atom, the stronger the oxyacid HClO > HIO acidity of oxyacids decreases down a group  same trend as binary acids helps weakens the H–O bond The larger the oxidation number of the central atom, the stronger the oxyacid H 2 CO 3 > H 3 BO 3 acidity of oxyacids increases to the right across a period  opposite trend of binary acids The more oxygens attached to Y, the stronger the oxyacid further weakens and polarizes the H–O bond HClO 3 > HClO 2 201 Tro, Chemistry: A Molecular Approach, 2/e

202 Copyright  2011 Pearson Education, Inc. Relationship Between Electronegativity and Acidity Acid H─O─Y Electronegativity of Y KaKa H─O─Cl3.02.9 x 10 −8 H─O─Br2.82.0 x 10 −9 H─O─IH─O─I2.52.3 x 10 −11 202 Tro, Chemistry: A Molecular Approach, 2/e

203 Copyright  2011 Pearson Education, Inc. Relationship Between Number of Oxygens on the Central Atom and Acidity 203 Tro, Chemistry: A Molecular Approach, 2/e

204 Copyright  2011 Pearson Education, Inc. Practice – Order the Following By Acidity (Least to Most) H 3 PO 4 HNO 3 H 3 PO 3 H 3 AsO 3 By Acidity (Least to Most) HClHBrH 2 SHS − By Basicity (Least to Most) CO 3 2− NO 3 − HCO 3 − BO 3 3− Tro, Chemistry: A Molecular Approach, 2/e204

205 Copyright  2011 Pearson Education, Inc. Practice – Order the Following By Acidity (Least to Most) H 3 PO 4 HNO 3 H 3 PO 3 H 3 AsO 3 H 3 AsO 3 < H 3 PO 3 < H 3 PO 4 < HNO 3 By Acidity (Least to Most) HClHBrH 2 SHS − HS − < H 2 S < HCl < HBr By Basicity (Least to Most) CO 3 2− NO 3 − HCO 3 − BO 3 3− NO 3 − < HCO 3 − < CO 3 2− < BO 3 3− Tro, Chemistry: A Molecular Approach, 2/e205

206 Copyright  2011 Pearson Education, Inc. Lewis Acid–Base Theory Lewis Acid–Base theory focuses on transferring an electron pair lone pair  bond bond  lone pair Does NOT require H atoms The electron donor is called the Lewis Base electron rich, therefore nucleophile The electron acceptor is called the Lewis Acid electron deficient, therefore electrophile 206 Tro, Chemistry: A Molecular Approach, 2/e

207 Copyright  2011 Pearson Education, Inc. Lewis Bases Lewis Base has electrons it is willing to give away to or share with another atom Lewis Base must have lone pair of electrons on it that it can donate Anions are better Lewis Bases than neutral atoms or molecules N: < N: − Generally, the more electronegative an atom, the less willing it is to be a Lewis Base O: < S: 207 Tro, Chemistry: A Molecular Approach, 2/e

208 Copyright  2011 Pearson Education, Inc. Lewis Acids Electron deficient, either from being attached to electronegative atom(s) not having an octet Must have empty orbital willing to accept the electron pair H + has empty 1s orbital B in BF 3 has empty 2p orbital and an incomplete octet Many small, highly charged metal cations have empty orbitals they can use to accept electrons Atoms that are attached to highly electronegative atoms and have multiple bonds can be Lewis Acids 208 Tro, Chemistry: A Molecular Approach, 2/e

209 Copyright  2011 Pearson Education, Inc. Lewis Acid–Base Reactions The base donates a pair of electrons to the acid Generally results in a covalent bond forming H 3 N: + BF 3  H 3 N─BF 3 The product that forms is called an adduct Arrhenius and Brønsted-Lowry acid–base reactions are also Lewis 209 Tro, Chemistry: A Molecular Approach, 2/e

210 Copyright  2011 Pearson Education, Inc. Examples of Lewis Acid–Base Reactions 210 Tro, Chemistry: A Molecular Approach, 2/e

211 Copyright  2011 Pearson Education, Inc. Examples of Lewis Acid–Base Reactions Ag + (aq) + 2 :NH 3(aq)  Ag(NH 3 ) 2 + (aq) Lewis Acid Lewis Base Adduct 211 Tro, Chemistry: A Molecular Approach, 2/e

212 Copyright  2011 Pearson Education, Inc. Practice – Identify the Lewis Acid and Lewis Base in Each Reaction 212 Lewis Base Lewis Base Lewis Acid Lewis Acid Tro, Chemistry: A Molecular Approach, 2/e

213 Copyright  2011 Pearson Education, Inc. U.S. Fuel Consumption Over 85% of the energy use in the United States comes from the combustion of fossil fuels oil, natural gas, coal Combustion of fossil fuels produces CO 2 CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(g) Natural fossil fuels also contain small amounts of S that burn to produce SO 2 (g) S(s) + O 2 (g) → SO 2 (g) The high temperatures of combustion allow N 2 (g) in the air to combine with O 2 (g) to form oxides of nitrogen N 2 (g) + 2 O 2 (g) → 2 NO 2 (g) 213 Tro, Chemistry: A Molecular Approach, 2/e

214 Copyright  2011 Pearson Education, Inc. 214 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, 2/e

215 Copyright  2011 Pearson Education, Inc. 215 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 (g) + H 2 O(l)  H 2 CO 3 (aq) 2 SO 2 (g) + O 2 (g) + 2 H 2 O(l)  2 H 2 SO 4 (aq) 4 NO 2 (g) + O 2 (g) + 2 H 2 O(l)  4 HNO 3 (aq) 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 Tro, Chemistry: A Molecular Approach, 2/e

216 Copyright  2011 Pearson Education, Inc. pH of Rain in Different Regions Tro, Chemistry: A Molecular Approach, 2/e216

217 Copyright  2011 Pearson Education, Inc. Weather Patterns The prevailing winds in the United States travel west to east Weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced Much of the northeast United States has rain of very low pH, even though it has very low sulfur emissions, due in part to the general weather patterns 217 Tro, Chemistry: A Molecular Approach, 2/e

218 Copyright  2011 Pearson Education, Inc. Sources of SO 2 from Utilities Tro, Chemistry: A Molecular Approach, 2/e 218

219 Copyright  2011 Pearson Education, Inc. 219 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 affects aquatic life Soil acidity causes more dissolving of minerals and leaching more minerals from soil making it difficult for trees Tro, Chemistry: A Molecular Approach, 2/e

220 Copyright  2011 Pearson Education, Inc. 220 Damage from Acid Rain Tro, Chemistry: A Molecular Approach, 2/e

221 Copyright  2011 Pearson Education, Inc. 221 Acid Rain Legislation 1990 Clean Air Act attacks acid rain forces utilities to reduce SO 2 Result is acid rain in the Northeast stabilized and beginning to be reduced Tro, Chemistry: A Molecular Approach, 2/e


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