1 Chapter 16: Acids, Bases, and Salts Key topics: ph scale; acid-base properties of water K a = acid ionization constant; K b = base ionization constant Polyprotic acids BrØnsted Acids and Bases Acid: Base: proton donor proton acceptor acid/base reaction: proton transfer Each side of the reaction has an acid and a base! Conjugate base: what remains after the acid donates a proton Conjugate acid: formed when the base accepts a proton
2 an acid and its conjugate base: a base and its conjugate acid: conjugate acid-base pair conjugate acid-base pair The two species in a conjugate pair differ only by H + There are two conjugate pairs in any acid-base reaction e.g., HSO 3 is the conjugate acid of what species? HSO 3 is the conjugate base of what species? Solution: SO 3 2, H 2 SO 3 e.g., (a) Write an equation in which HSO 4 reacts (with water) to form its conjugate base. (b) Write an equation in which HSO 4 reacts (with water) to form its conjugate acid. Solution: HSO 4 + H 2 O SO H 3 O + HSO 4 + H 2 O H 2 SO 4 + OH
3 Molecular Structure and Acid Strength The strength of an acid is measured by how easily it ionizes: HX! H + +X or HX + H 2 O! H 3 O + +X The two factors that matter are: 1. Strength of the H X bond 2. Polarity of the H X bond [show movie] Hydrohalic Acids: binary acids containing H and a halogen. o Strength increases left to right (same row) because the electronegativity of X increases (H X bond more polar) o Strength increases top to bottom because the H X bond strength decreases as we go down the group
4 Oxoacids (O H bond): contain H, O, and a central non-metal Acidity determined by the polarity of the O H bond: 1. central atom from the same group of the periodic table, same oxidation number Relative strength increases with increased electronegativity of the central atom. HClO 3 > HBrO 3 H 2 SO 4 > H 2 SeO 4 2. same central atom, different oxidation number (# of O s) Higher oxidation number = stronger acid. e.g., H 2 SO 4 > H 2 SO 3 e.g., HClO 4 > HClO 3 > HClO 2 > HClO
5 e.g., CH 3 COOH > CH 3 CH 2 OH 3. different central atom, different oxidation number The stronger the acid, the weaker the conjugate base. Compare sulfuric and phosphoric acid: H 2 SO 4 HSO 4 + H + H 3 PO 4 H 2 PO 4 + H + The lone oxygen atoms in HSO 4 attract H + less, making it a weaker base. Therefore sulfuric acid is the stronger acid.
6 Carboxylic acids ( COOH group): R COOH R COO + H + The carboxylate anion is resonance stabilized. The more electronegative the R group, the stronger the acid because it increases the polarity of the O-H bond. e.g., Arrange the following organic acids in order of increasing strength: bromoacetic acid (CH 2 BrCOOH), chloroacetic acid (CH 2 ClCOOH), fluoroacetic acid(ch 2 FCOOH), iodoacetic acid (CH 2 ICOOH). Answer: bond polarity is what matters here, so fluoroacetic acid is the strongest, iodoacetic acid the weakest.
7 The Acid-Base Properties of Water Water is amphoteric: it can act as an acid or a base. Water also autoionizes to a very small degree The equilibrium constant K w does not include the liquid, so K w =[H + ][OH ] or K w =[H 3 O + ][OH ] at 25 C, [H + ]= =[OH ] ) K w = For any temperature, pure water is neutral ([OH ] = [H + ]) but the concentrations vary slightly which is reflected in K w. For dilute aqueous solutions, K w still applies. If [H + ] = [OH ] the solution is neutral If [H + ] > [OH ] the solution is acidic If [H + ] < [OH ] the solution is basic e.g., at 25ºC, [H + ] = 4.6 x 10-8 M. Find [OH ]. Is the solution acidic or basic or neutral?
8 Solution: [H + ][OH ]= ) [OH ]= Therefore the solution is basic = The ph and poh Scales ph = log[h + ] and poh = log[oh ] and px = log X K w =[H 3 O + ][OH ] ) log K w = log[h 3 O + ] + log[oh ] ) pk w = ph + poh (= 14 at 25 C) For pure water at 25 C, ph = poh = 7 (neutral). When ph < 7, poh > 7 and the solution is acidic. When ph > 7, poh < 7 and the solution is basic. (from chemwiki.ucdavis.edu)
9 Strong Acids and Bases Strong acids and bases ionize / dissociate completely. Therefore the ph is easy to calculate: the concentration of [H + ] or [OH ] is just given by the starting concentration of the strong acid or base, and the stoichiometry. e.g., HCl H + + Cl Given 0.1 M HCl, the ph is log(0.1) = 1. e.g., Ba(OH) 2 Ba OH Given 0.1 M Ba(OH) 2, the poh is log(0.2) = 0.7 e.g., H 2 SO 4 H + + HSO 4 We will do this later in the chapter. For now, note that H 2 SO 4 is a strong acid while HSO 4 is a weak acid. List of strong acids.
10 List of strong bases. Note: the Group 2A hydroxides are completely dissociated in solutions of 0.01 M or less. These are insoluble bases which ionize 100% (the tiny amount that is soluble dissociates completely). e.g., An aqueous solution of a strong base has ph at 25 C. Calculate the concentration of base in the solution (a) if the base is NaOH and (b) if the base is Ba(OH) 2. Answer: ph = means that poh = = 1.76 Therefore [OH ] = = With NaOH, we must have [NaOH] = M With Ba(OH) 2, we have [Ba(OH) 2 ] = (0.017 / 2) = 8.7 x 10 3 M
11 Weak Acids and Acid Ionization Constants Weak acids do not fully dissociate in water. The extent to which a weak acid ionizes in water depends on 1. the concentration of the acid 2. the equilibrium constant for the ionization reaction, K a We will only consider monoprotic acids in this section: HA(aq) H + (aq)+a (aq) K a = [H+ ][A ] [HA] K a is called the acid ionization constant. pk a = -log K a The larger the value of K a, the stronger the acid. We can calculate ph from K a, or K a from ph. The math is simple if: [H + ] eq < (0.05) [HA] initial or [HA] initial > 400 K a Otherwise, we will need the quadratic formula or the method of successive approximations.
12 e.g., ph of a 0.50 M HF solution at 25 C? K a = 7.1 x 10-4 Answer: 400 K a = Therefore we can make the math simple. Set up an ICE table: HF H + + F Initial * 0 Change -x +x +x Equilibrium x x x * we have neglected the auto-ionization of water because it will have a negligible effect (x will be much larger than 10-7 )
13 At equilibrium we have x 2 K a = 0.50 x This is a quadratic equation in x. The simple math is to put (0.50 x) 0.50 Then we have K a = x ) x2 =(0.50)( ) ) x =0.019 The ph is then = -log(0.019) = 1.72 (redo with the) Quadratic formula approach: We rewrite K x 2 a = 0.50 x as x 2 + K a x (0.50)K a =0 and use the quadratic formula (page A-4) to write x =( K a ± p Ka 2 + 4(0.5)K a )/2 = This is more accurate because it gives K a = (0.0185)2 4 = whereas if we use x = we get K a = 7.5 x 10-4 (redo with the) Successive Approximations approach (page A-4): This approach is more accurate than the simple approach but less complicated than using the quadratic formula. We begin like the simple approach: Starting from x 2 K a = 0.50 x we put (0.50 x) 0.50 to get K a = x ) x2 =(0.50)( ) ) x =0.019
14 Next, we use this value of x in the (0.50 x) part of K a, giving x 2 K a = = x ) x = Then we do it again: x 2 K a = = x ) x = We stop when the answer stops changing. (The simple approach was good enough in this case because the simple math assumptions held.) Percent Ionization For a monoprotic acid (HA), percent ionization = [H+ ] eq [HA] 0 100% A weak acid ionizes more the weaker its concentration. (data for acetic acid)
15 e.g., Determine the ph and percent ionization for acetic acid solutions at 25ºC with concentrations of (a) 0.15 M, (b) M, (c) M (K a = 1.8 x 10-5 ) Solution: (a) We begin with an ICE table: CH3COOH H + + CH3COO Initial Change -x +x +x Equilibrium x x x The equilibrium expression is K a = x x Now test for math: 400 K a = < 0.15 So we can do 0.15 x 0.15 and get x 2 =(0.15)K a ) x = M ) ph = 2.78 with % ionization = 1.07 % (b) We still have 400 K a = < so we get x 2 =(0.015)K a ) x = M ) ph = 3.28 with % ionization = 3.47 % (c) The simple math fails because 400 K a = > K a = x x we will use successive approximations:
16 First we set x to get x 2 =(0.0015)K a ) x = M Then we put this value of x in the denominator: x 2 K a = = x which gives x = 1.55 x We repeat to get x 2 K a = = x which gives x = 1.56 x 10 4 The answer has stopped changing so we are done. This results in x = 1.56 x 10 4 M with ph = 3.81 and % ionization = 10.4 % Using ph to Determine K a We can use the ph of a weak acid to determine the equilibrium concentrations, which gives us the equilibrium constant. e.g., Calculate the K a of a weak acid if a M solution of the acid has a ph of 5.03 at 25ºC. Solution: ph = 5.03 = log[h + ] ) [H + ] = = M HA H + + A Initial Change x x x 10 6 Equilibrium x x 10 6 K a = ( ) =
17 Weak Bases and Base Ionization Constants To define the base ionization constant K b we write: B(aq) + H2O(l) HB + (aq) + OH (aq) K b = [HB+ ][OH ] [B] where HB + is the conjugate acid of base B All these Bronsted bases have a lone pair of electrons on a nitrogen atom which can accept a proton.
18 Calculating ph from K b We proceed as for a weak acid, using ph + poh = 14. e.g., Calculate the ph at 25ºC of a 0.16 M solution of a weak base with a K b of 2.9 x Solution: Since K b is so small we can make the math simple. B (+ H 2 O) HB + + OH Initial Change -x +x +x Equilibrium x x x K b = [HB+ ][OH ] [B] = x x x which gives x 2 = (0.16)K b or x = 2.15 x 10 6 poh = 5.67 Therefore ph = = 8.33 Using ph to Determine K b This is very similar the weak acid calculation. e.g., Determine the K b of a weak base if a 0.35 M solution of the base has a ph of at 25ºC. Solution: poh = 14 ph = 2.16 [OH ] = = 6.92 x 10 3 B (+ H 2 O) HB + + OH Initial Change x x x 10 3 Equilibrium x x 10 3 K b = ( ) =
19 Conjugate Acid-Base Pairs Consider dissolving NaCl in water. Since Cl is the conjugate base of a strong acid (HCl), it is called a weak conjugate base and it will not generate HCl: Cl +H 2 O 6! HCl + OH Next consider dissolving NaF in water. Since F is the conjugate base of a weak acid (HF), it is called a strong conjugate base and it will generate HF: F +H 2 O HF + OH Let us examine HF in more detail: F +H 2 O HF + OH K b = [HF][OH ] [F ] HF + H 2 O F +H 3 O + K a = [F ][H 3O + ] [HF] Add these two together: 2H 2 O OH +H 3 O + K = K a K b =[OH ][H 3 O + ]=K w Therefore pk a +pk b =pk w = 14 (at 25 C) This is the conjugate acid-base pair relationship. e.g., Determine (a) K b of the weak base B whose conjugate acid HB + has K a = 8.9 x 10 4 and (b) K a of the weak acid HA whose conjugate base has K b = 2.1 x Solution: We just use K a x K b = K w = 1.00 x (a) K b = K w / K a = 1.00 x / 8.9 x 10 4 = 1.12 x (b) K a = K w / K b = 1.00 x / 2.1 x 10 8 = 4.76 x 10 7
20 Diprotic and Polyprotic Acids Some acids can undergo two or three ionizations: (from Note that the conjugate base in the first ionization serves as the acid in the second ionization. The second ionization constant is very small compared to the first one (usually at least 1000x smaller). This is because it is harder to remove a proton (H + ) from a negatively charged species. This lets us make some simple approximations. e.g., Calculate the concentrations of H 2 SO 4, HSO 4, SO 4 2, and H + ions in a 0.14 M sulfuric acid solution at 25ºC. K a1 = very large; K a2 = 1.3 x 10 2 = 0.013
21 Solution: There are two ICE tables to set up: H 2 SO 4 H+ + HSO 4 Initial Change -x +x +x Equilibrium 0.14-x x x HSO 4 H+ + 2 SO 4 Initial x x 0 Change -y +y +y Equilibrium x-y x+y y First, (0.14 x) 0 since sulfuric acid is a strong acid. This gives us x = Now, (400)K a2 = 5.2 > 0.14; thus the simple math fails. We can use the quadratic equation formula: K a2 = [H+ ][SO 2 4 ] ( y)(y) = [HSO 4 ] 0.14 y y y = 0 ) y =( ± p (0.153) 2 + 4( ))/2 which gives y = Therefore at equilibrium, [H 2 SO 4 ] = 0, [HSO 4 ] = 0.13 M, [SO 4 2 ] = M, and [H + ] = 0.15 M
22 We can also use the successive approximations method: We begin with the equilibrium expression K a2 = [H+ ][SO 2 4 ] ( y)(y) = [HSO 4 ] 0.14 y and put ( y) 0.14 (0.14 y) 0.14 K a2 = y ) y =0.013 Then we put ( y) ( ) and (0.14 y) ( ) K a2 = y ) y = Then we put ( y) ( ) and (0.14 y) ( ) K a2 = y ) y = which is identical to the quadratic formula.
24 Acid-Base Properties of Salt Solutions When salts dissolve in water, the resulting anions and cations may be able to act as acids or bases. We already considered what happens when we dissolve NaF in water. Since F is the conjugate base of a weak acid (HF), it will generate HF and result in a solution with a basic ph: F +H 2 O HF + OH How basic is the solution? We can use the K a of the acid: e.g., Calculate the ph of a 0.10 M solution of NaF at 25ºC. Solution: K a for HF = 7.1 x 10 4, so K b for F is / K a = 1.41 x F (+ H 2 O) HF + OH Initial Change -x +x +x Equilibrium 0.10-x x x Since K b is so small, we can write (0.10 x) 0.10, giving K b = x 2 /0.10 which yields x = 1.2 x Since this is close to the concentration of OH in pure water, we can add the pure water value into the result: [OH ] = 1.2 x x 10 7 = 1.3 x Then the ph = 14 poh = (or if we didn t add the pure water OH concentration, ph = 8.08)
25 On the other hand, when the cation of a salt is the conjugate acid of a weak base, the salt solution will be acidic. For example, NH 4 Cl(s) H 2O NH + 4 (aq)+cl (aq) NH 3(aq)+H + (aq)+cl (aq) Remember that Cl is the conjugate base of a strong acid so it will not generate HCl. e.g., Determine the concentration of a solution of ammonium chloride that has a ph of 5.37 at 25ºC. Solution: Since Cl is a spectator ion, the reaction is fundamentally just NH + 4 NH 3 +H + Also, ph = 5.37 [H + ] = 4.27 x NH 4 NH 3 + H + Initial y 0 0 Change -x +x +x Equilibrium y-x 4.27 x x 10 6 We have x = 4.27 x Also, K a for NH 4 + is K w /K b for NH 3, which is K a = / (1.8 x 10 5 ) = 5.6 x Therefore we have = ( ) 2 y ) y =0.0326
26 Hydrated metal ions can make a solution acidic because they polarize the OH bond of water molecules, making it easier for H+ to ionize. For this to happen we need a metal ion of high charge density. o Group 1A metals: non-acids o Group 2A metals except Be 2+ : non-acids o all other metals: weak acids The acidity increases as the metal size decreases and the charge increases. In general: A cation that will make a solution acidic is the conjugate acid of a weak base NH + 4, CH 3NH + 3, C 2H 5 NH + 3 a small highly charged metal ion Al 3+, Cr 3+, Fe 3+, Bi 3+ An anion that will make a solution basic is the conjugate base of a weak acid CN, NO 2, CH 3 COO
27 A cation that will not affect the ph is a group 1A or heavy group 2A cation Li +, Na +, Ba 2+ An anion that will not affect the ph is the conjugate base of a strong acid Cl, NO 3, ClO 4 Salts in which both the cation and the anion hydrolyze The ph depends on the relative strength of the conjugate acid and the conjugate base: o when K b > K a the solution is basic o when K b < K a the solution is acidic o when K b = K a the solution is neutral e.g., NH 4 NO 2 dissociates to give NH 4 + (K a = 5.6 x ) and NO 2 (K b = 2.2 x ). Because K a for the ammonium ion > K b for the nitrite ion, the ph will be slightly acidic. Acid-Base Properties of Oxides and Hydroxides Alkali metal oxides and hydroxides and alkali earth metal oxides and hydroxides (except BeO) are basic Na 2 O(s)+H 2 O(l) 2NaOH(aq) BaO(s)+H 2 O(l) Ba(OH) 2 (aq) The non-metal oxides tend to be acidic CO 2 (g)+h 2 O(l) H 2 CO 3 (aq) N 2 O 5 (g)+h 2 O(l) 2HNO 3 (aq)
28 A few oxides are amphoteric: they can act as an acid or a base. For example, aluminum oxide can neutralize either: Al2 O3 (s) + 6HCl(aq) 2AlCl3 (aq) + 3H2 O(l) Al2 O3 (s) + 2NaOH(aq) + 3H2 O(l) 2NaAl(OH)4 (aq) Oxides of the main group elements in their highest oxidation states. Be(OH)2, Al(OH)3, Sn(OH)2, Pb(OH)2, Cr(OH)3, Cu(OH)2, Zn(OH)2, and Cd(OH)2, are amphoteric and insoluble in water but they can neutralize both acids and bases: Be(OH)2 (s) + 2H+ (aq) Be2+ (aq) + 2H2 O(l) Be(OH)2 (s) + 2OH (aq) Be(OH)24 (aq)
29 Lewis Acids and Bases Lewis acids and bases generalize the concept of what it means to be an acid or a base. A Lewis base is a substance that can donate a pair of electrons. A Lewis acid is a substance that can accept a pair of electrons. Every Lewis acid-base reaction results in the formation of a coordinate covalent bond (both electrons donated by the same species). Lewis acid-base reactions are different from redox reactions because a physical transfer of one or more electrons from donor to acceptor does not occur. BF 3 (Lewis acid) and NH 3 (Lewis base) react to form an adduct compound.
30 Simple acid-base neutralization. (from chemwiki.ucdavis.edu) F acts as a Lewis base. (from chemwiki.ucdavis.edu) An electron-donating species forms a coordination complex with a transition-metal ion. (from chemwiki.ucdavis.edu)
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Aqueous Ions and Reactions (ions, acids, and bases) Demo NaCl(aq) + AgNO 3 (aq) AgCl (s) Two clear and colorless solutions turn to a cloudy white when mixed Demo Special Light bulb in water can test for
WEAK ACIDS AND BASES [MH5; Chapter 13] Recall that a strong acid or base is one which completely ionizes in water... In contrast a weak acid or base is only partially ionized in aqueous solution... The
Bronsted Lowry Acid Base Chemistry Just a few reminders What makes an acid an acid? A BronstedLowry Acid is a compound that donates a proton (a hydrogen ion with a positive charge, H + ) o Think of the
ph: Measurement and Uses One of the most important properties of aqueous solutions is the concentration of hydrogen ion. The concentration of H + (or H 3 O + ) affects the solubility of inorganic and organic
Dr Mike Lyons School of Chemistry Trinity College Dublin. email@example.com ACID-BASE REACTIONS/ THE PH CONCEPT. Chemistry Preliminary Course 2011 1 Lecture topics. 2 lectures dealing with some core chemistry
Acids, Bases & Redox 1 Practice Problems for Assignment 8 1. A substance which produces OH ions in solution is a definition for which of the following? (a) an Arrhenius acid (b) an Arrhenius base (c) a
4. Acid Base Chemistry 4.1. Terminology: 4.1.1. Bronsted / Lowry Acid: "An acid is a substance which can donate a hydrogen ion (H+) or a proton, while a base is a substance that accepts a proton. B + HA
Acids and Bases Chapter 16 The Arrhenius Model An acid is any substance that produces hydrogen ions, H +, in an aqueous solution. Example: when hydrogen chloride gas is dissolved in water, the following
Name: Class: Date: Chapter 13 & 14 Practice Exam Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Acids generally release H 2 gas when they react with a.
Chemical Reactions in Water Ron Robertson r2 f:\files\courses\1110-20\2010 possible slides for web\waterchemtrans.doc Properties of Compounds in Water Electrolytes and nonelectrolytes Water soluble compounds
Chapter 17 2) a) HCl and CH 3 COOH are both acids. A buffer must have an acid/base conjugate pair. b) NaH 2 PO 4 and Na 2 HPO 4 are an acid/base conjugate pair. They will make an excellent buffer. c) H
AP Chemistry A. Allan Chapter 4 Notes - Types of Chemical Reactions and Solution Chemistry 4.1 Water, the Common Solvent A. Structure of water 1. Oxygen's electronegativity is high (3.5) and hydrogen's
UNIT (6) ACIDS AND BASES 6.1 Arrhenius Definition of Acids and Bases Definitions for acids and bases were proposed by the Swedish chemist Savante Arrhenius in 1884. Acids were defined as compounds that
CHEM 101/105 Aqueous Solutions (continued) Lect-07 aqueous acid/base reactions a. a little bit more about water Water is a polar substance. This means water is able to "solvate" ions rather well. Another
John W. Moore Conrad L. Stanitski Peter C. Jurs http://academic.cengage.com/chemistry/moore Chapter 15 Acids and Bases Arrhenius Definition Arrhenius: any substance which ionizes in water to produce: Protons
Acid/Base Basics How does one define acids and bases? In chemistry, acids and bases have been defined differently by two sets of theories. One is the Arrhenius definition, which revolves around the idea
CHEM 102: Sample Test 5 CHAPTER 17 1. When H 2 SO 4 is dissolved in water, which species would be found in the water at equilibrium in measurable amounts? a. H 2 SO 4 b. H 3 SO + 4 c. HSO 4 d. SO 2 4 e.
Review for Solving ph Problems: Acid Ionization: HA H 2 O A - H 3 O CH 3 COOH H 2 O CH 3 COO - H 3 O Base Ionization: B H 2 O BH OH - 1) Strong Acid complete dissociation [H ] is equal to original [HA]