Chapter 7 Acids and Bases

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1 Chapter 7 Acids and Bases 7.1 The Nature of Acids and Bases 7.2 Acid Strength 7.3 The ph Scale 7.4 Calculating the ph of Strong Acid Solutions Midterm Exam Calculating the ph of Weak Acid Solutions 7.6 Bases 7.7 Polyprotic Acids 7.8 Acid-Base Properties of Salts 7.9 Acid Solutions in Which Water Contributes to the H + Concentration 7.10 Strong Acid Solutions in Which Water Contributes to the H + Concentration 7.11 Strategy for solving Acid-Base Problems: A Summary 1 Arrhenius (or Classical) Acid-Base Definition An acid is a substance that contains hydrogen and dissociates in water to yield a hydronium ion: H 3 O + (H + ) A base is a substance that contains the hydroxyl group and dissociates in water to yield a hydroxide ion: OH - Neutralization is the reaction of an H + (H 3 O + ) ion from the acid and the OH - ion from the base to form water, H 2 O. An indicator is a substance that changes color with acidity. The neutralization reaction is exothermic and releases approximately 56 kj per mole of acid and base that react. H + (aq) + OH- (aq) H 2 O (l) + 56 kj/mol 2 1

2 Brønsted-Lowry Acid-Base Definition An acid is a proton donor, any species that donates an H + ion. An acid must contain H in its formula; HNO 3 and H 2 PO 4 - are two examples; all Arrhenius acids are Brønsted-Lowry acids. A base is a proton acceptor, any species that accepts an H + ion. A base must contain a lone pair of electrons to bind the H + ion; a few examples are NH 3, CO 3 2-, F -, as well as OH -. Brønsted-Lowry bases are not Arrhenius bases, but all Arrhenius bases produce the Brønsted-Lowry base OH -. Therefore, in the Brønsted-Lowry perspective, one species donates a proton and another species accepts it: an acid-base reaction is a proton transfer process. Acids donate a proton to water Bases accept a proton from water 3 Acids donate a proton to water Bases accept a proton from water Water molecules can behave as a base or as an acid! 4 2

3 Identify the Brønsted acids and bases in the following equation (A = Brønsted acid, B = Brønsted base): HPO HSO 4 H 2 PO 4 + SO B, A, B, A 2. B, A, A, B 3. A, B, A, B 4. A, B, B, A 5 Molecular Model of Conjugate Acids & Bases The reaction of an acid HA with water to form H 3 O+ and a conjugate base HA A Acid Base Conjugate Conjugate acid base Conjugate acid is formed when a proton is transferred to the base. Conjugate base is everything that remains of the acid molecule after a proton is lost. Conjugated acid-base pair consists of two substances related to each other by donating and accepting a single proton. 6 3

4 The Conjugate Pairs in Some Acid-Base Reactions Conjugate Pair Acid + Base Base + Acid Conjugate Pair Reaction 1 HF + H 2 O F + H 3 O + Reaction 2 HCOOH + CN HCOO + HCN Reaction 3 NH CO 2 3 NH 3 + HCO 3 Reaction 4 H 2 PO 4 + OH HPO H 2 O Reaction 5 H 2 SO 4 + N 2 H + 5 HSO 4 + N 2 H 2+ 6 Reaction 6 HPO SO 3 2 PO HSO 3 7 Identifying Conjugate Acid-Base Pairs Problem: The following chemical reactions are important for industrial processes. Identify the conjugate acid-base pairs. (a) HSO - 4 (aq) + CN - (aq) SO 4 2- (aq) + HCN (aq) (b) ClO - (aq) + H 2 O (l) HClO (aq) + OH- (aq) (c) S 2- (aq) + H 2 O (aq) HS- (aq) + OH- (aq) Plan: To find the conjugate acid-base pair, we find the species that donates H + and the one that accepts it. The acid (or base) on the left becomes the conjugate base (or acid) on the right. Solution: (a) The proton is transferred from the sulfate to the cyanide so: HSO 4 - (aq) /SO 4 2- (aq) and CN - (aq) /HCN (aq ) (b) The water gives up one proton to the hypochlorite anion so: ClO - (aq) /HClO (aq) and H 2 O (l) / OH- (aq ) (c) One of water s protons is transferred to the sulfide ion so: S 2- (aq) /HS- (aq) and H 2 O (l) /OH- (aq) 8 4

5 The conjugate acid and conjugate base of bicarbonate ion (HCO 3 ) are, respectively, 1. H 3 O + and OH 2. H 3 O + and CO H 2 CO 3 and OH 4. H 2 CO 3 and CO CO 2 3 and OH 9 The Acid-Dissociation Constant (K a ) Strong acids dissociate completely in water to form ions: HA (aq) + H 2 O (l) H 3 O + (aq) + A- (aq) In a dilute solution of a strong acid, almost no HA molecules exist: [H 3 O + ] = [HA] init or [HA] eq = 0 K = K a = [H 3 O+ ][A - ] [HA] at equilibrium, and K a >> 1 Nitric acid is an example: HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) Weak acids dissociate very slightly into ions in water: HA (aq) + H 2 O (l) H 3 O + (aq) + A- (aq) In a dilute solution of a weak acid, the great majority of HA molecules are not dissociated: [H 3 O + ] << [HA] init or [HA] eq = [HA] init K = K a = [H 3 O + ][A - ] [HA] at equilibrium, and K a <<

6 The Meaning of K a, the Acid Dissociation Constant For the ionization of an acid, HA: HA (aq) + H 2 O (l) H 3 O + (aq) + A- (aq) Therefore: K a = [H 3 O+ ] [A - ] [HA] The stronger the acid, the higher the [H 3 O + ] at equilibrium, and the larger the K a : Stronger acid higher [H 3 O + ] larger K a For a weak acid with a relative high K a (~10-2 ), a 1 M solution has ~10% of the HA molecules dissociated. For a weak acid with a moderate K a (~10-5 ), a 1 M solution has ~ 0.3% of the HA molecules dissociated. For a weak acid with a relatively low K a (~10-10 ), a 1 M solution has ~ 0.001% of the HA molecules dissociated. 11 The Extent of Dissociation for Strong and Weak Acids 12 6

7 Review Acids Selected Acids and Bases Bases Strong: H + (aq) + A - (aq) Strong: M + (aq) + OH - (aq) hydrochloric, HCl lithium hydroxide, LiOH hydrobromic, HBr sodium hydroxide, NaOH hydroiodoic, HI potassium hydroxide, KOH nitric acid, HNO 3 calcium hydroxide, Ca(OH) 2 sulfuric acid, H 2 SO 4 strontium hydroxide, Sr(OH) 2 perchloric acid, HClO 4 barium hydroxide, Ba(OH) 2 *(M is Group I or II metal) Weak Weak hydrofluoric, HF ammonia, NH 3 phosphoric acid, H 3 PO 4 accepts proton from water to make acetic acid, CH 3 COOH NH + 4 (aq) and OH - (aq) (or HC 2 H 3 O 2 ) 13 Classifying the Relative Strengths of Acids and Bases I Strong acids. There are two types of strong acids: 1. The hydrohalic acids: HCl, HBr, and HI 2. Oxyacids in which the number of O atoms exceeds the number of ionizable H atoms by two or more, such as HNO 3, H 2 SO 4, HClO 4 Weak acids. There are many more weak acids than strong ones. Four types, with examples, are: 1. The hydrohalic acid, HF 2. Those acids in which H is NOT bound to O or to a halogen, such as HCN and H 2 S 3. Oxyacids in which the number of O atoms equals or exceeds by one the number of ionizable H atoms, such as HClO, HNO 2, and H 3 PO 4 4. Organic acids (general formula RCOOH), such as CH 3 COOH and C 6 H 5 COOH 14 7

8 Classifying the Relative Strengths of Acids and Bases - II Strong bases. Soluble compounds containing O 2- or OH - ions are strong bases. The cations are usually those of the most active metals: 1) M 2 O or MOH, where M= Group 1A(1) metals (Li, Na, K, Rb, Cs) 2) MO or M(OH) 2, where M = Group 2A(2) metals (Ca, Sr, Ba) (MgO and Mg(OH) 2 are only slightly soluble, but the soluble portion dissociates completely.) Weak bases. Many compounds with an electron-rich nitrogen are weak bases (none are Arrhenius bases). The common structural feature is an N atom that has a lone electron pair in its Lewis structure. 1) Ammonia (:NH 3 ) 2) Amines (general formula RNH 2, R 2 NH, R 3 N), such as CH 3 CH 2 NH 2, (CH 3 ) 2 NH, (C 3 H 7 ) 3 N, and C 5 H 5 N : : : : : : :

9 Relative strengths of acid/conjugate base partners Examples: HCl and Cl - CH 3 COOH and CH 3 COO - HCN and CN

10 Trends in Acid/Base Strength Among Proton Donors Strong acid Conj. acid of weak base Water Conj. acid of strong base Higher K a, stronger acid Strength Among Proton Acceptors Strong base Conj. base of weak acid Water Conj. base of strong acid Higher K b, stronger base How do we know for sure? Compare the K a and K b values

11 OXYACIDS most acids are oxyacids acidic proton is attached to oxygen! organic acids contain carboxyl (C-O) group organic acids are weak acids many inorganic acids are strong acids acids can be mono- or polyprotic monoprotic acid dissociates one proton polyprotic acid dissociates several protons

12 23 Acid and Base Character and the ph Scale To handle the very large variations in the concentrations of the hydrogen ion in aqueous solutions, a scale called the ph scale is used for which: ph = - log [H + ] What is the ph of a solution that is 10-2 M in hydronium ion? ph = -log[h + ] = (-1)log 10-2 = 2 ph of a neutral solution = 7.00 ph of an acidic solution < 7.00 ph of a basic solution >

13 Figure 7.3: The ph scale and ph values of some common substances 25 The definition of ph ph = -log[h 3 O + ] = -log[h + ] Significant Figures for ph The number of decimal places in the log is equal to the number of significant figures in the original number

14 Calculating ph from its definition ph = - log[h + ] What is the ph of a solution that is 1 x M in hydronium ion? ph = -log[h 3 O + ] = -log (1 x ) = -(-12.0) = 12.0 What is the ph of a solution that is 7.3 x 10-9 M in H +? ph = -log (7.3 x 10-9 ) = -(-8.14) = 8.14 What is the ph of a 1 x 10-1 M H 3 O + (aq) solution? ph = -log[h 3 O + ] = -log (0.1) = -(-1.0) = ph of a Strong Acid Calculate ph of 1.0 M HClO 4 major species : H +, ClO 4 - and H 2 O perchloric acid is a strong acid: FULLY dissociated Therefore, 1.0 M HClO M H + ph = - log [H + ] = - log [1.0] = 0.00 Calculate ph of 0.01 M HClO 4 ph = - log [H + ] = - log [0.01] = 2.0 Calculate ph of 0.02 M HClO 4 ph = - log [H + ] = - log [0.02] =

15 Water itself is a weak acid and weak base: Two water molecules react to form H 3 O + and OH - (but only slightly) H 2 O H 2 O H 3 O + OH - 29 Autoionization of Water H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH- (aq) [H K = 3 O + ][OH - ] [H 2 O] 2 The ion-product for water, K w : = [H + ][OH - ] = K w Set = 1.0 since nearly pure liquid water even when salts, acids, or bases present. K = K w = [H 3 O + ][OH - ] = 1.0 x (at 25 C) For pure water the concentration of hydroxyl and hydronium ions must be equal: [H 3 O + ] = [OH - ] = 1.0 x = 1.0 x 10-7 M (at 25 C) 30 15

16 What is the ph of (a) pure water at 25 C ([H + ] = 1.0 x 10-7 M) and (b) M HCl? 1. a) -7.00, b) a) 7.00, b) a) -7.00, b) a) 7.00, b) a) -7.00, b) a) 7.00, b) ph Calculations What is the ph of (a) pure water at 25 o C and (b) M HCl? (a) For pure water, [H + ] = 1.0 x 10-7 M, so ph = -log (1.0 x10-7 ) = 7.00 (b) The ph of M HCl (a STRONG ACID) is ph = -log (1 x 10-2 ) =

17 The Meaning of K b, the Base Dissociation Constant For the generalized reaction between a base, B(aq), and water: B (aq) + H 2 O (l) BH + (aq) + OH- (aq) Equilibrium constant: K = K b = [BH + ][OH - ] [B] The stronger the base, the higher the [OH - ] at equilibrium and the larger the K b. 33 The Relation Between K a and K b for a Conjugate Acid-Base Pair Acid HA + H 2 O H 3 O + + A - Base A - + H 2 O HA + OH - 2 H 2 O H 3 O + + OH - [H 3 O + ] [A - ] [HA] [OH - ] x = [H 3 O + ] [OH - ] [HA] [A - ] K a x K b = K w For HNO 2 : K a = 4.5 x 10-4 and for NO 2 - : K b = 2.2 x K a x K b = (4.5 x 10-4 )(2.2 x ) = 9.9 x or ~ 10 x = 1 x = K w 34 17

18 Two New Definitions Remember: K w = [H+][OH-] so that [OH-] = K w / [H+] 1) poh = -log [OH-] = - log (K w /[H + ]) = - log K w (- log [H + ]) = pk w ph = - log (1.0 x ) ph poh = ph 2) It is convenient to express equilibrium constants for acid dissociation in the form pk a = - log K a 35 ph of a Strong Base Calculate ph of 1.0 M NaOH major species : Na +, OH - and H 2 O sodium hydroxide is a strong base: FULLY dissociated Therefore, 1.0 M NaOH 1.0 M [OH - ] ph = - log [H + ]; poh= - log [OH - ]; pk w = ph + poh = 14 ph = 14 poh ph = 14 log 1.0 = 14.0 Calculate ph of 0.01 M NaOH ph= 14 log 0.01 = 14 2 = 12.0 Calculate ph of 0.02 M NaOH ph= 14 log 0.02 = =

19 Calculating [H 3 O + ], ph, [OH - ], and poh Problem: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) M. Calculate the [H 3 O + ], ph, [OH - ], and poh of the two solutions at 25 o C. Plan: HCl is Strong Acid Complete dissociation: [H + ] = initial HCl concentration Use [H + ] to calculate ph, OH - and poh 37 Calculating [H 3 O + ], ph, [OH - ], and poh Problem: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) M. Calculate the [H 3 O + ], ph, [OH - ], and poh of the two solutions at 25 o C. Solution: (a) [H + ] = 3.0 M ph = -log[h + ] = -log(3.0) = [OH - ] = K w = 1 x = 3.3 x M [H + ] 3.0 poh = - log(3.3 x ) = (b) [H + ] = M ph = -log[h + ] = -log(0.0024) = 2.62 [OH - ] = K w = 1 x = 4.2 x M [H + ] poh = -log(4.2 x ) =

20 What is the ph of a 2.0 x 10-3 M solution of NaOH? x x Review Problem - ph of a Strong Base Calculate the ph of a 2.0 x 10-3 M solution of NaOH. Since NaOH is a strong base, it will dissociate 100% in water. NaOH (aq) Na + (aq) + OH- (aq) Since [NaOH] = 2.0 x 10-3 M, [OH - ] = 2.0 x 10-3 M The concentration of [H + ] can be calculated from K w at 25 o C: K w [H + ] = = 1.0 x = 5.0 x M [OH - ] 2.0 x 10-3 ph = - log [H + ] = - log(5.0 x ) = OR ph = pk w poh ph = (-log [OH - ]) = =

21 ph= -log[h + ] poh= - log [OH - ] pk w = 14 = ph + poh 41 K w = [H + ][OH - ] ph Box ph 14-pOH 14-pH poh 10 -ph -log([h 3 O + ]) -log([oh - ]) 10 -poh K w /[OH - ] [H 3 O + ] [OH - ] K w /[H 3 O + ] 42 21

22 The Acid-Dissociation Constant (K a ) Remember Acids dissociate into ions in water: HA (aq) + H 2 O (l) HA (aq) H 3 O + (aq) + A- (aq) H + (aq) + A- (aq) K a = [H 3 O + ][A - ] [HA] = [H + ][A - ] [HA] In a dilute solution of a WEAK acid, most of HA remains undissociated and therefore [HA] init = [HA] eq K a = [H + ][A - ]/[HA] = [H + ] 2 / [HA] init 43 Calculate the ph of a 1.50 M HNO 2 Solution Problem: Calculate the ph of a 1.00 M solution of nitrous acid HNO 2. Solution: HNO 2 (aq) H + (aq) + NO 2 - (aq) K a = 4.0 x 10-4 Initial concentrations: [H + ] = 0, [NO 2- ] = 0, [HNO 2 ] = 1.00 M Final concentrations: [H + ] = x, [NO 2 - ] = x, [HNO 2 ] = 1.00 M - x [H + ] [NO - K a = 2 ] = 4.0 x 10-4 = [HNO 2 ] (x) (x) x Assume 1.00 x 1.00 to simplify the problem. x x 10-4 or x x x = 2.0 x 10-2 = 0.02 M = [H + ] = [NO 2 - ] ph = - log[h + ] = - log(2.0 x 10-2 ) =

23 Hypochlorous acid is a weak acid formed in laundry bleach. What is the [H + ] of a M HClO solution? K a = 3.5 x 10-8, K b = 2.9 x x 10-9 M x 10-5 M x 10-8 M x 10-4 M 45 Determining Concentrations from K a and Initial [HA] Problem: Hypochlorous acid is a weak acid formed in laundry bleach. What is the [H + ] and ph of a M HClO solution? K a = 3.5 x 10-8 Plan: We need to find [H + ]. First we write the balanced equation and the expression for K a and solve for the hydronium ion concentration. Solution: HClO (aq) + H 2 O (l) H 3 O + (aq) + ClO- (aq) [H K a = + ] [ClO - ] = 3.5 x 10-8 [HClO] Concentration (M) HClO H 2 O H + ClO - Initial Change -x x +x Equilibrium x ---- x x (x)(x) K a = = 3.5 x x Assume (0.125 x) x 2 = x 10-8 x = 0.66 x 10-4 ph = -log(h + ) = -log(0.66 x 10-4 ) =

24 Problem: The weak acid hypochlorous acid is formed in bleach solutions. If the ph of a 0.12 M solution of HClO is 4.19, what is the value of the K a of this weak acid? Plan: We are given [HClO] initial and the ph, which will allow us to find [H 3 O + ] and then the hypochlorite anion concentration, so we can write the reaction and expression for K a and solve directly. Solution: Finding K a for a Weak Acid from the ph I Calculating [H 3 O + ] : [H 3 O + ] = 10 -ph = = 6.46 x 10-5 M Concentration (M) HClO (aq) + H 2 O (l) H 3 O + (aq) + ClO - (aq) Initial x Change -x x +x Equilibrium x ---- x + 1 x 10-7 x Assumptions: [H 3 O + ] = [H 3 O + ] HClO 47 since HClO is a weak acid, we assume (0.12 M x) 0.12 M Finding the K a for a Weak Acid from the ph II HClO (aq) + H 2 O (l) H 3 O + (aq) + ClO - (aq) x = [H 3 O + ] = [ClO - ] = 6.46 x 10-5 M [H 3 O + ] [ClO - ] (6.46 x 10-5 M) (6.46 x 10-5 M) K a = = = 3.48 x 10-8 [HClO] 0.12 M K a = 3.48 x 10-8 In text books it is found to be: 3.5 x 10-8 Checking: 1. For [H 3 O + 1 x 10 ] from water : -7 M x 100 = 0.155% 6.46 x 10-5 M assumptions are OK 6.46 x For [HClO] dissoc : -5 M x 100 = % 0.12 M 48 24

25 Overview: Solving Weak Acid Equilibrium Problems List the major species in the solution. Choose the species that can produce H + and write balanced equations for the reactions producing H +. Comparing the values of the equilibrium constants for the reactions you have written, decide which reaction will dominate in the production of H +. Write the equilibrium expression for the dominant reaction. List the initial concentrations of the species participating in the dominate reaction. Define the change needed to achieve equilibrium; that is, define x. Write the equilibrium concentrations in terms of x. Substitute the equilibrium concentrations into the equilibrium expression. Solve for x the easy way assume that [HA] 0 x [HA] 0 Verify the approximation is valid (5% rule). Calculate [H + ] and ph. 49 Mixtures of Several Acids - I Calculate the ph of a solution that contains 1.00 M HF (K a = 7.2 x 10-4 ) and 5.00 M HOCl (K a = 3.5 x 10-8 ). Also calculate the concentrations of the F- and OCl- ions at equilibrium. Three components produce H + : HF (aq) H + (aq) + F- (aq) K a = 7.2 x 10-4 HOCl (aq) H + (aq) + OCl- (aq) K a = 3.5 x 10-8 H 2 O (aq) H + (aq) + OH- (aq) K a = 1.0 x Even though HF is a weak acid, it has by far the greatest K a, so it will be the dominant producer of H +. Solve for its I.C.E. table first, then use those concentrations in the second most important equilibrium. Move down the list, in each case setting K a = [H + ] [A - ] [HA] = specific value 50 25

26 Mixtures of Several Acids - II Initial Concentration (mol/l) Equilibrium Concentration (mol/l) [HF] 0 = 1.00 [HF] = 1.00 x [F - ] = 0 x mol HF [F - ] = x [H + ] = ~ 0 dissociates [H + ] = x [H + ] [F - ] K a = = 7.2 x 10-4 (x) (x) = [HF] 1.00-x x Therefore, x = 2.7 x 10-2 [F - ] = [H + ] = x = 2.7 x 10-2 and ph = Mixtures of Several Acids - III The concentration of H + comes from the first part of this problem: Initial Concentration (mol/l) Equilibrium Concentration (mol/l) [HOCl] 0 = 5.00 [HOCl] = 5.00 x [OCl - ] = 0 x mol HOCl [OCl - ] = x [H + ] = 2.7 x 10-2 dissociates [H + ] = 2.7 x x [H + ] [OCl - ] (2.7 x x) (x) K a = = 3.5 x 10-8 = [HOCl] 5.00-x (2.7 x 10-2 ) x 5.00 x = 6.5 x 10-6 M = [OCl - ] Therefore, ph = 1.56 [F - ] = 2.7 x 10-2 M [OCl - ] = 6.5 x 10-6 M 52 26

27 PERCENT DISSOCIATION amount dissociated (mol/l) % dissociation = x 100 initial concentration (mol/l) x % dissociation = x 100 [HA] o for a given acid, % dissociation increases as the acid is diluted for a weak acid, [H+] decreases with increasing dilution as [HA] does, but % dissociation increases as dilution increases 53 Figure 7.5: Effect of dilution on the percent dissociation and [H+] K a = [H+ ] [A - ] [HA] 54 27

28 Calculate the percent dissociation of a 1.00 M hydrocyanic acid solution, K a = 6.20 x x 10-5 % x 10-3 % x % x 10-8 % 55 Problem: Calculate the percent dissociation of a 1.00 M hydrocyanic acid solution, K a = 6.20 x HCN (aq) + H 2 O (l) H 3 O + (aq) + CN- (aq) HCN H 3 O + CN - [H 3 O + ][CN - ] Initial 1.00 M 0 0 K a = [HCN] Change -x +x +x Final 1.00 x x x (x)(x) K a = = 6.20 x 10 (1.00-x) -10 Assume 1.00-x 1.00 K a = = 6.2 x x = 2.49 x x 10-5 % dissociation = x 100 = % 1.00 x

29 Weak Bases Many compounds with an electron-rich nitrogen are weak bases. The common structural feature is an N atom that has a lone electron pair in its Lewis structure. EXAMPLES: 1) Ammonia (:NH 3 ) 2) Amines (general formula RNH 2, R 2 NH, R 3 N) H 3 C.. N H CH 3 Dimethylamine H 3 C.. N CH 3 CH 3 Trimethylamine H H 3 C.. N H.. N H Ethylamine 57 H Methylamine C 2 H 5 General reaction (R is a hydrocarbon group or a H): NR 3 (aq) + H 2 O(l) HNR 3+ (aq) + OH - (aq) K b = [HNR 3+ ] [OH - ] [NR 3 ] 58 29

30 Determining ph from K b and Initial [B] I Problem: Ammonia is commonly used cleaning agent and is a weak base, with a K b of 1.8 x What is the ph of a 1.5 M NH 3 solution? Plan: Ammonia reacts with water to form [OH - ]. Calculate [H 3 O + ] and the ph. The balanced equation and K b expression are: NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) K b = [NH 4 + ] [OH - ] [NH 3 ] Concentration (M) NH 3 H 2 O NH 4 + OH - Initial Change -x x +x Equilibrium x ---- x x making the assumption: since K b is small, (1.5 M x) 1.5 M 59 Determining ph from K b and Initial [B] II Substituting into the K b expression and solving for x: [NH + 4 ] [OH - ] (x)(x) K b = = = 1.8 x 10-5 [NH 3 ] 1.5 x 2 = 2.7 x 10-5 x = 5.20 x 10-3 = [OH - ] = [NH + 4 ] 60 30

31 What is the ph of the 1.5 M NH 3 solution? [OH - ]=5.20 x 10-3 M Determining ph from K b and Initial [B] III Calculating ph: [H 3 O + ] = K w 1.0 x = [OH - ] 5.20 x 10-3 = 1.92 x ph = -log[h 3 O + ] = - log (1.92 x ) ph =

32 Polyprotic Acids 63 The Stepwise Dissociation of Phosphoric Acid Calculating ions concentrations in a certain molarity of phosphoric acid H 3 PO 4 (aq) + H 2 O (l) H 2 PO 4 - (aq) + H 3 O + (aq) K a1 = 7.5x10-3 H 2 PO 4 - (aq) + H 2 O (l) HPO 4 2- (aq) + H 3 O + (aq) K a2 = 6.2x10-8 HPO 4 2- (aq) + H 2 O (l) PO 4 3- (aq) + H 3 O + (aq) K a3 = 4.8x10-13 Net: H 3 PO 4 (aq) + 3 H 2 O (l) PO 4 3- (aq) + 3 H 3 O + (aq) Order from biggest K a to smallest: Just like in the example with multiple weak acids, each subsequent K a is so much smaller then the one above it that its contribution can be neglected with respect to changing any concentrations of previous products!! 64 32

33 Dissociation of Phosphoric Acid - Problem I Calculate the ph of a 5.0 M H 3 PO 4 solution and determine the equilibrium concentrations of the species: H 3 PO 4, H 2 PO 4 -, HPO 4-2, and PO 4-3 Solution: H 3 PO 4 (aq) H + (aq) + H 2 PO 4 - (aq) K a1 = 7.5 x 10-3 = [H + ] [H 2 PO 4 - ] [H 3 PO 4 ] Initial Concentration (mol/l) Equilibrium Concentration (mol/l) [H 3 PO 4 ] 0 = 5.0 [H 3 PO 4 ] = x [H 2 PO - 4 ] 0 = 0 [H 2 PO - 4 ] = x [H + ] 0 = 0 [H + ] = x K a1 = 7.5 x 10-3 = (x)(x) 5.0-x x x = 1.9 x 10-1 M 65 Dissociation of Phosphoric Acid - Problem II Since 1.9 x 10-1 is less than 5% of 5.0, the approximation is acceptable: x = [H + ] = [H 2 PO 4- ] = 0.19 M ph = 0.72, [H 3 PO 4 ] = 5.0 x = 4.8 M The concentration of HPO 2-4 can be obtained from K a2 : where: K [H + ] [HPO 2-4 ] a2 = 6.2 x 10-8 = = [H 2 PO - 4 ] (0.19 M) [HPO 4 2- ] (0.19 M) [H + ] = [H 2 PO 4 - ] = 0.19 M, so [HPO 4 2- ] = K a2 = 6.2 x 10-8 M To calculate [PO 4 3- ], we use the expression for K a3, and the values obtained from the other calculations: K [H + ] [PO 3-4 ] a3 = 4.8 x = = [HPO 2-4 ] [PO 4 3- ] = (4.8 x )(6.2 x 10-8 M) (0.19 M) (0.19 M) [PO 4 3- ] (6.2 x 10-8 M) = 1.6 x M 66 33

34 Calculate the ph of 3.00 x 10-3 M Sulfuric Acid Polyprotic acid, more than one proton to lose!!! Multiple K a s H 2 SO 4 (aq) HSO 4 - (aq) + H + (aq) K a1 = LARGE HSO 4 - (aq) SO 4 - (aq) + H + (aq) K a2 = 1.2 x 10-2 Because first dissociation has a very large K a it can be treated as a strong acid (~100% dissociated). 67 Calculate the ph of 3.00 x 10-3 M Sulfuric Acid HSO 4- (aq) SO 4- (aq) + H + (aq) K a = 1.2x10-2 Initial Concentration (mol/l) Equilibrium Concentration (mol/l) [HSO - 4 ] 0 = x mol/l HSO - [HSO ] = x [SO 2-4 ] 0 = 0 dissociates [SO 2-4 ] = x [H + ] 0 = to reach [H + ] = x equilibrium From dissociation of H 2 SO 4 [H + ][SO 2-4 ] K a2 = 1.2 x 10-2 ( x)(x) = = [HSO - 4 ] ( x) No obvious approximation looks valid, so we must solve with the quadratic formula. 0 = x x 3.6 x b + a = 1 x = - b 2 4ac b = x = 2.10 x a c = -3.6 x 10-5 [H + ] = x = ph = Note: Change in [H+] does not violate huge K a for 1 st dissociation: [H 2 SO 4 ] = 0. 34

35 SALTS and their ph sodium acetate (CH 3 COONa) solution is basic CH 3 COO - + H 2 O CH 3 COOH + OH - ammonium chloride (NH 4 Cl) solution is acidic NH H 2 O NH 3 + H 3 O + sodium chloride solution is neutral NaCl + H 2 O Na + + Cl - + H 2 O 69 Effects of Salts on ph and Acidity Salts that consist of cations of strong bases and the anions of strong acids have no effect on the [H + ] when dissolved in water. Examples: NaCl, KNO 3, Na 2 SO 4, NaClO 4, KBr, etc. A salt whose cation alone has neutral properties (such as Na + or K + ) and whose anion is the conjugate base of a weak acid, produces a basic solution in water, since anion captures some H +. Examples: NaF, KCN, NaC 2 H 3 O 2, Na 3 PO 4, Na 2 CO 3, K 2 S, Na 2 C 2 O 4, etc. A salt whose anion alone has neutral properties and whose cation is the conjugate acid of a weak base OR a small and highly charged metal ion will produce an acidic solution in water. Examples: NH 4 Cl, AlCl 3, Fe(NO 3 ) 3, etc

36 See Table If the solid sodium cyanide (NaCN) is dissolved in pure water, will the resulting solution be acidic, basic, or neutral? 1. Acidic 2. Basic 3. Neutral 72 36

37 Example 7.15: Additional complex salts Predict whether an aqueous solution of each of the following salts will be acidic, basic, or neutral. a) NH 4 C 2 H 3 O 2 b) NH 4 CN c) Al 2 (SO 4 ) 3 Solution: get needed K from K a = K w /K b or K b = K w /K a a) ammonium ion (NH + 4 ): K a from K b K a (for NH + K 4 ) = w = 5.6 x K b (for NH 3 ) acetate ion (C 2 H 3 O - 2 ): K b from K a is 5.6 x Since K a = K b the solution will be neutral and the ph close to 7. b) ammonium ion (NH 4 + ): K a is 5.6 x cyanide ion (CN - ): K b from K a is 1.6 x 10-5 Since K b >> K a the solution will be basic (ph > 7). c) hydrated aluminum ion: K a for Al(H 2 O) 6 3+ = 1.4 x 10-5 sulfate ion: K b for sulfate (from K a for HSO 4 - ) = 8.3 x Since K a >> K b the solution will be acidic (ph < 7). 73 Determining the ph of a Solution of A - I Problem: Sodium cyanide in water will produce a basic solution. What is the ph of a 0.25 M solution of NaCN? K a of HCN = 4.0 x Plan: We have to find the ph of a solution of the cyanide ion, CN -, which acts as a base in water: CN - (aq) + H 2 O (l) HCN (aq) + OH- (aq) K b = [HCN] [OH- ] Solution: Setting up the reaction table: [CN - ] Concentration (M) CN - H 2 O HCN OH - Initial Change -x x +x Equilibrium x ---- x x 74 37

38 Determining the ph of a Solution of A - II Solving for K b : K w 1.0 x K b = = = 2.5 x 10-5 K a 4.0 x Substituting into the expression for K b and solving for x. Make the assumption: since K b is small, (0.25 M x) 0.25 M [HCN] [OH - ] K b = = 2.5 x 10-5 = [CN - ] (x)(x) 0.25 x 2 = 6.25 x 10-6 x = 2.50 x 10-3 = [OH - ] solving for ph: ph = (-log [OH - ]) ph = (-log 2.50 x 10-3 ) ph = = Small and highly charged cations produce acidic solutions in water. Examples: Al 3+, Fe 3+, etc

39 ph of a small highly charged cation - I Calculate the ph of a M AlCl 3 solution. The K a value for the Al(H 2 O) 6 3+ ion is 1.4 x Solution: Since the Al(H 2 O) 6 3+ ion is a stronger acid than water, the dominate equilibrium will be: Al(H 2 O) 6 3+ (aq) Al(OH)(H 2 O) 5 2+ (aq) + H + (aq) 1.4 x 10-5 = K a = [Al(OH)(H 2 O) 2+ 5 ][H + ] [Al(H 2 O) 3+ 6 ] Initial Concentration (mol/l) Equilibrium Concentration (mol/l) [Al(H 2 O) 3+ 6 ] 0 = [Al(H 2 O) 3+ 6 ] = x [Al(OH)(H 2 O) 2+ 5 ] = 0 [Al(OH)(H 2 O) 2+ 5 ] = x [H + ] 0 = ~ 0 x mol/l [H + ] = x 77 ph of a small highly charged cation - II Thus: [Al(OH)(H K a = 2 O) 2+ 5 ] [H + ] = 1.4 x 10-5 [Al(H 2 O) 3+ 6 ] K a = (x) (x) = x 2 = 1.4x x x = 3.7 x 10-4 Since the approximation is valid by the 5% rule: [H + ] = x = 3.7 x 10-4 M and ph =

40 Predicting the Relative Acidity of Salt Solutions Problem: Determine whether an aqueous solution of iron(iii) nitrite, Fe(NO 2 ) 3, is acidic, basic, or neutral. Plan: The formula consists of the small, highly charged, and therefore weakly acidic, Fe 3+ cation and the weakly basic NO - 2 (anion of the weak acid HNO 2 ). To determine the relative acidity of the solution, we write equations that show the reactions of the ions with water, and then find K a and K b of the ions to see which ion reacts to form H + or OH - to a greater extent. Solution: Writing the reactions with water: Fe(H 2 O) 3+ 6 (aq) + H 2 O (l) Fe(H 2 O) 5 OH 2+ (aq) + H 3 O+ (aq) NO 2 - (aq) + H 2 O (l) HNO 2(aq) + OH - (aq) Obtaining K a and K b of the ions: for Fe 3+ (aq) K a = 6 x 10-3 for NO 2 - (aq), K b must be determined: K b of NO - K 2 = w 1.0 x 10 = -14 = 2.5 x K a of HNO x 10-4 Since K a of Fe 3+ > K b of NO - 2, the solution is acidic. 79 If the solid solid ammonium fluoride (NH 4 F) is dissolved in pure water, will the resulting solution be acidic, basic, or neutral? NH 3, K b = HF, K a = Acidic 2. Basic 3. Neutral 80 40

41 Summary: Solving Acid-Base Equilibria Problems List the major species in solution. Look for reactions that can be assumed to go to completion, such as a strong acids/bases dissociating or H + reacting with OH -. For a reaction that can be assumed to go to completion: a) Determine the concentrations of the products. b) Write down the major species in solution after the reaction. Look at each major component of the solution and decide whether it is an acid or a base. Pick the equilibrium that will control the ph. Use known values of the dissociation constants for the various species to determine the dominant equilibrium. a) Write the equation for the reaction and the equilibrium expression. b) Set up the I.C.E. table and define x. c) Compute the equilibrium concentrations in terms of x. d) Substitute the [ ] eq into the equilibrium expression and solve for x. e) Check the validity of the approximation. f) Calculate the ph and other concentrations as required. 81 Summary: General Strategies for Solving Acid-Base Problems Think Chemistry. Focus on the solution components and their reactions. It will almost always be possible to choose one reaction that is the most important. Be systematic. Write down all the things you know, including things like the equilibrium expression, list what you can calculate, and list what s needed. Be flexible. Although all acid-base problems are similar in many ways, important differences do occur. Treat each problem as a separate entity. Do not try to force a given problem to match any you have solved before. Look for both the similarities and the differences. Be patient. The complete solution to a complicated problem cannot be seen immediately in all its detail. Pick the problem apart into its workable steps. Be confident. Look within the problem for the solution, and let the problem guide you. Assume that you can think it out. Do not rely on memorizing solutions to problems. In fact, memorizing solutions is usually detrimental because you tend to try to force a new problem to be the same as one you have seen before. Understand and think - don t t just memorize

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