Chapter 18 Acid-Base Equilibria - University of Washington

Chapter 7 Acids and Bases Chapter 7 Acids and Bases 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 The Nature of Acids and Bases Acid Strength

The pH Scale Calculating the pH of Strong Acid Solutions Calculating the pH of weak Acid Solutions Bases Polyprotic Acids Acid-Base Properties of Salts 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 A circle of shiny pennies is created by the reaction between the citric acid of the lemon and the tarnish on the surface of the copper.

Source: Fundamental Photos Arrhenius (or Classical) Acid-Base Definition An acid is a substance that contains hydrogen and dissociates in water to yield a hydronium ion : H3O+ A base is a substance that contains the hydroxyl group and dissociates in water to yield : OH Neutralization is the reaction of an H+ (H3O+) ion from the acid and the OH - ion from the base to form water, H2O. The neutralization reaction is exothermic and releases approximately 56 kJ per mole of acid and base. H+(aq) + OH-(aq) H2O(l) H0rxn = -55.9 kJ

Brnsted-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; HNO3 and H2PO4- are two examples, all Arrhenius acids are Brnsted-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 NH3, CO32-, F -, as well as OH -. Brnsted-Lowry bases are not Arrhenius bases, but all Arrhenius bases contain the Brnsted-Lowry base OH-. Therefore in the Brnsted-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 Molecular model: Two water molecules

react to form H3O+ and OH- Molecular model: The reaction of an acid HA with water to form H3O+ and a conjugate base. Acid Base Conjugate acid Conjugate base The Acid-Dissociation Constant (Ka)

Strong acids dissociate completely into ions in water: HA(g or l) + H2O(l) H3O+(aq) + A-(aq) In a dilute solution of a strong acid, almost no HA molecules exist: [H3O+] = [HA]init or [HA]eq = 0 [H3O+][A-] Qc = at equilibrium, Qc = Kc >> 1 [HA][H2O] Nitric acid is an example: HNO3 (l) + H2O(l) H3O+(aq) + NO3-(aq) Weak acids dissociate very slightly into ions in water: HA(aq) + H2O(aq) H3O+(aq) + A-(aq) In a dilute solution of a weak acid, the great majority of HA molecules are undissociated: [H3O+] << [HA]init or [HA]eq = [HA]init

[H3O+][A-] Qc = [HA][H2O] at equilibrium, Qc = Kc << 1 The Meaning of Ka, the Acid Dissociation Constant For the ionization of an acid, HA: HA(aq) + H2O(l) H3O+(aq) + A-(aq) Since the concentration of water is + [H O ] [A

] 3 Kc = high, and does not change significantly [HA] [H2O] during the reaction, its value is absorbed Therefore: into the constant. [H3O+] [A-] The stronger the acid, the higher the [H3O+] Kc = [HA] at equilibrium, and the larger the Ka: Stronger acid higher [H3O+] larger Ka For a weak acid with a relative high Ka (~10-2 ), a 1 M solution

has ~10% of the HA molecules dissociated. For a weak acid with a moderate Ka (~10-5 ), a 1 M solution has ~ 0.3% of the HA molecules dissociated. For a weak acid with a relatively low Ka (~10-10 ), a 1 M solution has ~ 0.001% of the HA molecules dissociated. Figure 7.1: Graphical representation of the behavior of acids of different strengths in aqueous solution. A Strong Acid A Weak Acid The Extent of

Dissociation for Strong and Weak Acids Figure 7.2: Relationship of acid strength and conjugate base strength The Six Strong Acids Hydrogen Halides HCl

HBr HI Hydrochloric Acid Hydrobromic Acid HydroIodioic Acid Oxyacids H2SO4 HNO3 HClO4 Sulfuric Acid Nitric Acid Perchloric Acid

Molecular model: Sulfuric acid Molecular model: Nitric acid Molecular model: Perchloric acid The Stepwise Dissociation of Phosphoric Acid Phosphoric acid is a weak acid, and normally only looses one proton in solution, but it will lose all three when reacted with a strong base with heat. The ionization constants are given for comparison. H3PO4 (aq) + H2O(l) H2PO4-(aq) + H3O+(aq) H2PO4-(aq) + H2O(l)

HPO42-(aq) + H3O+(aq) HPO42-(aq) + H2O(l) PO43-(aq) + H3O+(aq) H3PO4 (aq) + 3 H2O(l) PO43-(aq) + 3 H3O+(aq) The Conjugate Pairs in Some Acid-Base Reactions Conjugate Pair Acid +

Base Base + Acid Conjugate Pair Reaction 1 Reaction 2 HF +


HCN Reaction 3 NH4+ + CO32 NH3 + HCO3

Reaction 4 H2PO4 + OH HPO42 + H2O Reaction 5

H2SO4 + N2H5+ HSO4 2 2 3 +

N2H62+ Identifying Conjugate Acid-Base Pairs Problem: The following chemical reactions are important for industrial processes. Identify the conjugate acid-base pairs. (a) HSO4-(aq) + CN-(aq) SO42-(aq) + HCN(aq) (b) ClO-(aq) + H2O(l) HClO(aq) + OH-(aq) (c) S2-(aq) + H2O(aq) HS-(aq) + OH-(aq) Plan: To find the conjugate acid-base pairs, we find the species that donate H+ and those that accept it. The acid (or base) on the left becomes its conjugate base (or acid) on the right. Solution:

(a) The proton is transferred from the sulfate to the cyanide so: HSO4-(aq)/SO42-(aq) and CN-(aq)/HCN(aq ) are the two acid-base pairs. (b) The water gives up one proton to the hypochlorite anion so: ClO-(aq)/HClO(aq) and H2O(l) / OH-(aq ) are the two acid-base pairs. (c) One of waters protons is transferred to the sulfide ion so: S2-(aq)/HS-(aq) and H2O(l)/OH-(aq) are the two acid-base pairs. Autoionization of Water H2O(l) + H2O(l) Kc = H3O+ + OH[H3O+][OH-] [H2O]2 The ion-product for water, Kw: Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25C)

For pure water the concentration of hydroxyl and hydronium ions must be equal: [H3O+] = [OH-] = 1.0 x 10-14 = 1.0 x 10 -7 M (at 25C) The molarity of pure water is: 1000g/L = 18.02 g/mol M Figure 7.3: The pH scale and

pH values of some common substances The pH Values of Some Familiar Aqueous Solutions [H3O+] [OH ] = - KW [H3O+]

[OH-] [H3O+]> [OH-] [H3O+]< [OH-] acidic solution neutral solution basic solution [H3O+] =

[OH-] The Relationship Between Ka and pKa Acid Name (Formula) Hydrogen sulfate ion (HSO4-) Ka at 25oC 1.02 x 10-2 pKa 1.991 Nitrous acid (HNO2) 7.1 x 10-4

3.15 Acetic acid (CH3COOH) 1.8 x 10-5 4.74 Hypobromous acid (HBrO) 2.3 x 10-9 8.64 Phenol (C6H5OH)

1.0 x 10-10 10.00 Acid and Base Character and the pH Scale In acidic solutions, the protons that are released into solution will not remain alone due to their large positive charge density and small size. They are attracted to the negatively charged electrons on the oxygen atoms in water, and form hydronium ions. H+(aq) + H2O(l) = H3O+(l) [H+] = [H3O+] To handle the very large variations in the concentrations of the hydrogen ion in aqueous solutions, a scale called the pH scale is used which is: pH = - log[H3O+] What is the pH of a solution that is 10-12 M in hydronium ion ? pH = -log[H3O+] = (-1)log 10-12 = (-1)(-12) = 12

What is the pH of a solution that is 7.3 x 10-9 M in H3O+ ? pH = -log(7.3 x 10-9) = -1(log 7.3 + log 10-9) = -1[(0.863)+(-9)] = 8.14 pH of a neutral solution = 7.00 pH of an acidic solution < 7.00 pH of a basic solution > 7.00 Classifying the Relative Strengths of Acids and BasesI Strong acids. There are two types of strong acids: 1. The hydrohalic acids HCl, HBr, and HI 2. Oxoacids in which the number of O atoms exceeds the number of ionizable H atoms by two or more, such as HNO3, H2SO4, HClO4 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 bounded to O or to halogen, such as HCN and H2S

3. Oxoacids in which the number of O atoms equals or exceeds by one the number of ionizable H atoms, such as HClO, HNO2, and H3PO4 4. Organic acids (general formula RCOOH), such as CH3COOH and C6H5COOH Classifying the Relative Strengths of Acids and BasesII Strong bases. Soluble compounds containing O2- or OH- ions are strong bases. The cations are usually those of the most active metals: 1) M2O 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 (:NH3) 2) Amines (general formula RNH2, R2NH, R3N), such as CH3CH2NH2, (CH3)2NH, (C3H7)3N, and C5H5N : : :

: Figure 7.4: (a) Measuring the pH of vinegar. (b) Measuring the pH of aqueous ammonia. Methods for Measuring the pH of an Aqueous Solution (a) pH paper (b) Electrodes of a pH meter Summary: (P 233)

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, Acid-Base problems require a step-by-step approach. 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 sloution 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; dont

just memorize. Calculating [H3O+], pH, [OH-], and pOH Problem: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H3O+], pH, [OH-], and pOH of the two solutions at 25C. Plan: We know that hydrochloric acid is a strong acid, so it dissociates completely in water; therefore [H3O+] = [HCl]init.. We use the [H3O+] to calculate the [OH-] and pH as well as pOH. Solution: (a) [H3O+] = 3.0 M pH = -log[H3O+] = -log(3.0) = ________ -14 K 1 x

10 w [OH ] = = = _________________ M + 3.0 [H3O ] pOH = - log(3.333 x 10-15) = 15.000 - 0.477 = _______ (b) [H3O+] = 0.0024 M pH = -log[H3O+] = -log(0.0024) = _______ Kw 1 x 10-14 [OH ] = = = ________________ M

+ [H3O ] 0.0024 pOH = -log(4.167 x 10-12) = 12.000 - 0.6198 = __________ Calculate the pH of a 1.00 M HNO2 Solution Problem: Calculate the pH of a 1.00 M Solution of Nitrous acid HNO2. Solution: HNO2 (aq) H+(aq) + NO2-(aq) Ka = 4.0 x 10-4 Initial concentrations = [H+] = 0 , [NO2-] = 0 , [HNO2] = 1.00 M Final concentrations = [H+] = x , [NO2-] = x , [HNO2] = 1.00 M - x [H+] [NO2-] (x) (x) -4 Ka =

= 4.0 x 10 = [HNO2] 1.00 - x Assume 1.00 x = 1.00 to simplify the problem. x2 = 4.0 x 10-4 or x2 = 4.0 x 10-4 1.00 x = 2.0 x 10-2 = 0.02 M = [H+] = [NO2-] pH = - log[H+] = - log(2.0 x 10-2) = 2.00 0.30 = ___________ Molecular model: Nitrous acid Molecular model: HF and H2O Summary:

(P 237) 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-that is, by assuming that [HA] 0 x = [HA]0 Verify whether the approximation is valid ( the 5% rule is the test in this case). Calculate [H+] and pH. Like Example 7.3 (P 237)-I Calculate the pH of a solution that contains 1.00 M HF (Ka = 7.2 x 10-4) and 5.00 M HOCl (Ka = 3.5 x 10-8). Also calculate the concentrations of the Fluoride and Hypochlorite ions at equilibrium. Three components produce H+: HF(aq) HOCl(aq) H2O(aq) H+(aq) + F-(aq) Ka = 7.2 x 10-4

H+(aq) + OCl-(aq) Ka = 3.5 x 10-8 H+(aq) + OH-(aq) Ka = 1.0 x 10-14 Even though HF is a weak acid, it has by far the greatest Ka, therefore it will be the dominate producer of H+. [H+] [F-] Ka = = 7.2 x 10-4 [HF]

Like Example 7.3 (P 236)-II Initial Concentration (mol/L) [HF]0 = 1.00 [F-] = 0 ~0 [H+] = Equilibrium Concentration (mol/L) x mol HF dissociates [HF] = 1.00 x [F-] = x [H+] = x

[H+] [F-] (x) (x) x2 -4 Ka = = 7.2 x 10 = = [HF] 1.00-x 1.00 x = 2.7 x 10-2 using the 5% rule: Therefore, [F- ] = [H+] = x = 2.7 x 10-2 x = 2.7 x 10-2

x = 2.7 x 10-2 x 100% = 2.7% [HF]0 1.00 and pH = ___________ Like Example 7.3 (P 236)-III [H+] [OCl-] Ka = = 3.5 x 10-8 [HOCl] The concentration of H+ comes from the first part of this problem = 2.7 x 10-2 M

[HOCl] = 5.00 M ; [OCl-] = x (2.7 x 10-2)[OCl-] 3.5 x 10 = (5.00 - x) -8 -8 5.00 ( 3.5 x 10 ) [OCl ] = 2.7 x 10-2

Assume: 5.00 x = 5.00 = 6.48 x 10-6 M pH = 1.56 [F-] = 2.7 x 10-2 M ; [OCl-] = 6.48 x 10-6 M Molecular model: Hypochlorous acid (HOC1) Molecular model: HCN, HNO2, and H2O Figure 7.5: Effect of dilution on the percent dissociation and [H+] Problem: Calculate the Percent dissociation of a 0.0100M Hydrocyanic acid solution, Ka = 6.20 x 10-10.

HCN(aq) + H2O(l) HCN Initial 0.0100M Change -x Final 0.0100 x H3O+(aq) + CN- (aq) H3O+ CN- 0 +x x

0 +x x Assume 0.0100-x = 0.0100 Ka = 2.49 x 10-6 0.0100 % dissociation = [H3O+][CN-] Ka = [HCN] (x)(x) Ka = (0.0100-x) = 6.20 x 10-10

x2 0.0100 = 6.2 x 10-10 x = 2.49 x 10-6 x 100% = ______________ Runner struggles to top of a hill Source: Corbis Molecular model: HC3H5O3 and H2O Finding the Ka of a Weak Acid from the pH of its SolutionI

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 Ka of this weak acid. Plan: We are given [HClO]initial and the pH which will allow us to find [H3O+] and, hence, the hypochlorite anion concentration, so we can write the reaction and expression for Ka and solve directly. Solution: Calculating [H3O+] : [H3O+] = 10-pH = 10-4.19 = 6.46 x 10-5 M Concentration (M) HClO(aq) + H2O(l) H3O+(aq) + ClO -(aq) Initial Change Equilibrium 0.12

-x 0.12 -x ---------- ------+x +x ------+x +x Assumptions: [H3O+] = [H3O+]HClO since HClO is a weak acid, we assume 0.12 M - x = 0.12 M Finding the Ka of a Weak Acid from the pH of its SolutionII

HClO(aq) + H2O(l) H3O+(aq) + ClO -(aq) x = [H3O+] = [ClO-] = 6.46 x 10-5 M Ka = [H3O+] [ClO-] [HClO] = (6.46 x 10-5 M) (6.46 x 10-5 M) Ka = 3.48 x 10-8 Checking:

1. For [H3O+]from water : 0.12 M = 348 x 10-10 In text books it is found to be: 3.5 x 10-8 1 x 10-7 M x 100 = 0.155% 6.46 x 10-5 M assumption is OK 6.46 x 10-5 M 2. For [HClO]dissoc : x 100 = 0.0538 % 0.12 M

Molecular model: Acetic acid Molecular model: Benzoic acid Determining Concentrations from Ka and Initial [HA] Problem: Hypochlorous acid is a weak acid formed in laundry bleach. What is the [H3O+] of a 0.125 M HClO solution? Ka = 3.5 x 10-8 Plan: We need to find [H3O+]. First we write the balanced equation and the expression for Ka and solve for the hydronium ion concentration. + HClO + H O H O

+ ClO (aq) 2 (l) 3 (aq) (aq) Solution: + [H O ] [ClO ] = 3.5 x 10-8 Ka = 3 [HClO]

Concentration (M) Initial Change Equilibrium HClO H 2O H3O+ ClO- 0.125 -x

0.125 - x ---------- 0 +x x 0 +x x Ka = (x)(x) = 3.5 x 10-8 0.125-x assume 0.125 - x = 0.125

x2 = 0.4375 x 10-8 x = 0.661 x 10-4 Solving Problems Involving Weak-Acid EquilibriaI There are two general types of equilibrium problems involving weak acids and their conjugate bases: 1. Given equilibrium concentrations, find Ka. 2. Given Ka and some concentration information, find the other equilibrium concentrations. The problem-solving approach. 1. Write the balanced equation and Ka expression; these will tell you what to find. 2. Define x as the unknown concentration that changes during the reaction. Frequently, x = [HA]dissoc., the concentration of HA that dissociates which, through the use of certain assumptions, also

equals [H3O+] and [A-] at equilibrium. 3. Construct a reaction table that incorporates the unknown. 4. Make assumptions that simplify the calculation, usually that x is very small relative to the initial concentration. Solving Problems Involving Weak-Acid EquilibriaII 5. Substitute the values into the Ka expression and solve for x. 6. Check that the assumptions are justified. We normally apply the 5% rule; if the value of x is greater than 5% of the value it is compared with, you must use the quadratic formula to find x. The notation system. Molar concentrations of species are indicated by using square brackets around the species of interest. Brackets with no subscript refer to the molar concentration of the species at equilibrium. The assumptions. The two key assumptions to simplify the arithmetic are: 1. The [H3O+] from the autoionization of water is negligible. In fact,

the presence of acid from whatever is put into solution will hinder the autoionization of water, and make it even less important. 2. A weak acid has a small Ka. Therefore, it dissociates to such a small extent that we can neglect the change in its concentration to find its equilibrium concentration. Tanks in Miami, Florida Source: Visuals Unlimited Like Example 7.5 (P 243) 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 Kw: Kw [H ] = = [OH ] + 1.0 x 10-14 2.0 x 10-3 = 5.0 x 10-12 M pH = - log [H+] = - log( 5.0 x 10-12) =12.00 0.70 = _________

Amines: Bases with the Nitrogen Atom .. .. N N H3C H H H3C H CH3 .. N Methylamine Dimethylamine .. N H3C CH CH3 Pyridine

.. N 3 Trimethylamine H H C2H5 Ethylamine Determining pH from Kb and Initial [B]I Problem: Ammonia is commonly used cleaning agent in households and is a weak base, with a Kb of 1.8 x 10-5. What is the pH of a 1.5 M NH3 solution?

Plan: Ammonia reacts with water to form [OH-] and then calculate [H3O+] and the pH. The balanced equation and Kb expression are: Kb = NH3 (aq) + H2O(l) [NH4+] [OH-] [NH3] Concentration (M) Initial Change Equilibrium NH4+(aq) + OH-(aq)

NH3 H2O NH4+ OH- 1.5 -x 1.5 - x ---------- 0 +x

x 0 +x x making the assumption: since Kb is small: 1.5 M - x = 1.5 M Determining pH from Kb and Initial [B]II Substituting into the Kb expression and solving for x: Kb = [NH4+] [OH-]

(x)(x) = 1.5 [NH3] = 1.8 x 10-5 x2 = 2.7 x 10-5 = 27 x 10-6 x = 5.20 x 10-3 = [OH-] = [NH4+] Calculating pH: [H3O+] = Kw [OH-]

1.0 x 10-14 -12 = = 1.92 x 10 5.20 x 10-3 pH = -log[H3O+] = - log (1.92 x 10-12) = 12.000 - 0.283 pH = ___________ Molecular model: Na+ and OH- and H2O Molecular model: CH3NH2 and H2O

The Relation Between Ka and Kb of a Conjugate Acid-Base Pair Acid HA + H2O H3O+ + A- Base A- + H2O HA + OH- 2 H2O

H3O+ + OH- [H3O+] [OH-] = Kw = [H3O+] [A-] [HA] Ka x x [HA] [OH-] [A-] Kb

Ka = 4.5 x 10-4 -4 -11 -15 K x K = (4.5 x 10 )(2.2 x 10 ) =

9.9 x 10 For HNO2 a b -15 -14 -11 or ~ 10 x 10 = 1

x 10 = Kw Kb = 2.2 x 10 Molecular model: H+, HSO4-, and H2O Molecular model: Phosphoric acid Molecular model: H3PO4 and H2O Like Example 7.7 (P247-8)-I Calculate the pH of a 5.0 M H3PO4 solution and determine equilibrium concentrations of the species: H3PO4 , H2PO4-, HPO4-2, and PO4-3. Solution:

H3PO4 (aq) H+(aq) + H2PO4-(aq) [H+][H2PO4-] Ka = 7.5 x 10-3 = [H3PO4] Initial Concentration (mol/L) [H3PO4]0 = 5.0 [H2PO4-]0 = 0 [H+]0 = 0 Ka1 = 7.5 x 10 = -3 Equilibrium Concentration (mol/L) [H3PO4] = 5.0 - x [H2PO4-] =x

[H+] = x [H+][H2PO4-] (x)(x) ~ = = 5.0-x [H3PO4] x2 5.0 Like Example 7.7 (P247-8) - II Solving for x ~ = 1.9 x 10-1 Since 1.9 x 10-1 is less than 5% of 5.0, the approximation is acceptable and: [H+] = x = 0.19 M = [H2PO4-] ,

pH = 0.72 [H3PO4] = 5.0 x = 4.8 M The concentration of HPO42- can be obtained from Ka2: [H+][HPO42-] Ka2 = 6.2 x 10 = [H2PO4-] -8 where: [H+] = [H2PO4-] = 0.19 M ; [HPO42-] = Ka2 = 6.2 x 10-8 M To calculate [PO43-], we use the expression for Ka3 , and the values obtained from the other calculations: +

33[H ][PO ] 0.19[PO ] 4 4 -13 Ka3 = 4.8 x 10 = = 2[HPO4 ] 6.2 x 10-8 -13 -8 (4.8 x

10 )(6.2 x 10 ) 3[PO4 ] = = _______________ M 0.19 A Drexel University Chemist shows that a new form of concrete called ZeoTech (on the right) can withstand soaking in sulfuric acid for 30

days Source: AP/Wide World Photos Like Example 7.9 (P251-2) Calculate the pH of a 3.00 x 10-3 M Sulfuric acid solution. Initial Concentration (mol/L) [HSO4-]0 = 0.00300 [SO42-]0 = 0 [H+]0= 0.00300 From dissociation of H2SO4 Equilibrium Concentration (mol/L) X mol/L HSO4dissociates

to reach equilibrium [HSO4-] = 0.00300 x [SO42-] = x [H+] = 0.00300 + x Assume [H+][SO42-] (0.00300 + x)(x) Ka2 = 1.2 x 10 = = x << 0.00300 (0.00300 x) [HSO4 ] When we solve for x we get x = 1.2 x 10-2, which is close to 0.00300 therefore the approximation is not valid, and we must solve with

the quadratic formula. Multiplying the expression out we get: 0 = x2 + 0.015x 3.6 x 10-5 -b +- b2 4ac x = 2.10 x 10-3 x= 2a a = 1, b = 0.015 c = -3.6 x 10-5 [H+] = 0.00300 + x = 0.00510 pH = 2.29 -2 Figure 7.6: A plot of the fractions of H2CO3, HCO-3 and CO32- The Effect of Atomic

and Molecular Properties on Nonmetal Hydride Acidity 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, KNO3, Na2SO4, NaClO4, KBr, etc. For any salt whose cation has neutral properties(such as Na+ or K+) and whose anion is the conjugate base of a weak acid, the aqueous solution will be basic. Examples: NaF, KCN, NaC2H3O2, Na3PO4, Na2CO3, K2S, Na2C2O4, etc. A salt whose cation is the conjugate acid of a weak base produces an acidic solution when dissolved in water.

Examples: NH4Cl, AlCl3, Fe(NO3)3, etc. Molecular model: Na+, C2H3O2-, and H2O Like Example 7.11 (P255) - I Calculate the pH of a 0.45M NaCN solution. The Ka value for HCN is 6.2 x 10-10. Solution: Since HCN is a weak acid, the Cyanide ion must have significant affinity for protons. CN-(aq) + H2O(l) HCN(aq) + OH-(aq) [HCN][OH-] Kb =

[CN-] Kw Kb = = Ka (for HCN) Initial Concentration (mol/L) [CN-]0 = 0.45 [HCN]0 = 0 [OH-]0 = 0 The value of Kb can be calculated from Kw and the Ka value for HCN. 1.0 x 10-14 6.2 x 10-10 = 1.61 x 10-5

Equilibrium Concentration (mol/L) X mol/L CN- reacts with H2O to reach equilibrium [CN-] = 0.45 x [HCN] = x [OH-] = x Like Example 7.11 (P255) - II Thus: [HCN][OH-] (x)(x) ~ x2 Kb = 1.61 x 10 =

= = [CN ] 0.45 - x 0.45 -5 ~ 2.69 x 10-3 x= Although this is not exactly valid by the 5% rule, it is only off by 1%, so we will use it for now! x = [OH-] = 2.69 x 10-3 M pOH = -log[OH-] = 3 0.43 = 2.57

pH = 14.00 2.57 =____________ Like Example 7.12 (P 255-6) - I Calculate the pH of a 0.010 M AlCl3 solution. The Ka value for the Al(H2O)63+ ion is 1.4 x 10-5. Solution: Since the Al(H2O)63+ ion is a stronger acid than water, the dominate equilibrium will be: Al(H2O)63+(aq) Al(OH)(H2O)52+(aq) + H+(aq) 2+ + [Al(OH)(H O) ][H ]

2 5 -5 1.4 x 10 = Ka = [Al(H2O)63+] Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [Al(H2O)63+]0 = 0.010 [Al(OH)(H2O)52+] = 0 ~ + [H ]0 = 0 X mol/L Al(H2O)63+

Dissociates to reach equilibrium [Al(H2O)63+] = 0.010 x [Al(OH)(H2O)52+] = x [H+] = x Like Example 7.12 (P 255-6) - II Thus: 1.4 x 10-5 = Ka = Ka = [Al(OH)(H2O)52+][H+] [Al(H2O)63+]

(x) (x) 0.010 - x = = x2 0.010 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 = ___________

Ka Values of Some Hydrated Metal Ions at 25oC Ion Fe3+ (aq) Sn2+ (aq) Cr3+ (aq) Al3+ (aq) Be2+ (aq) Cu2+ (aq) Pb2+ (aq) Zn2+ (aq) Co2+ (aq) Ni2+ (aq) Ka 6 x 10-3 4 x 10-4

1 x 10-4 1 x 10-5 4 x 10-6 3 x 10-8 3 x 10-8 1 x 10-9 2 x 10-10 1 x 10-10 Increasing acidity Molecular model: C1-, A1(H2O)63+, H2O A pH meter showing that the pH of 0.1 M AICI3 is 2.93

Predicting the Relative Acidity of Salt Solutions Problem: Determine whether an aqueous solution of iron(III) nitrite, Fe(NO2)3, is acidic, basic, or neutral. Plan: The formula consists of the small, highly charged, and therefore weakly acidic, Fe3+ cation and the weakly basic NO2- anion of the weak acid HNO2, To determine the relative acidity of the solution, we write equations that show the reactions of the ions with water, and then find Ka and Kb of the ions to see which ion reacts to form to a greater extent. Solution: Writing the reactions with water: 3+ Fe(H2O)6 (aq) + H2O(l) Fe(H2O)5OH2+(aq) + H3O+(aq) NO2-(aq) + H2O(l) HNO2(aq) + OH -(aq)

Obtaining Ka and Kb of the ions: For Fe3+(aq) Ka = 6 x 10-3. For NO2-(aq), Kb must be determined: -14 K 1.0 x 10 -11 w Kb of NO2 = = = 1.4 x 10

Ka of HNO2 7.1 x 10-4 Since Ka of Fe3+ > Kb of NO2-, the solution is acidic. Electron-Pair Donation and the Lewis Acid-Base Definition The Lewis acid-base definition : A base is any species that donates an electron pair. An acid is any species that accepts an electron pair. Protons act as Lewis acids in that they accept an electron pair in all reactions: .. B +

H+ B H+ The product of any Lewis acid-base reaction is called an adduct, a single species that contains a new covalent bond: A Lewis base is a lone pair of electrons to donate. A Lewis acid is a vacant orbital Metal Cations as Lewis Acids M2+ +

4 H2O(l) M(H2O)42+ Metal ions can accept electron pairs from water molecules to form complexes. An example is nickel which forms an adduct with water to form the hexa aqua complex: Ni2+ + 6 H2O(l) Ni(H2O)62+(aq) Ammonia is a stronger Lewis base than water because it displaces water from hydrated ions when aqueous ammonia is added. Ni(H2O)62+(aq) + 6 NH3 (aq) Ni(NH3)62+(aq) + 6 H2O(aq) Many essential biomolecules are Lewis adducts with central metal

ions. Chlorophyll is a Lewis adduct of a central Mg2+ ion. Vitamin B12 has a similar structure with a central Co3+, as does heme with a central Fe2+ ion. Other metals such as Zn2+, Mo2+, and Cu2+ are bound to the active site in enzymes and participate in the catalytic action by virtue of The Mg2+ Ion as a Lewis Acid in the Chlorophyll Molecule Identifying Lewis Acids and Bases Problem: Identify the Lewis acids and bases in the following reactions: (a) F- + BF3 (b) Co2+ + 6 H2O (c) NH3 + H+ BF4Co(H2O)62+

NH4+ Plan: We examine the species to see which species accepts the electron pair (Lewis acid) and which donates it (Lewis base) in the reactions. Solution: (a) The BF3 accepted an electron pair from the fluoride ion. BF3 is the acid and F- is the base. (b) The Co2+ ion accepted the electron pairs from the water molecules. Co2+ is the acid and H2O is the base. (c) The H+ ion accepted the electron pair from the ammonia molecule. H+ is the acid and water is the base. Example 7.14 (P259) Predict whether an aqueous solution of each of the following salts will be acidic, basic, or neutral.

a) NH4C2H3O2 b) NH4CN c) Al2(SO4)3 lution: The ions are the ammonium and acetate ions, Ka for NH4+ is 5.6 x 10-10, and Kb for C2H3O2- is 5.6 x 10-10 . Since the are equal the solution will be neutral and the pH close to 7. The solution will contain the ammonium and cyanide ions, the Ka value for NH4+ is 5.6 x 10-10, and Kw Kb (for CN ) = = 1.6 x 10-5 Ka(for HCN) Since Kb for CN- is much larger than Ka for NH4+, this solution

will be basic. This solution copntains the hydrated Aluminum ion and the sulfate ion. Ka for Al(H2O)63+ = 1.4 x 10-5, for sulfate, Kb = 8.3 x 10-13 ; Molecular model: Na+, F-, H2O Molecular model: C1-, NH4+, and H2O A pH meter showing that the pH of 0.1 M HN4CI is 5.13 Table 7.6 (P 259) - I Acid Base Properties of Aqueous Solutions of Various Types of Salts Type of Salt Cation is from strong base;

anion is from strong acid Examples KCl, KNO3 NaCl, NaNO3 Cation is from strong base; anion is from weak acid Cation is conjugate acid of weak base; anion is from NaC2H3O2

KCN, NaF NH4Cl, NH4NO3 Comment Neither acts as an acid or a base Anion acts as a base; cation has no effect on pH Cation acts as an acid; anion

has no effect pH of Solution Neutral Basic Acidic Table 7.6 (P 259) - II Acid Base Properties of Aqueous Solutions of Various Types of Salts Type of Salt Cation is conjugate acid of weak base anion is conjugate

base of weak acid Cation is highly charged metal ion; anion is from strong acid Examples NH4C2H3O2 NH4CN Al(NO3)3, FeCl3 Comment Cation acts as

an acid; anion acts as a base Hydrated cation acts as an acid; anion has no effect on pH pH of Solution Acidic if : K a > Kb Basic if : K b > Ka Neutral if : K a = Kb

Acidic Value of Ka - I We start with the expression for the value of Ka , for the weak acid HA: From the conservation of charge equation: [H+] [A-] [H+] = [A-] + [OH-] Ka = [HA] From the Kw expression for water: Kw [OH ] = + [H ] The charge balance equation becomes: Kw

[Kw] + + or: [H ] = [A ] + [A ] = [H ] + [H ] [H+] The material balance equation is: [HA]0 = [HA] + [A-] Since: or Kw +

[A ] = [H ] [H+] [HA] = [HA]0 [A-] We have: Kw [HA] = [HA]0 ([H ] - + ) [H ] + Value of Ka - II Now we substitute the expressions for [A-] and [HA] into Ka: Kw [H ]( [H ] ) + + [H ] [A ]

[H ] Ka = = [HA] Kw [HA]0 ( [H+] ) + [H ] + + [H+]2 - Kw Ka = [HA]0 -

[H+]2 - Kw [H+] Simplified, this equation becomes: [H+]2 Ka = [HA] [H+] 0 = Like Example 7.16 (P264-5) -I Calculate the [H+] in: a) 1.0 M HOCl and b) 1 x 10-4 M HOCl for hypochlorous acid HOCl , Ka = 3.5 x 10-8 a) First do the weak acid problem in the normal way. 2

x2 x -8 ~ = 3.5 x 10 = 1.0 x 1.0 X = 1.87 x 10-4 M = [H+] b) First we do the weak acid problem in the normal way. 2 x2 x -8 ~ = 3.5 x 10 = -4

1.0 x 10 x 1.0 x 10-4 X = 1.87 x 10-6 M = [H+] In this very dilute solution of HOCl we should use the full equation to obtain the correct H+ concentration. Ka = 3.5 x 10 = -8 [H+]2 10-14 + 2 -14 [H ]

10 -4 1.0 x 10 [H+] Like Example 7.16 (P264-5) -II To solve this we will use successive approximations, first substituting the value we obtained in the normal way: 1.87 x 10-6 M To do this we add in the correction for water ionization, 1.0 x 10-7 M, giving as an approximation: 1.97 x 10-6M for H+. [H+]2 10-14 Ka = 3.5 x 10 = -6 2 -14 (1.97 x 10

) 1.0 x 10 -4 1.0 x 10 1.97 x 10-6 + 2 -14 + 2 -14 [H ] 10 [H

] 10 3.5 x 10-8 = = 9.8 x 10-5 1.0 x 10-4 1.97 x 10-6 -8 [H+]2 = 3.44 x 10-12 [H+] = 1.85 x 10-6 Substituting 1.85 x 10-6 in to the equation in place of 1.97 x 10-6 yields 1.85 x 10-6 M so the approximation yields the same answer, so the final answer is 1.85 x 10-6 M.

pH = - log(1.85 x 10-6) = __________ Summary: The pH Calculations for an Aqueous Solution of a Weak Acid HA (major species HA and H2O) The full equation for this case is: [H+]2 - Kw Ka = [H+]2 Kw [HA]0 [H+] When the weak acid by itself produces [H+] > 10-6 M,the full equation becomes:

This corresponds to the typical [H+]2 Ka = weak acid case: [HA]0 [H+] When: [H+]2 - Kw the full equation [H+]2 - Kw [HA]0>> Ka = + becomes: [H ] [HA]0

Which gives: [H+] = Ka[HA]0 + Kw 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 acid 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) Compute the initial concentrations (assuming that the dominant equilibrium has not yet occurred-for example, there has been no acid dissociation). c) Define x. d) Compute the equilibrium concentrations in terms of x. e) Substitute the concentrations into the equilibrium expression, and solve for x. f) Check the validity of the approximation. g) Calculate the pH and other concentrations as required.

Recently Viewed Presentations

  • Multimodality Therapy for High Risk Disease - The Surgical ...

    Multimodality Therapy for High Risk Disease - The Surgical ...

    MULTI-MODALITY TREATMENT. IMPACT testing (on biopsy): 34 somatic mutations. ... Establishing a Cure Paradigm for Locally Advanced and Low Volume Metastatic Disease. Multimodality Therapy. Clinical trials with neoadjuvant enzalutamide and/or abiraterone.
  • Data Liberation Initiative Please Use Our Data! IASSIST

    Data Liberation Initiative Please Use Our Data! IASSIST

    Thanks to: Chuck Humphrey, University of Alberta Wendy Watkins, Carleton University Nancy Lemay, University of Ottawa IASSIST 2007 Overview Background The Outcome The Survival Kit, section by section Next Steps A Sneak Peak Background Data Liberation Initiative (DLI) Statistics Canada...
  • Slide 1

    Slide 1

    Specifically, the proposed development at Clifton Gate is for approximately 3400 dwellings, rather than 100-odd, and is intended to be delivered as a sustainable new settlement. It therefore includes local facilities, social infrastructure and convenience retail, as well as a...
  • Metathesis Problems (and Some Solutions) Identified Through ...

    Metathesis Problems (and Some Solutions) Identified Through ...

    Times Helvetica Arial Symbol Microsoft Office 98 Chapter 11: Chemical Bonding Chemical bonds Charges on ions Covalent bonds Electronegativity Dipole moment Lewis structures Writing Lewis structures Multiple-bonds A few exceptions to the octet rule 3-dimensional molecular structure Electron pair arrangements...
  • Chapter 3

    Chapter 3±P. 3.2e+5 is equivalent to 320,000 . 1.76e-3 is equivalent to .00176 . OR in standard decimal notation. Default type is double . Table 2-10 . Values and sizes for floating-point types. C# Programming: From Problem Analysis to Program Design.
  • * Veraz proprietary information notice: This document and

    * Veraz proprietary information notice: This document and

    * Veraz proprietary information notice: This document and the contents therein are the property of Veraz Networks Inc. Any duplication, reproduction, or transmission to unauthorized parties without prior written permission of Veraz Networks Inc. is prohibited.
  • Enterprise Resource Planning April 25, 2007

    Enterprise Resource Planning April 25, 2007

    Enterprise Resource Planning Wisnu Cahyono Wenbin Li Erica Price Jurlian Sitanggang April 25, 2007 Overview Introduction Suppliers of ERP Implementation of ERP Case Studies ERP Evolution What is ERP?
  • Sustainable Transport and Key Infrastructure in BEST METROPOLISES

    Sustainable Transport and Key Infrastructure in BEST METROPOLISES

    Activity 5 Transport, job accessibility and daily mobility Some results (comparison between three metropolises): Major trends in modal split New solutions Public transport at the boundary of the city Paris Successful in reducing the car share and increasing the public...