Chapter 18 Acid-Base Equilibria
Acid-Base Equilibria 18.1 Acids and Bases in Water 18.2 Autoionization of Water and the pH Scale 18.3 Proton Transfer and the Brønsted-Lowry Acid-Base Definition 18.4 Solving Problems Involving Weak-Acid Equilibria 18.5 Weak Bases and Their Relation to Weak Acids 18.6 Molecular Properties and Acid Strength 18.7 Acid-Base Properties of Salt Solutions 18.8 Generalizing the Brønsted-Lowry Concept: The Leveling Effect 18.9 Electron-Pair Donation and the Lewis Acid-Base Definition
Arrhenius Acid-Base Definition This is the earliest acid-base definition, which classifies these substances in terms of their behavior in water. An acid is a substance with H in its formula that dissociates to yield H3O+. A base is a substance with OH in its formula that dissociates to yield OH-. When an acid reacts with a base, they undergo neutralization: H+(aq) + OH-(aq) → H2O(l) DH°rxn = -55.9 kJ
Strong and Weak Acids A strong acid dissociates completely into ions in water: HA(g or l) + H2O(l) → H3O+(aq) + A-(aq) A dilute solution of a strong acid contains no HA molecules. A weak acid dissociates slightly to form ions in water: HA(aq) + H2O(l) H3O+(aq) + A-(aq) In a dilute solution of a weak acid, most HA molecules are undissociated. [H3O+][A-] [HA][H2O] Kc = has a very small value.
Figure 18.1A The extent of dissociation for strong acids. Strong acid: HA(g or l) + H2O(l) → H3O+(aq) + A-(aq) There are no HA molecules in solution.
Figure 18.1B The extent of dissociation for weak acids. Weak acid: HA(aq) + H2O(l) H3O+(aq) + A-(aq) Most HA molecules are undissociated.
Figure 18.2 Reaction of zinc with a strong acid (left) and a weak acid (right). Zinc reacts rapidly with the strong acid, since [H3O+] is much higher. 1 M HCl(aq) 1 M CH3COOH(aq)
The Acid Dissociation Constant, Ka HA(aq) + H2O(l) H3O+(aq) + A-(aq) [H3O+][A-] [HA][H2O] Kc = The value of Ka is an indication of acid strength. [H3O+][A-] [HA] Kc[H2O] = Ka = Stronger acid Weaker acid higher [H3O+] lower % dissociation of HA larger Ka smaller Ka
Classifying the Relative Strengths of Acids • Strongacids include • the hydrohalic acids (HCl, HBr, and HI) and • oxoacids in which the number of O atoms exceeds the number of ionizable protons by two or more (eg., HNO3, H2SO4, HClO4.) • Weak acids include • the hydrohalic acid HF, • acids in which H is not bonded to O or to a halogen (eg., HCN), • oxoacids in which the number of O atoms equals or exceeds the number of ionizable protons by one (eg., HClO, HNO2), and • carboxylic acids, which have the general formula RCOOH (eg., CH3COOH and C6H5COOH.)
Classifying the Relative Strengths of Bases • Strongbases include • water-soluble compounds containing O2- or OH- ions. • The cations are usually those of the most active metals: • M2O or MOH, where M = Group 1A(1) metal (Li, Na, K, Rb, Cs) • MO or M(OH)2 where M = group 2A(2) metal (Ca, Sr, Ba). • Weak bases include • ammonia (NH3), • amines, which have the general formula • The common structural feature is an N atom with a lone electron pair.
PROBLEM: Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base. (a) KOH (b) (CH3)2CHCOOH (c) H2SeO4(d) (CH3)2CHNH2 PLAN: We examine the formula and classify each acid or base, using the text descriptions. Particular points to note for acids are the numbers of O atoms relative to ionizable H atoms and the presence of the –COOH group. For bases, note the nature of the cation or the presence of an N atom that has a lone pair. Sample Problem 18.1 Classifying Acid and Base Strength from the Chemical Formula SOLUTION: (a) Strong base: KOH is one of the group 1A(1) hydroxides.
Sample Problem 18.1 (b)Weak acid: (CH3)2CHCOOH is a carboxylic acid, as indicated by the –COOH group. The –COOH proton is the only ionizable proton in this compound. (c)Strong acid: He2SO4 is an oxoacid in which the number of atoms exceeds the number of ionizable protons by two. (d)Weak base: (CH3)2CHNH2 has a lone pair of electrons on the N and is an amine.
Autoionization of Water Water dissociates very slightly into ions in an equilibrium process known as autoionization or self-ionization. 2H2O (l) H3O+(aq) + OH- (aq)
The Ion-Product Constant for Water (Kw) [H3O+][A-] [H2O]2 Kc = Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0x10-14 (at 25°C) = 1.0x10-7 (at 25°C) In pure water, [H3O+] = [OH-] = Both ions are present in all aqueous systems. 2H2O (l) H3O+(aq) + OH- (aq)
A change in [H3O+] causes an inverse change in [OH-], and vice versa. We can define the terms “acidic” and “basic” in terms of the relative concentrations of H3O+ and OH- ions: In an acidic solution, [H3O+] > [OH-] In a neutral solution, [H3O+] = [OH-] In basic solution, [H3O+] < [OH-] Higher [H3O+] Higher [OH-] lower [OH-] lower [H3O+]
Figure 18.3 The relationship between [H3O+] and [OH-] and the relative acidity of solutions.
Sample Problem 18.2 Calculating [H3O+] or [OH-] in an Aqueous Solution PROBLEM: A research chemist adds a measured amount of HCl gas to pure water at 25°C and obtains a solution with [H3O+] = 3.0x10-4 M. Calculate [OH-]. Is the solution neutral, acidic, or basic? PLAN: We use the known value of Kw at 25°C (1.0x10-14) and the given [H3O+] to solve for [OH-]. We can then compare [H3O+] with [OH-] to determine whether the solution is acidic, basic, or neutral. SOLUTION: 1.0x10-14 3.0x10-4 Kw [H3O+] = Kw = 1.0x10-14 = [H3O+] [OH-] so [OH-] = = 3.3x10-11 M [H3O+] is > [OH-] and the solution is acidic.
The pH Scale pH = -log[H3O+] The pH of a solution indicates its relative acidity: In an acidic solution, pH < 7.00 In a neutral solution, pH = 7.00 In basic solution, pH > 7.00 The higher the pH, the lowerthe [H3O+] and the less acidic the solution.
Figure 18.4 The pH values of some familiar aqueous solutions. pH = -log [H3O+]
Table 18.3 The Relationship between Ka and pKa pKa = -logKa A low pKacorresponds to a high Ka.
pH, pOH, and pKw Kw = [H3O+][OH-] = 1.0x10-14 at 25°C pH = -log[H3O+] pOH = -log[OH-] pKw = pH + pOH = 14.00 at 25°C pH + pOH = pKw for any aqueous solution at any temperature. • Since Kw is a constant, the values of pH, pOH, [H3O+], and [OH-] are interrelated: • If [H3O+] increases, [OH-] decreases (and vice versa). • If pH increases, pOH decreases (and vice versa).
Figure 18.5 The relations among [H3O+], pH, [OH-], and pOH.
PLAN: HNO3 is a strong acid so it dissociates completely, and [H3O+] = [HNO3]init. We use the given concentrations and the value of Kw at 25°C to find [H3O+] and [OH-]. We can then calculate pH and pOH. Sample Problem 18.3 Calculating [H3O+], pH, [OH-], and pOH PROBLEM: In an art restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated HNO3 to 2.0 M, 0.30 M, and 0.0063 M HNO3. Calculate [H3O+], pH, [OH-], and pOH of the three solutions at 25°C. SOLUTION: Calculating the values for 2.0 M HNO3: [H3O+] = 2.0 M pH = -log[H3O+] = -log(2.0) = -0.30 = 5.0x10-15M [OH-] = = 1.0x10-14 2.0 Kw [H3O+] pOH = -log[OH-] = -log(5.0x10-15) = 14.30
Sample Problem 18.3 Calculating the values for 0.30 M HNO3: [H3O+] = 0.30 M pH = -log[H3O+] = -log(0.30) = 0.52 = 3.3x10-14M pOH = -log[OH-] = -log(3.3x10-14) = 13.48 Calculating the values for 0.0063 M HNO3: [H3O+] = 0.0063 M pH = -log[H3O+] = -log(0.30) = 2.20 = 1.6x10-12M pOH = -log[OH-] = -log(1.6x10-12) = 11.80 [OH-] = [OH-] = = = 1.0x10-14 0.30 1.0x10-14 0.0063 Kw [H3O+] Kw [H3O+]
pH paper pH meter Figure 18.6 Methods for measuring the pH of an aqueous solution.
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. • A base is a proton acceptor, any species that accepts an H+ ion. • A base must contain a lone pair of electrons to bond to H+. An acid-base reaction is a proton-transfer process.
Figure 18.7 Dissolving of an acid or base in water as a Brønsted-Lowry acid-base reaction. Lone pair binds H+ (acid, H+ donor) (base, H+ acceptor) Lone pair binds H+ (base, H+ acceptor) (acid, H+ donor)
Conjugate Acid-Base Pairs In the forward reaction: NH3accepts a H+ to form NH4+. H2S + NH3HS- + NH4+ H2S + NH3HS- + NH4+ H2S donates a H+ to form HS-. In the reverse reaction: NH4+donates a H+ to form NH3. HS-accepts a H+ to form H2S.
Conjugate Acid-Base Pairs H2S + NH3HS- + NH4+ acid1 + base2base1 + acid2 H2S and HS- are a conjugate acid-base pair: HS- is the conjugate base of the acid H2S. NH3 and NH4+ are a conjugate acid-base pair: NH4+ is the conjugate acid of the base NH3. A Brønsted-Lowry acid-base reaction occurs when an acid and a base react to form their conjugate base and conjugate acid, respectively.
Acid + Base Base + Acid Reaction 1 HF + H2O F- + H3O+ Reaction 2 HCOOH + CN- HCOO- + HCN Reaction 3 NH4+ + CO32- NH3 + HCO3- Reaction 4 H2PO4- + OH- HPO42- + H2O Reaction 5 H2SO4 + N2H5+ HSO4- + N2H62+ Reaction 6 HPO42- + SO32- PO43- + HSO3- Table 18.4 The Conjugate Pairs in some Acid-Base Reactions Conjugate Pair Conjugate Pair
PROBLEM: The following reactions are important environmental processes. Identify the conjugate acid-base pairs. PLAN: To find the conjugate pairs, we find the species that donated an H+ (acid) and the species that accepted it (base). The acid donates an H+ to becomes its conjugate base, and the base accepts an H+ to becomes it conjugate acid. Sample Problem 18.4 Identifying Conjugate Acid-Base Pairs (a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq) (b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq) SOLUTION: (a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq) acid1 base2 base1 acid2 The conjugate acid-base pairs are H2PO4-/HPO42- and CO32-/HCO3-.
Sample Problem 18.4 (b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq) acid1 base2 base1 acid2 The conjugate acid-base pairs are H2O/OH- and SO32-/HSO3-.
Net Direction of Reaction The net direction of an acid-base reaction depends on the relative strength of the acids and bases involved. H2S + NH3HS- + NH4+ A reaction will favor the formation of the weaker acid and base. stronger acid weaker base weaker acid stronger base This reaction favors the formation of the products.
Figure 18.8 Strengths of conjugate acid-base pairs. The stronger the acid is, the weaker its conjugate base. When an acid reacts with a base that is farther down the list, the reaction proceeds to the right (Kc > 1).
PROBLEM: Predict the net direction and whether Kc is greater or less than 1 for each of the following reactions (assume equal initial concentrations of all species): PLAN: We identify the conjugate acid-base pairs and consult figure 18.8 to see which acid and base are stronger. The reaction favors the formation of the weaker acid and base. Sample Problem 18.5 Predicting the Net Direction of an Acid-Base Reaction (a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq) (b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq) SOLUTION: (a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq) stronger acid stronger base weaker base weaker acid The net direction for this reaction is to the right, so Kc > 1.
Sample Problem 18.5 (b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq) weaker acid weaker base stronger base stronger acid The net direction for this reaction is to the left, so Kc < 1.
PLAN: A stronger acid and base yield a weaker acid and base, so we have to determine the relative acid strengths of HX and HY to choose the correct molecular scene. The concentrations of the acid solutions are equal, so we can recognize the stronger acid by comparing the pH values of the two solutions. Sample Problem 18.6 Using Molecular Scenes to Predict the Net Direction of an Acid-Base Reaction PROBLEM: Given that 0.10 M of HX (blue and green) has a pH of 2.88, and 0.10 M HY (blue and orange) has a pH 3.52, which scene best represents the final mixture after equimolar solutions of HX and Y- are mixed?
Sample Problem 18.6 SOLUTION: The HX solution has a lower pH than the HY solution, so HX is the stronger acid and Y- is the stronger base. The reaction of HX and Y- has a Kc> 1, which means the equilibrium mixture will contain more HY than HX. Scene 1 has equal numbers of HX and HY, which could occur if the acids were of equal strength. Scene 2 shows fewer HY than HX, which would occur if HY were the stronger acid. Scene 3 is consistent with the relative acid strengths, because it contains more HY than HX.
Solving Problems Involving Weak-Acid Equilibria Problem-solving approach Write a balanced equation. Write an expression for Ka. Define x as the change in concentration that occurs during the reaction. Construct a reaction table in terms of x. Make assumptions that simplify the calculation. Substitute values into the Ka expression and solve for x. Check that the assumptions are justified.
Solving Problems Involving Weak-Acid Equilibria • The notation system • Molar concentrations are indicated by [ ]. • A bracketed formula with no subscript indicates an equilibrium concentration. • The assumptions • [H3O+] from the autoionization of H2O is negligible. • A weak acid has a small Ka and its dissociation is negligible. [HA] ≈ [HA]init.
PROBLEM: Phenylacetic acid (C6H5CH2COOH, simplified here as HPAc) builds up in the blood of persons with phenylketonuria, an inherited disorder that, if untreated, causes mental retardation and death. A study of the acid shows that the pH of 0.12 M HPAc is 2.62. What is the Ka of phenylacetic acid? PLAN: We start with the balanced dissociation equation and write the expression for Ka. We assume that [H3O+] from H2 is negligible and use the given pH to find [H3O+], which equals [PAc-] and [HPAc]dissoc. We assume that [HPAc] ≈ [HPAc]init because HPAc is a weak acid. SOLUTION: HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq) [H3O+][PAc-] Ka = [HPAc] Sample Problem 18.7 Finding Ka of a Weak Acid from the Solution pH
Initial 0.12 - 0 0 Change - x - + x +x Equilibrium 0.12 - x - x x (2.4x10-3) (2.4x10-3) 0.12 1x10-7 M [H3O+] = x 100 2.4x10-3 M 2.4x10-3 M [HPAc]dissoc = x 100 0.12 M Sample Problem 18.7 Concentration (M) HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq) [H3O+] = 10-pH = 2.4x10-3 M which is >> 10-7 (the [H3O+] from water) x ≈ 2.4x10-3 M ≈ [H3O+] ≈ [PAc-] [HPAc] = 0.12 - x ≈ 0.12 M So Ka = = 4.8x10-5 Checking the assumptions by finding the percent error in concentration: = 4x10-3 % (<5%; assumption is justified). from H2O = 2.0% (<5%; assumption is justified).
PLAN: We write a balanced equation and the expression for Ka. We know [HPr]init but not [HPr] (i.e., the concentration at equilibrium). We define x as [HPr]dissoc and set up a reaction table. We assume that, since HPr has a small Ka value, it dissociates very little and therefore [HPr] ≈ [HPr]init. Ka = [H3O+][Pr−] [HPr] Sample Problem 18.8 Determining Concentration from Ka and Initial [HA] PROBLEM: Propanoic acid (CH3CH2COOH, which we simplify as HPr) is a carboxylic acid whose salts are used to retard mold growth in foods. What is the [H3O+] of 0.10 M HPr (Ka = 1.3x10−5)? HPr(aq) + H2O(l) H3O+(aq) + Pr−(aq) SOLUTION:
[H3O+][Pr-] x2 Ka = 1.3x10-5 = = 0.10 [HPr] HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq) Initial 0.10 - 0 0 Change −x - +x +x Equilibrium 0.10 - x - x x Sample Problem 18.8 Concentration (M) Since Ka is small, we will assume that x<< 0.10 and [HPr] ≈ 0.10 M. = 1.1x10-3 M = [H3O+] x = 1.1x10-3M 0.10 M Check: [HPr]diss = x 100 = 1.1% (< 5%; assumption is justified.)
[HA]dissoc Percent HA dissociated = x 100 [HA]init Concentration and Extent of Dissociation HA(aq) + H2O(l) H3O+(aq) + A-(aq) As the initial acid concentration decreases, the percent dissociation of the acid increases. A decrease in [HA]init means a decrease in [HA]dissoc = [H3O+] = [A-], causing a shift towards the products. The fraction of ions present increases, even though the actual [HA]dissoc decreases.
PLAN: We are given the percent dissociation of the original HA solution (33%), and we know that the percent dissociation increases as the acid is diluted. Thus, we calculate the percent dissociation of each diluted sample and see which is greater than 33%. Sample Problem 18.9 Using Molecular Scenes to Determine the Extent of HA Dissociation PROBLEM: A 0.15 M solution of acid HA (blue and green) is 33% dissociated. Which scene best represents a sample of the solution after it is diluted with water?
[H3O+] x 100 % dissociation = [HA] + [H3O+] Sample Problem 18.9 SOLUTION: In each case, remember that each unit of H3O+ is produced from one unit of HA, so [HA]init = [HA] + [H3O+]. Solution 1. % dissociated = 4/(5 + 4) x 100 = 44% Solution 2. % dissociated = 2/(7 + 2) x 100 = 22% Solution 3. % dissociated = 3/(6 + 3) x 100 = 33% Scene 1 represents the diluted solution.
[H3O+][HPO42-] [H3O+][H2PO4-] [H3O+][PO43-] Ka1 = Ka3 = Ka2 = [HPO42-] [H3PO4] [H2PO4-] Polyprotic Acids A polyprotic acid is an acid with more than one ionizable proton. In solution, each dissociation step has a different value for Ka: H3PO4(aq) + H2O(l) H2PO4-(aq) + H3O+(aq) = 7.2x10-3 = 6.3x10-8 H2PO4-(aq) + H2O(l) HPO42-(aq) + H3O+(aq) HPO42-(aq) + H2O(l) PO43-(aq) + H3O+(aq) Ka1 > Ka2 > Ka3 = 4.2x10-13 We usually neglect [H3O+] produced after the first dissociation.