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 Relations 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

The Nature of Acids and Bases: Acids: Acids taste sour. React with metals and produce H gas. turns blue litmus red pH < 7 * Bases: Bases taste bitter. They are slippery. turns red litmus blue. pH >7

Arrhenius concept Acids produce H+ ions. Bases produce OH- ions. This concept is limited because it applies to only aqueous solution and defines only OH containing bases.

Neutralization: acid + base _______salt + water.

Acid Dissociation Constant (Ka) Write Ka expression for strong and weak acids.

The extent of dissociation for strong acids. Figure 18.1 The extent of dissociation for strong acids. Strong acid: HA(g or l) + H2O(l) H2O+(aq) + A-(aq)

The extent of dissociation for weak acids. Figure 18.2 The extent of dissociation for weak acids. Weak acid: HA(aq) + H2O(l) H2O+(aq) + A-(aq)

Reaction of zinc with a strong and a weak acid. Figure 18.3 Reaction of zinc with a strong and a weak acid. 1M HCl(aq) 1M CH3COOH(aq)

The Acid-Dissociation Constant Strong acids dissociate completely into ions in water. HA(g or l) + H2O(l) H3O+(aq) + A-(aq) Kc >> 1 Weak acids dissociate very slightly into ions in water. HA(aq) + H2O(l) H3O+(aq) + A-(aq) Kc << 1 The Acid-Dissociation Constant Kc = [H3O+][A-] [H2O][HA] stronger acid higher [H3O+] larger Ka Kc[H2O] = Ka = [H3O+][A-] [HA] smaller Ka lower [H3O+] weaker acid

Classifying the Relative Strengths of acids and Bases: Strong acids: HCl, HBr, HI Oxo acids. HNO3, H2SO4, HClO4 Weak acids: HF HCN , H2S (H not bonded to O or halogen) Oxo acids. HClO, HNO2, H3PO4 Carboxylic acids. CH3COOH

Classifying the Relative Strengths of acids and Bases: Strong bases: M2O, MOH M= (group 1A metal) MO , M(OH)2 M=group 2A metal Weak bases: (N atom and lone pair of electrons) NH3 Amines.

ACID STRENGTH

SAMPLE PROBLEM 18.1: Classifying Acid and Base Strength from the Chemical Formula PROBLEM: Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base. (a) H2SeO4 (b) (CH3)2CHCOOH (c) KOH (d) (CH3)2CHNH2 PLAN: Pay attention to the text definitions of acids and bases. Look at O for acids as well as the -COOH group; watch for amine groups and cations in bases. SOLUTION: (a) Strong acid - H2SeO4 - the number of O atoms exceeds the number of ionizable protons by 2. (b) Weak acid - (CH3)2CHCOOH is an organic acid having a -COOH group. (c) Strong base - KOH is a Group 1A(1) hydroxide. (d) Weak base - (CH3)2CHNH2 has a lone pair of electrons on the N and is an amine.

Autoionization of water and the pH scale Water dissociates into its ions.

Autoionization of Water and the pH Scale H2O(l) H2O(l) + OH-(aq) H3O+(aq) +

H2O(l) + H2O(l) H3O+(aq) + OH-(aq) Kc = [H3O+][OH-] [H2O]2 The Ion-Product Constant for Water Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0 x 10-14 at 250C A change in [H3O+] causes an inverse change in [OH-]. In an acidic solution, [H3O+] > [OH-] In a basic solution, [H3O+] < [OH-] In a neutral solution, [H3O+] = [OH-]

The relationship between [H3O+] and [OH-] and the relative acidity of solutions. Figure 18.4 Divide into Kw [H3O+] [OH-] [H3O+] > [OH-] [H3O+] = [OH-] [H3O+] < [OH-] ACIDIC SOLUTION NEUTRAL SOLUTION BASIC SOLUTION

SAMPLE PROBLEM 18.2: Calculating [H3O+] and [OH-] in an Aqueous Solution PROBLEM: A research chemist adds a measured amount of HCl gas to pure water at 250C and obtains a solution with [H3O+] = 3.0x10-4M. Calculate [OH-]. Is the solution neutral, acidic, or basic? PLAN: Use the Kw at 250C and the [H3O+] to find the corresponding [OH-]. SOLUTION: Kw = 1.0x10-14 = [H3O+] [OH-] so [OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 = 3.3x10-11M [H3O+] is > [OH-] and the solution is acidic.

pH scale pH  log[H+] pH in water ranges from 0 to 14. Kw = 1.00  1014 = [H+] [OH] pKw = 14.00 = pH + pOH As pH rises, pOH falls (sum = 14.00). pH=7-neutral, pH>7-basic, pH<7-acidic

pH scale Acidic solns have a higher pOH. pK=-log K equation reaches equ, mostly products present, low pK (high K) Reverse of the above statement is also true. pH is measure using pH meter, pH paper or acid-base indicator.

Methods for measuring the pH of an aqueous solution Figure 18.7 Methods for measuring the pH of an aqueous solution pH (indicator) paper pH meter

The pH values of some familiar aqueous solutions Figure 18.5 The pH values of some familiar aqueous solutions pH = -log [H3O+]

Table 18.3 The Relationship Between Ka and pKa Acid Name (Formula) Ka at 250C pKa Hydrogen sulfate ion (HSO4-) 1.02x10-2 1.991 3.15 Nitrous acid (HNO2) 7.1x10-4 4.74 Acetic acid (CH3COOH) 1.8x10-5 2.3x10-9 8.64 Hypobromous acid (HBrO) Phenol (C6H5OH) 1.0x10-10 10.00

The relations among [H3O+], pH, [OH-], and pOH. Figure 18.6 The relations among [H3O+], pH, [OH-], and pOH.

Problems P.3.Calculate pH and pOH at 25 C for: 1.0 M H+ P.4.pH=6.88, Calculate [ H+] and [ OH-] for this sample. P.5.Calculate pH of 0.10 M HNO3 P.6.Calculate pH of 1.0 x 10-10 HCl.

Bronsted-Lowry model: An acid is a proton (H+ ) donor. A base is a proton acceptor. * A conjugate acid-base pair only differs by one H. HCl + H2O _________ H3O+ Cl- Acid base conj acid conj base HA(aq) + H2O (l)  H3O+ (aq) +A-(aq) CA1 CB2 CA2 CB1

Bronsted-Lowry model: conjugate base: everything that remains of the acid molecule after a proton is lost. Has one H less and one more minus charge than the acid. conjugate acid: formed when the proton is transferred to the base. Has one more H and one less charge than the base. Do Follow up problem-18.4-Page.779

Brønsted-Lowry Acid-Base Definition An acid is a proton donor, any species which donates a H+. A base is a proton acceptor, any species which accepts a H+. An acid-base reaction can now be viewed from the standpoint of the reactants AND the products. An acid reactant will produce a base product and the two will constitute an acid-base conjugate pair.

Table 18.4 The Conjugate Pairs in Some Acid-Base Reactions + Base Acid + Conjugate Pair 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- +

Proton transfer as the essential feature of a Brønsted- Lowry acid-base reaction. Figure 18.8 HCl H2O + Lone pair binds H+ Cl- H3O+ + (acid, H+ donor) (base, H+ acceptor) NH3 H2O + Lone pair binds H+ NH4+ OH- + (base, H+ acceptor) (acid, H+ donor)

SAMPLE PROBLEM 18.4: Identifying Conjugate Acid-Base Pairs PROBLEM: The following reactions are important environmental processes. Identify the conjugate acid-base pairs. (a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq) (b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq) PLAN: Identify proton donors (acids) and proton acceptors (bases). conjugate pair2 conjugate pair1 SOLUTION: (a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq) proton donor proton acceptor proton acceptor proton donor conjugate pair2 conjugate pair1 (b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq) proton donor proton acceptor proton acceptor proton donor

SAMPLE PROBLEM 18.5: Predicting the Net Direction of an Acid-Base Reaction PROBLEM: Predict the net direction and whether Ka is greater or less than 1 for each of the following reactions (assume equal initial concentrations of all species): (a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq) (b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq) PLAN: Identify the conjugate acid-base pairs and then consult Figure 18.10 (button) to determine the relative strength of each. The stronger the species, the more preponderant its conjugate. SOLUTION: (a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq) stronger acid weaker acid stronger base weaker base Net direction is to the right with Kc > 1. (b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq) stronger acid weaker acid stronger base weaker base Net direction is to the left with Kc < 1.

Strengths of conjugate acid-base pairs Figure 18.9 Strengths of conjugate acid-base pairs

Solving Problems Involving Weak-Acid Equilibria. Write balanced equation and Ka expression. Make an ICE table. Make required assumptions. Substitute values and solve for x. Verify assumptions by calculating % error.

Problems P.7. Calculate the pH of 1.00 M aqueous solution of HF. Ka=7.2 x 10-4 P.8 Calculate the pH of 0.100 M aq solution of HOCl. Ka=3.5 x 10-8

SAMPLE PROBLEM 18.6: Finding the Ka of a Weak Acid from the pH of Its Solution 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 0.12M HPAc is 2.62. What is the Ka of phenylacetic acid? PLAN: Write out the dissociation equation. Use pH and solution concentration to find the Ka. Assumptions: With a pH of 2.62, the [H3O+]HPAc >> [H3O+]water. [PAc-] ≈ [H3O+]; since HPAc is weak, [HPAc]initial ≈ [HPAc]initial - [HPAc]dissociation SOLUTION: HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq) Ka = [H3O+][PAc-] [HPAc]

SAMPLE PROBLEM 18.6: Finding the Ka of a Weak Acid from the pH of Its Solution continued Concentration(M) HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq) Initial 0.12 - 1x10-7 Change - -x +x Equilibrium - 0.12-x x x +(<1x10-7) [H3O+] = 10-pH = 2.4x10-3 M which is >> 10-7 (the [H3O+] from water) x ≈ 2.4x10-3 M ≈ [H3O+] ≈ [PAc-] [HPAc]equilibrium = 0.12-x ≈ 0.12 M (2.4x10-3) (2.4x10-3) 0.12 So Ka = = 4.8 x 10-5 [H3O+]from water; 1x10-7M 2.4x10-3M x100 Be sure to check for % error. = 4x10-3 % x100 [HPAc]dissn; 2.4x10-3M 0.12M = 2.0 %

SAMPLE PROBLEM 18.7: Determining Concentrations from Ka and Initial [HA] PROBLEM: Propanoic acid (CH3CH2COOH, which we simplify and HPr) is an organic acid whose salts are used to retard mold growth in foods. What is the [H3O+] of 0.10M HPr (Ka = 1.3x10-5)? PLAN: Write out the dissociation equation and expression; make whatever assumptions about concentration which are necessary; substitute. Assumptions: For HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq) x = [HPr]diss = [H3O+]from HPr= [Pr-] Ka = [H3O+][Pr-] [HPr] SOLUTION: Concentration(M) HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq) Initial 0.10 - Change - -x +x Equilibrium - 0.10-x x Since Ka is small, we will assume that x << 0.10

SAMPLE PROBLEM 18.7: Determining Concentrations from Ka and Initial [HA] continued 1.3x10-5 = [H3O+][Pr-] [HPr] = (x)(x) 0.10 = 1.1x10-3 M = [H3O+] Check: [HPr]diss = 1.1x10-3M/0.10 M x 100 = 1.1%

Percent dissociation: For a weak acid % dissociation increases as the acid becomes more dilute. For solution of any weak acid, [ H+] decreases as [HA]0 decreases, but % dissociation increases as [HA]0 increases.

Problems P.11.Calculate the % dissociation of acetic acid (Ka= 1.8 x 10-5 ) 1.00 M and 0.100 M solutions. P.12.In a 0.100 M HC3H5O3 aqueous solution, lactic acid is 3.7 % dissociated. Calculate ka for the acid.

Percent HA dissociation = [HA]dissociated [HA]initial x 100 Polyprotic acids acids with more than more ionizable proton Ka1 = [H3O+][H2PO4-] [H3PO4] H3PO4(aq) + H2O(l) H2PO4-(aq) + H3O+(aq) = 7.2x10-3 Ka2 = [H3O+][HPO42-] [H2PO4-] H2PO4-(aq) + H2O(l) HPO42-(aq) + H3O+(aq) = 6.3x10-8 Ka3 = [H3O+][PO43-] [HPO42-] HPO42-(aq) + H2O(l) PO43-(aq) + H3O+(aq) = 4.2x10-13 Ka1 > Ka2 > Ka3

ACID STRENGTH

Polyprotic Acids They can furnish more than one proton (H+) to the solution. Characteristics: 1. Ka values are much smaller than the first value , so only the first dissociation step makes a significant contribution to equilibrium concentration of H+ . 2. For sulfuric acid, it behaves as a strong acid in the first dissociation step and a weak acid in the second step, where its contribution of H+ ions can be neglected. Use quadratic equation to solve such kind of a problem.

Problems P.13.Calculate the pH of a 5.0 M H3PO4 solution and the equilibrium concentration of the species H3PO4 , H2PO4-, HPO42-, PO43- P.14.Calculate the pH of 1.0 M sulfuric acid solution.

SAMPLE PROBLEM 18.8: Calculating Equilibrium Concentrations for a Polyprotic Acid PROBLEM: Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1 = 1.0x10-5 and Ka2 = 5x10-12) found in citrus fruit. Calculate [H2Asc], [HAsc-], [Asc2-], and the pH of 0.050M H2Asc. PLAN: Write out expressions for both dissociations and make assumptions. Ka1 >> Ka2 so the first dissociation produces virtually all of the H3O+. Ka1 is small so [H2Asc]initial ≈ [H2Asc]diss After finding the concentrations of various species for the first dissociation, we can use them as initial concentrations for the second dissociation. SOLUTION: Ka1 = [HAsc-][H3O+] [H2Asc] H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq) = 1.0x10-5 Ka2 = [Asc2-][H3O+] [HAsc-] HAsc-(aq) + H2O(l) Asc2-(aq) + H3O+(aq) = 5x10-12

SAMPLE PROBLEM 18.8: Calculating Equilibrium Concentrations for a Polyprotic Acid continued Concentration(M) H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq) Initial 0.050 - Change - x - + x Equilibrium 0.050 - x - x Ka1 = [HAsc-][H3O+]/[H2Asc] = 1.0x10-5 = (x)(x)/0.050 M x x = 7.1x10-4 M pH = -log(7.1x10-4) = 3.15 HAsc-(aq) + H2O(l) Asc2-(aq) + H3O+(aq) Concentration(M) Equilibrium Change Initial 7.1x10-4M - - x - + x 7.1x10-4 - x - x x = 6x10-8 M

Bases “Strong” and “weak” are used in the same sense for bases as for acids. strong = complete dissociation (hydroxide ion supplied to solution) NaOH(s)  Na+(aq) + OH(aq) weak = very little dissociation (or reaction with water) H3CNH2(aq) + H2O(l)  H3CNH3+(aq) + OH(aq)

BASE STRENGTH [BH+][OH-] [B] Kb =

Problems P.15.Calculate pH of 5.0 x 10-2 M NaOH solution. P.16.Calculate the pH for a 1.50 M solution of ammonia (Kb=1.8 x 10-5 )

Abstraction of a proton from water by methylamine. Figure 18.11 Abstraction of a proton from water by methylamine. Lone pair binds H+ + CH3NH2 H2O Methylamine Base in water + CH3NH3+ OH- methylammonium ion

SAMPLE PROBLEM 18.9: Determining pH from Kb and Initial [B] PROBLEM: Dimethylamine, (CH3)2NH, a key intermediate in detergent manufacture, has a Kb of 5.9x10-4. What is the pH of 1.5M (CH3)2NH? PLAN: Perform this calculation as you did those for acids. Keep in mind that you are working with Kb and a base. (CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH-(aq) Assumptions: Kb >> Kw so [OH-]from water is neglible [(CH3)2NH2+] = [OH-] = x ; [(CH3)2NH2+] - x ≈ [(CH3)2NH]initial SOLUTION: (CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH-(aq) Concentration Initial 1.50M - Change - x - + x Equilibrium 1.50 - x - x

SAMPLE PROBLEM 18.9: Determining pH from Kb and Initial [B] continued Kb = 5.9x10-4 = [(CH3)2NH2+][OH-] [(CH3)2NH] 5.9x10-4 = (x) (x) 1.5M x = 3.0x10-2M = [OH-] Check assumption: 3.0x10-2M/1.5M x 100 = 2% [H3O+] = Kw/[OH-] = 1.0x10-14/3.0x10-2 = 3.3x10-13M pH = -log 3.3x10-13 = 12.48

SAMPLE PROBLEM 18.10: Determining the pH of a Solution of A- PROBLEM: Sodium acetate (CH3COONa, or NaAc for this problem) has applications in photographic development and textile dyeing. What is the pH of 0.25M NaAc? Ka of acetic acid (HAc) is 1.8x10-5. PLAN: Sodium salts are soluble in water so [Ac-] = 0.25M. Write the association equation for acetic acid; use the Ka to find the Kb. SOLUTION: Ac-(aq) + H2O(l) HAc(aq) + OH-(aq) Concentration Initial 0.25M - Change -x +x - Equilibrium - 0.25M-x x Kb = [HAc][OH-] [Ac-] = Kw Ka Kb = 1.0x10-14 1.8x10-5 = 5.6x10-10M

SAMPLE PROBLEM 18.10: Determining the pH of a Solution of A- continued [Ac-] = 0.25M-x ≈ 0.25M Kb = [HAc][OH-] [Ac-] 5.6x10-10 = x2/0.25M x = 1.2x10-5M = [OH-] Check assumption: 1.2x10-5M/0.25M x 100 = 4.8x10-3 % [H3O+] = Kw/[OH-] = 1.0x10-14/1.2x10-5 = 8.3x10-10M pH = -log 8.3x10-10M = 9.08

Molecular Properties and Acid strength: Trends in acid strength of Nonmetal Hydrides: Across a period nonmetal hydride acid strength increases. Down a group nonmetal hydride acid strength increases.

The effect of atomic and molecular properties on Figure 18.12 The effect of atomic and molecular properties on nonmetal hydride acidity. 6A(16) H2O H2S H2Se H2Te 7A(17) HF HCl HBr HI Electronegativity increases, acidity increases Bond strength decreases, acidity increases

Trends in Acid strength of Oxoacids 1. For oxoacids with same number of O around E , acid strength increases with electronegativity of E. 2. For oxoacids with different number of O around E, acid strength increases with number of O atoms.

<<   The relative strengths of oxoacids. H O I Br Cl >  Figure 18.13 The relative strengths of oxoacids. H O I Br Cl >       H O Cl H O Cl   <<  

Table 18.7 Ka Values of Some Hydrated Metal Ions at 250C Free Ion Hydrated Ion Ka ACID STRENGTH Fe3+ Fe(H2O)63+(aq) 6 x 10-3 Sn2+ Sn(H2O)62+(aq) 4 x 10-4 Cr3+ Cr(H2O)63+(aq) 1 x 10-4 Al3+ Al(H2O)63+(aq) 1 x 10-5 Cu2+ Cu(H2O)62+(aq) 3 x 10-8 Pb2+ Pb(H2O)62+(aq) 3 x 10-8 Zn2+ Zn(H2O)62+(aq) 1 x 10-9 Co2+ Co(H2O)62+(aq) 2 x 10-10 Ni2+ Ni(H2O)62+(aq) 1 x 10-10

The acidic behavior of the hydrated Al3+ ion. Figure 18.13 The acidic behavior of the hydrated Al3+ ion. Electron density drawn toward Al3+ Nearby H2O acts as base Al(H2O)63+ Al(H2O)5OH2+ H3O+ H2O

Acid-Base properties of Salt: Salt is an ionic compound. Ka xKb=Kw

SAMPLE PROBLEM 18.11: Predicting Relative Acidity of Salt Solutions PROBLEM: Predict whether aqueous solutions of the following are acidic, basic, or neutral, and write an equation for the reaction of any ion with water: (a) Potassium perchlorate, KClO4 (b) Sodium benzoate, C6H5COONa (c) Chromium trichloride, CrCl3 (d) Sodium hydrogen sulfate, NaHSO4 PLAN: Consider the acid-base nature of the anions and cations. Strong acid-strong base combinations produce a neutral solution; strong acid-weak base, acidic; weak acid-strong base, basic. SOLUTION: (a) The ions are K+ and ClO4- , both of which come from a strong base(KOH) and a strong acid(HClO4). Therefore the solution will be neutral. (b) Na+ comes from the strong base NaOH while C6H5COO- is the anion of a weak organic acid. The salt solution will be basic. (c) Cr3+ is a small cation with a large + charge, so it’s hydrated form will react with water to produce H3O+. Cl- comes from the strong acid HCl. Acidic solution. (d) Na+ comes from a strong base. HSO4- can react with water to form H3O+. So the salt solution will be acidic.

SAMPLE PROBLEM 18.12: Predicting the Relative Acidity of Salt Solutions from Ka and Kb of the Ions PROBLEM: Determine whether an aqueous solution of zinc formate, Zn(HCOO)2, is acidic, basic, or neutral. PLAN: Both Zn2+ and HCOO- come from weak conjugates. In order to find the relatively acidity, write out the dissociation reactions and use the information in Tables 18.2 and 18.7. SOLUTION: Zn(H2O)62+(aq) + H2O(l) Zn(H2O)5OH+(aq) + H3O+(aq) HCOO-(aq) + H2O(l) HCOOH(aq) + OH-(aq) Ka Zn(H2O)62+ = 1x10-9 Ka HCOO- = 1.8x10-4 ; Kb = Kw/Ka = 1.0x10-14/1.8x10-4 = 5.6x10-11 Ka for Zn(H2O)62+ >>> Kb HCOO-, therefore the solution is acidic.

Molecules as Lewis Acids An acid is an electron-pair acceptor. A base is an electron-pair donor. acid base adduct H2O(l) M(H2O)42+(aq) M2+ adduct

The Mg2+ ion as a Lewis acid in the chlorophyll molecule. Figure 18.15 The Mg2+ ion as a Lewis acid in the chlorophyll molecule.

SAMPLE PROBLEM 18.13: Identifying Lewis Acids and Bases PROBLEM: Identify the Lewis acids and Lewis bases in the following reactions: (a) H+ + OH- H2O (b) Cl- + BCl3 BCl4- (c) K+ + 6H2O K(H2O)6+ PLAN: Look for electron pair acceptors (acids) and donors (bases). SOLUTION: acceptor (a) H+ + OH- H2O donor donor (b) Cl- + BCl3 BCl4- acceptor acceptor (c) K+ + 6H2O K(H2O)6+ donor