Lecture Slides on Acids and Bases - General Chemistry | CHEM 142, Study notes of Chemistry

Download Lecture Slides on Acids and Bases - General Chemistry | CHEM 142 and more Study notes Chemistry in PDF only on Docsity! 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 7.5 Calculating the pH of weak Acid Solutions 7.6 Bases STOP HERE FOR TEST 2 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 From Chapter 4: Acids and Bases An Acid is a substance that produces H+ (H3O+) ions when dissolved in water, and is a proton donor A Base is a substance that produces OH - ions when dissolved in water: Example: NaOH(aq) → Na+(aq) + OH-(aq) The OH- ions react with the H+ ions (if an acid is present) to produce water, H2O, and are therefore proton acceptors. Acids and Bases are electrolytes. Their strength is categorized in terms of their degree of dissociation in water to make hydronium or hydroxide ions. Strong acids and bases dissociate completely, and are strong electrolytes. Weak acids dissociate partially (some small % of the molecules dissociate) and are weak electrolytes. Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two Hydronium ions, and one sulfate ion. What is the molarity of the sulfate and Hydronium ions in a solution prepared by dissolving 155g of concentrate sulfuric acid into sufficient water to produce 2.30 Liters of acid solution? Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity. Solution: Two moles of H+ are released for every mole of acid: H2SO4 (l) + 2 H2O(l) 2 H3O+(aq) + SO4- 2(aq) Molarity of H+ = 2 x 0.687 M = 1.37 Molar in H+ (or H3O+) Moles H2SO4 = = 1.58 moles H2SO4155 g H2SO4 1 mole H2SO4 98.09 g H2SO4 x 1.58 mol SO4-2 2.30 L solutionMolarity of SO4 - 2 = = 0.687 Molar in SO4- 2 The Nature of the Hydrated Proton a @ H,0* Acid - Base Reactions : Neutralization Rxns. The generalized reaction between an Acid and a Base is: HA(aq) + MOH(aq) MA(aq) + H2O(l) Acid + Base → Salt + Water The salt product can either be dissolved as ions or form a precipitate. 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 as written: H+(aq) + OH-(aq) → H2O(l) + 56 kJ / mol 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; HNO3 and H2PO4- 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 like to bind the H+ ion; a few examples are NH3, CO32-, 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 Molecular model: The reaction of an acid HA with water to form H3O+ and a conjugate base. Acid Base Conjugate Conjugate acid base A A A H,O%ag) + A (ag) HA HQT —_~ = dissociation B Weak acid: HAlag) + HOt) Alter Betore dissociation Ha SBPOW JO JE qWNU SANE] ay SS|OWW JO JBQWNM BANE] EY A Strong acid: HAfaq) + HOY!) —e HyOt(aq) + A fag) ion t ia for Strong and Weak ids f ISSOCI Ac The Extent Oo D  TABLE 7.1 Various Ways to Describe Acid Strength Property Strong Acid Weak Acid K, value Position of the dissociation equilibrium Equilibrium concentration of H' compared with original con- centration of HA Strength of conjugate base com- pared with that of water K, is large Far to the right [H*] ~ [HA]o A much weaker base than HO K, is small Far to the left [H"] < [HA] A much stronger base than H,O  K. a Values | Ka Values for Some Monoprotic Acids at 25°C Name (Formula) Lewis Structure* K, for Some .__ lodic acid (HIOs) ee 1.6107" Monoprotic ©: Acids Chlorous acid (HCIO,) 9 1.12%107 Nitrous acid (HNO,) . 7.1x107 Hydrofluoric acid (HF) Fi 6.8x107* 205 Formic acid (HCOOH) H-C-6—H 1.8x107 :0: Benzoic acid (C,H, COOH) O}-b-8—n 6.3x10° H :O: Acetic acid (CH,COOH) h—o-E-8-4 1.8x10° H H H 30: Propanoic acid Wt Ws -5 (CH,CH,COOH) a eee Lente Hypochlorous acid (HCIO) =4—6—-€: 2.9x10% Hypobromous acid (HBrO) #—6-ér: 2.3x107° Hydrocyanic acid (HCN) = H#—c=" 6.2x107° Phenol (C,H;OH) Oy-8-n 1.0x107"" Hypoiodous acid (HIO) H—-O-1: 2.3x10" *Red type indicates the ionizable proton; structures have zero formal charge. The Conjugate Pairs in Some Acid-Base Reactions Acid + Base Conjugate Base + Conjugate Acid Conjugate Pair 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+ • • • • • • • • • • • • • • • • • • • • 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 water’s protons is transferred to the sulfide ion so: S2-(aq)/HS-(aq) and H2O(l)/OH-(aq) are the two acid-base pairs. Strengths of Conjugate Acid-Base Pairs ACID rHCl H,SO, HNO, | H30* rHso,” H,SO, HPO, HF CH,COOH H,CO, Weak; H,S HSO, H,PO, NH," HCO,” HPO,” LH,O Negligibie{ ¥S OH™ Strong- BASE ci" HSO, no, H,0 so, HSO,” HPO, Es CH,COO™ HCO,” HS~ so?” ae HPO, NH, co,” Po,” OH" st Oo? + Negligible > Weak > Strong BASE STRENGTH Water itself is a weak acid and weak base: Two water molecules react to form H3O+ and OH- (but only slightly) K = Ka = Kw = 1.0 x 10-14 Autoionization of Water H2O(l) + H2O(l) H3O+(aq) + OH-(aq) Ka = = [H+][OH-] = Kw [H3O+][OH-] [H2O]2 The ion-product for water, Kw: Ka = Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25°C) 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 25°C) The molarity of pure water is: = 55.5 M1000g/L 18.02 g/mol Set = 1.0 since nearly pure liquid water even when salts, acids or bases present. The Definition of pH pH = -log10[H3O+] = -log10[H+] Significant Figures for Logarithms The number of decimal places in the log is equal to the number of significant figures in the original number. What is the pH of (a) pure water, (b) 0.10 M HCl, and (c) 0.1 M NaOH? (a)For pure water, [H+] = 1.0 x 10-7 M, so pH = -log10[1.0 x10-7] = 7.00 (b) The pH of 0.01 M HCl (a STRONG ACID) is pH = -log10[1 x 10-2] = 2.0 (c)The pH of 0.1 M NaOH (a STRONG BASE) Kw = 10-14 = [H+][OH-] = [H+](0.1) so [H+] = 10-14 / 0.1 pH = -log10(1.0 x 10-14/0.1) = -log10[1 x 10-13] = 13.0 The pH Values of Some Familiar Aqueous Solutions [H3O+] [OH-] [OH-] = KW [H3O+] neutral solution acidic solution basic solution [H3O+]> [OH-] [H3O+]< [OH-] [H3O+] = [OH-] Two new definitions Remember: Kw = [H+][OH-] so that [OH-] = Kw / [H+] 1) pOH = -log10[OH-] = - log10(Kw/[H+]) = - log10Kw - log10[H+]) = - log10 (1.0 x 10-14) – pH = 14.00 - pH pOH = 14.00 – pH 2) It is convenient to express equilibrium constants for acid dissociation in the form pKa = - log10Ka 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 25°C. 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) [H+] = 3.0 M pH = -log[H+] = -log(3.0) = -0.477 [OH-] = = = 3.333 x 10-15 MKw [H+] 1 x 10-14 3.0 pOH = - log(3.333 x 10-15) = 15.000 - 0.477 = 14.523 (b) [H+] = 0.0024 M pH = -log[H+] = -log(0.0024) = 2.62 [OH-] = = = 4.167 x 10-12 M pOH = -log(4.167 x 10-12) = 12.000 - 0.6198 = 11.38 Kw [H+] 1 x 10-14 0.0024 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. Solution: HClO(aq) + H2O(l) H3O+(aq) + ClO- (aq) Ka = = 3.5 x 10-8 [H3O+] [ClO-] [HClO] Concentration (M) HClO H2O H3O+ ClO- Initial 0.125 ---- 0 0 Change -x ---- +x +x Equilibrium 0.125 - x ---- x x Ka = = 3.5 x 10-8 (x)(x) 0.125-x Set 0.125 - x = 0.125 since x must be tiny. x2 = 0.4375 x 10-8 x = 0.661 x 10-4 Finding the Ka of a Weak Acid from the pH of its Solution–I Calculating [H3O+] : [H3O+] = 10-pH = 10-4.19 = 6.46 x 10-5 M = x Concentration (M) HClO(aq) + H2O(l) H3O+(aq) + ClO -(aq) Initial 0.12 ---- 0 0 Change -x ---- +x +x Equilibrium 0.12 -x ---- +x +x Assumed: [H3O+] = [H3O+]HClO since it’s >> than the 10-7 M H+ we get from water 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: Solving Problems Involving Weak-Acid Equilibria–II 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 common good 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. 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 (Table 7.2) Ka = = 4.0 x 10-4 = [H+] [NO2-] [HNO2] Initial concentrations = [H+] = 0 , [NO2-] = 0 , [HNO2] = 1.00 M Final concentrations = [H+] = x , [NO2-] = x , [HNO2] = 1.00 M - x (x) (x) 1.00 - x Assume 1.00 – x = 1.00 since left side so tiny. 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 = 1.70 Molecular model: Nitrous acid Nitrous acid (HNO,) Problem: Calculate the Percent dissociation of a 1.00 M Hydrocyanic acid solution, Ka = 6.20 x 10-10. HCN(aq) + H2O(l) H3O+(aq) + CN- (aq) HCN H3O+ CN- Initial 1.00 M 0 0 Ka = Change -x +x +x Final 1.00 –x x x Ka = = 6.20 x 10-10 Assume 0.0100-x = 0.0100 Ka = = 6.2 x 10-10 x = 2.49 x 10-5 % dissociation = x 100% = 0.00249% [H3O+][CN-] [HCN] (x)(x) (1.00-x) x2 1.00 2.49 x 10-5 1.00 BASES Like Example 7.5 (P 250) 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: [H+] = = = 5.0 x 10-12 MKw[OH-] 1.0 x 10-14 2.0 x 10-3 pH = - log [H+] = - log( 5.0 x 10-12) =12.00 – 0.70 = 11.30 Determining pH from Kb and Initial [B] – II Substituting into the Kb expression and solving for x: Kb = = = 1.8 x 10-5 [NH4+] [OH-] [NH3] (x)(x) 1.5-x x2 = 2.7 x 10-5 = 27 x 10-6 x = 5.20 x 10-3 = [OH-] = [NH4+] Calculating pH: [H3O+] = = = 1.92 x 10-12 Kw [OH-] 1.0 x 10-14 5.20 x 10-3 pH = -log[H3O+] = - log (1.92 x 10-12) = 12.000 - 0.28 pH = 11.72 Set 1.5 - x = 1.5 since right side is tiny. Amines: Bases with the Nitrogen Atom H3C N HH .. H3C N H3C N H .. N H H .. .. C2H5 Ethylamine Methylamine CH3 Dimethylamine CH3CH3 Trimethylamine .. N Pyridine General reaction: NR3(aq) + H2O(l) HNR3+(aq) + OH-(aq) Kb = [HNR3 +] [OH-] [NR3] Like Example 7.3 (P 245)-II Initial Concentration Equilibrium Concentration (mol/L) (mol/L) [HF]0 = 1.00 [HF] = 1.00 – x [F-] = 0 [F-] = x [H+] = 0 [H+] = x~ x mol HF dissociates Ka = = 7.2 x 10-4 = = [H+] [F-] [HF] (x) (x) 1.00-x x2 1.00 x = 2.7 x 10-2 Therefore, [F- ] = [H+] = x = 2.7 x 10-2 and pH = 1.56 Like Example 7.3 (P 245)-III Ka = = 3.5 x 10-8 [H+] [OCl-] [HOCl] The concentration of H+ comes from the first part of this problem = 2.7 x 10-2 M [HOCl]initial = 5.00 M ; [OCl-]eqbm = x [HOCl]eqbm = 5.00 - x 3.5 x 10-8 = (2.7 x 10 -2)( x) (5.00 - x) 5.00 ( 3.5 x 10-8) 2.7 x 10-2x = = 6.48 x 10 -6 M = [OCl-] pH = -log [H+] = -log [2.7x10-2] = 1.56 [F-] = 2.7 x 10-2 M ; [OCl-] = 6.48 x 10-6 M Left is tiny so x must be tiny. Assume: 5.00 – x = 5.00 Summary: Solving Weak Acid (P 244) 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.