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Acids and Bases

Acids and Bases. Acids and bases Acid-base properties of water (K w ) pH scale Strength of Acids and Bases Weak acid (K a ) Weak base (K b ) Relation between conjugate acid-base ionization constants (K a and K b ) Chemical structure and acidity Acid-base properties of salt solutions

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Acids and Bases

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  1. Acids and Bases • Acids and bases • Acid-base properties of water (Kw) • pH scale • Strength of Acids and Bases • Weak acid (Ka) • Weak base (Kb) • Relation between conjugate acid-base ionization constants (Ka and Kb) • Chemical structure and acidity • Acid-base properties of salt solutions • Lewis acid-base

  2. HCl(aq) +H2O(l) H3O+(aq) + Cl-(aq) HNO3(aq)+H2O(l) H3O+(aq) + NO3-(aq) NH3(aq) + H2O (l) NH4+(aq) + OH-(aq) I. Acid and Base A. Arrehenius acid - a compound that increases [H+] in water B. Arrehenius Base : a compound that increases [OH-] in water

  3. NH3(aq) + H2O (l) NH4+(aq) + OH-(aq) I. Acid and Base C. Brønsted acid : a proton donor D. Brønsted Base : a proton acceptor base acid acid base conjugateacid conjugate base base acid E. Conjugate acid-base pair : acid-base pair that are related by a loss or gain of a proton H+

  4. species Acid/base/both Conjugate acid Conjugate base Example: Conjugate acid-base pair • Classify each of the following species a Brønsted acid, base, or both. Give the conjugate acid or base. HCO3– CO32– base HSO4– H2SO4 acid H2O both OH– H3O+ H2PO4– both H3PO4 HPO42– • Amphoteric: a substance that can act either as an acid or a base

  5. H H O O - H H H O + [ ] + H H + O H H2O (l) + H2O (l) H3O+(aq) + OH-(aq) H2O (l) H+(aq) + OH-(aq) II. Acid-base properties of water A. Autoionization of water: water is a very weak electrolyte base conjugateacid conjugatebase acid

  6. H2O (l) H+(aq) + OH-(aq) B. Ion-product constant, Kw KC = Kw = [H+][OH-] (1) Kw: the product of the molar concentrations of H+ and OH- ions at a particular temperature. Kw = [H+][OH-] = 1.0 x 10-14 at 25oC (2) H+ and OH– ions are always present in aqueous solutions [H+] = [OH-] Neutral solution [H+] > [OH-] Acidic solution [H+] < [OH-] Basic solution

  7. = 1.0 x 10-14 Kw 2.0 x10-2 [OH–] [H+] = What is the concentration of H+ in a (a) neutral solution and (b) 2.0 x10-2 M NaOH solution at 25oC? Q: Kw = [H+][OH-] = 1.0 x 10-14 (a) Neutral solution [H+] = [OH–]= x Kw = [H+][OH-] = 1.0 x 10-14 = x2 = 1.0 x 10-7M (b) [OH–] = 2.0 x10-2M = 5.0 x 10-13 M (basic)

  8. [H+] pH pOH III. pH scale – a measure of acidity pH = -log [H+] pOH = -log [OH–] At 25oC [H+][OH-] = Kw = 1.0 x 10-14 pH + pOH = 14.00 -log [H+] – log [OH-] = 14.00 Solution is At 250C pH = pOH = 7 neutral [H+] = [OH-] [H+] = 1 x 10-7 pH < 7 pOH >7 [H+] > 1 x 10-7 acidic [H+] > [OH-] pH > 7 pOH <7 basic [H+] < [OH-] [H+] < 1 x 10-7 As the solution is more acidic when

  9. 16.1

  10. If a solution has a pOH value of 10.5, what is the H+ ion concentration? Q: pH = 14.0 - pOH = 14.0-10.5 = 3.5 pH = 3.5 = -log [H+] = 10-3.5 = 3.2 x 10-4M [H+] = 10-pH (acidic) • Q: The OH- ion concentration of a blood sample is 2.5 x 10-7 M. • What is the pH of the blood? pH + pOH = 14.00 pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60 (basic) pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

  11. NO2-(aq) + H2O (l) OH-(aq) + HNO2(aq) H2O NaCl (s) Na+ (aq) + Cl- (aq) H2O NaOH (s) Na+(aq) + OH-(aq) CH3COOH CH3COO- (aq) + H+ (aq) HClO4(aq) + H2O (l) H3O+(aq) + ClO4-(aq) IV. Strength of Acid and Base Strong Electrolyte – 100% dissociation (ionization) - soluble ionic compound, strong acid, strong base Weak Electrolyte – not completely dissociated - weak acid, weak base

  12. HCl, strong acid HF, weak acid

  13. HCl (aq) + H2O (l) H3O+(aq) + Cl-(aq) H2O (l) + HCl (aq) H3O+(aq) + Cl-(aq) H2SO4(aq) + H2O (l) H3O+(aq) + HSO4-(aq) • Strong electrolyte, completely ionize (dissociate) in water • HCl, HBr, HI, HNO3, H2SO4, HClO4 C. Strong acid Q: What is the pH of the solution when 8.4 g of HCl is dissolved in 1.2 L of water? HCl is a strong acid – 100% dissociation. = 0.19 M 0.19 M 0.00 M 0.00 M Initial -0.19 M 0.19 M 0.19 M Change 0.19 M 0.19 M Equilibrium 0.00 M pH = -log [H+] = -log [H3O+] = -log(0.19) = 0.72

  14. Ca(OH)2(aq) Ca2+(aq) + 2OH-(aq) Na2O (s) + H2O (l) 2Na+(aq) + 2OH-(aq) H2O Ba(OH)2(s) Ba2+(aq) + 2OH-(aq) • Hydroxide of alkali (IA) and alkaline earth (IIA) metal • Ionic metal oxide (Na2O, CaO), metal hydride, and nitrides D. Strong Base Ex: What is the pH of a 0.048M Ca(OH)2 solution? Ca(OH)2 is a strong base – 100% dissociation. 0.048 M 0.000 M 0.000 M Initial 0.048 M -0.048 M Change 2 x 0.048 M 0.048 M 0.096 M Equilibrium 0.000 M pOH = -log [OH-] = -log(0.096) = 1.02 pH = 14.00 - pOH = 12.98

  15. HCN (aq) + F– (aq) HF (aq) + CN– (aq) E. Conjugate acid-base pair • The conjugate base of a strong acid is extremely weak • The conjugate acid of a strong base is extremely weak. • The stronger the acid, the weaker the conjugate base • The stronger the base, the weaker the conjugate acid Ex: Predict the direction of the following reaction * Reaction goes to the side that produces gas, solid, or weak/nonelectrolyte Which is a stronger acid? HCN or HF Which is a stronger base? CN– or F– Equilibrium will proceed from right to left since HF is a stronger acid and CN– is a stronger base. KC < 1

  16. 16.2

  17. HA (aq) + H2O (l) H3O+(aq) + A-(aq) HA (aq) H+(aq) + A-(aq) [H+][A-] Ka = [HA] acid strength Ka V. Weak Acid Ka: acid ionization constant

  18. HF (aq) H+(aq) + F-(aq) = 7.1 x 10-4 = 7.1 x 10-4 = 7.1 x 10-4 [H+][F-] x2 x2 Ka Ka = Ka = 0.80 - x [HF] 0.80 HF (aq) H+(aq) + F-(aq) What is the pH of a 0.80M HFsolution (at 250C)? Ex: Initial (M) 0.80 0.00 0.00 Change (M) -x +x +x Equilibrium (M) 0.80 - x x x 0.80 – x 0.80 Ka << 1 x2 = 5.68 x 10-4 x = 0.024 M [H+] = [F-] = 0.024 M pH = -log [H+] = 1.62 [HF] = 0.80 – x = 0.78 M

  19. = 7.1 x 10-4 = 7.1 x 10-4 0.0027 M 0.024 M x2 x2 x 100% = 3.0% x 100% = 27% 0.80 M 0.010 M Ka Ka = 0.010 - x 0.010 When will this approximation work? 0.80 – x 0.80 Ka << 1 When x is less than 5% of the value from which it is subtracted. Less than 5% Approximation ok. x = 0.024 Ex. What is the pH of a 0.010M HFsolution (at 250C)? x = 0.0027 M More than 5% Approximation not ok. Must solve for x exactly using quadratic equation or other method. x2 + 7.1x10-4 x -7.1x10-6 = 0 pH = -log [H+] = 2.64 [H+] = x = 0.0023 M

  20. Steps to solve weak acid ionization problems: • Identify the major species that can affect the pH. • In most cases, you can ignore the autoionization of water. • Ignore [OH-] because it is determined by [H+]. • Use ICE to express the equilibrium concentrations in terms of single unknown x. • Write Kain terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly. • Calculate concentrations of all species and/or pH of the solution.

  21. ionized acid concentration at equilibrium x 100% x 100% % ionization = initial concentration of acid [H+] [HA]0 V. Weak Acid B. Percent ionization % ionization = For a monoprotic acid HA [HA]0 = initial concentration decreases *Percent ionization as initial concentration increases

  22. H2S (aq) H+(aq) + HS-(aq) Ka1= 9.5 x10-8 HS–(aq) H+(aq) + S2-(aq) Ka2= 1 x10-19 H3PO4(aq) H+(aq) + H2PO4-(aq) Ka1= 7.5 x10-3 H2PO4–(aq) H+(aq) + HPO42–(aq) Ka2= 6.2 x10-8 HPO42–(aq) H+(aq) + PO43–(aq) Ka3= 4.8 x10-13 < < V. Weak Acid Polyprotic acid Acids that have more than one ionizable H diprotic triprotic Ka3 Ka2 Ka1

  23. = 1.3 x 10-2 x2 Ka1 = 0.024 - x H2SO3(aq) H+(aq) + HSO3–(aq) What is the pH of a 0.024M H2SO3 solution at 25oC? (Ka1= 1.3 x 10-2 and Ka2=6.3 x 10-8) Ex: Treat this type problem with step-by-step ionization Initial (M) 0.024 0.00 0.00 Change (M) -x +x +x Equilibrium (M) 0.024 - x x x x = 0.0123 or -0.025 x2 + 1.3 x 10-2 - 3.12 x 10-4 = 0 [H+] = [HSO3-] = 0.0123 M [H2SO3] = 0.024 – x = 0.0117 M Not done yet! Next step, let us consider the second ionization of H

  24. = 6.3 x 10-8 (0.0123+y)y Ka2 = 0.0123 - y 6.3 x10-8M x 100% << 5 % 0.0123 M HSO3–(aq) H+(aq) + SO3–(aq) Initial (M) 0.0123 0.0123 0.00 Change (M) -y +y +y Equilibrium (M) 0.0123 - y 0.0123+y y Ka2<< 1 0.0123 – y 0.0123 0.0123 + y 0.0123 y= 6.3 x10-8 [H+] = 0.0123 + y = 0.012 M pH = -log[H+] =1.92 [HSO3–] = 0.0123 – y = 0.012 M [SO32–] = y = 6.3 x10-8M [H2SO3] = 0.012 M

  25. NH3(aq) + H2O (l) NH4+(aq) + OH-(aq) B (aq) + H2O (l) HB+(aq) + OH-(aq) [HB+][OH-] Kb = [B] base strength Kb VI. Weak Base Kb: base ionization constant Solve weak base problems like weak acids except solve for [OH-] instead of [H+].

  26. C2H5NH2(aq) +H2O(l)C2H5NH3+(aq) + OH-(aq) Initial (M) Change(M) = 5.6 x 10-4 = 5.6 x 10-4 Equilibrium(M) x2 x2 Kb Kb = 0.50 - x 0.50 0.017 M x 100% = 3.4 % assumption okay 0.50 M What is the pH of a 0.50M C2H5NH2 solution at 250C(Kb=5.6 x10-4)? Ex: 0.50 0.00 0.00 -x +x +x x x 0.50 - x Kb << 1 0.50 – x 0.50 x2 = 2.8 x 10-4 x = 0.017 M [OH–] = [C2H5NH3+] = 0.017 M [C2H5NH2] = 0.50 – x = 0.48 M pH = 14.00 - pOH = 14.00 + log [OH–] = 12.23

  27. 16.5

  28. HA (aq) H+(aq) + A-(aq) A–(aq) + H2O (l) OH- (aq) + HA (aq) H2O (l) H+(aq) + OH-(aq) Kw Kw Kw Ka= Kb= Kb= = Kb Ka Ka 1.0 x 10-14 C6H5COO– (aq) + H2O (l) OH- (aq) + C6H5COOH (aq) = 1.5 x 10-10 6.5 x 10-5 VII. Ionization Constants of Conjugate Acid-Base Pairs Ka Kb Kw KaKb = Kw Weak acid (base) and its conjugate base (acid) Ex. What is the reaction and the equilibrium constant of C6H5COO– in water? C6H5COO– is a base and is the conjugate base of C6H5COOH

  29. < < < < < < VIII. Molecular Structure and Acid Strength A. Factors affecting acidity - bond strength : weaker bond, stronger acid - bond polarity: more polar bond, stronger acid B. Binary acid : bond strength is more important - hydrohalic acidHF HCl HBr HI Ex: H2O H2S H2Se

  30. d- d+ O- Z + H+ O Z H C. Oxoacids • The O-H bond will be more polar and easier to break if: • Z is very electronegative or • Z is in a high oxidation state

  31. > O Cl is more electronegative than Br H O Cl O O H O Br O > > > • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number (O atom) • - Acid strength increases with increasing EN of Z C. Oxoacids > HBrO HIO • Oxoacids having the same central atom (Z) but different numbers of attached groups. • - Acid strength increases as the • oxidation no of Z increases HClO4 HClO3 HClO2 HClO

  32. withdraw electron O H H C C O H H can be ionized • • • • • • • • D. Carboxylic acids • Acids with functional group -COOH Ex. Predict the strength of the following acids. CF3COOH CH3COOH >

  33. H2O H2O KNO3 (s) K+ (aq) + NO3– (aq) NaBr (s) Na+ (aq) + Br- (aq) • Salts are strong electrolytes, completely ionizes in water. The acidity/basicity of salt solutions depends on the ions. • Hydrolysis: reaction of ions with H2O to produce OH– or H+ • NO2– + H2O <=> HNO2 + OH– • NH4+ + H2O <=> NH3 + H3O+ IX. Salt Solutions A. Neutral Solutions: Salts containing an alkali metal or alkaline earth metal ion (except Be2+) and the conjugate base of a strong acid (e.g. Cl–, Br–, I–, and NO3–).

  34. H2O NH4Cl (s) NH4+ (aq) + Cl- (aq) NH4+(aq) NH3(aq) + H+(aq) H2O CH3COONa (s) Na+ (aq) + CH3COO- (aq) 3+ 2+ Al(H2O)6(aq) Al(OH)(H2O)5(aq) + H+(aq) 2+ 3+ Al(H2O)6(aq) +H2O (l) Al(OH)(H2O)5(aq) + H3O+(aq) CH3COO-(aq) + H2O (l) CH3COOH (aq) + OH-(aq) B. Basic Solutions: Salts derived from a strong base and a weak acid. C. Acidic Solutions: a. Salts derived from a strong acid and a weak base. Salts with small, highly charged metal cations (e.g. Al3+, Cr3+, and Be2+) and the conjugate base of a strong acid. For example: AlCl3

  35. Acid hydrolysis of Al3+ Acidity of metal ion increases with smaller size and higher charge.

  36. 1.0 x 10-14 1.0 x 10-14 4.5 x 10-4 1.8 x 10-5 Kw Kw Kb= = = 2.2 x 10-11 Ka= = = 5.6 x 10-10 Kb Ka D. Solutions in which both the cation and the anion hydrolyze: • Kb for the anion > Ka for the cation, solution will be basic • Kb for the anion < Ka for the cation, solution will be acidic • Kb for the anion Ka for the cation, solution will be neutral Example: NH4NO2 NO2– and NH4+ Ka,HNO2 =4.5 x10-4 Kb,NH3 =1.8 x10-5 NO2– : Ka >Kb Weak acid NH4+ :

  37. H2O (l) + F–(aq) HF (aq) + OH-(aq) F–(aq) +H2O (l) HF (aq) + OH-(aq) = 1.4 x 10-11 [HF][OH-] x2 Kb Kb = 0.10 [F–] x2 0.10 - x 1.0 x 10-14 7.1 x 10-4 Kw Kb= = = 1.4 x 10-11= Ka What is the pH of a 0.10 M solution of a NaF? Acid or basic? Ex: Initial (M) 0.10 0.00 0.00 Change (M) -x +x +x Equilibrium (M) 0.10 - x x x Kb << 1 0.10 – x 0.10 [OH–] = x = 1.2 x10-6M x2 = 1.4 x 10-12 pH = 14.00 -pOH = 14.00 + log [OH–] = 8.08

  38. X. Acidic, basic, and amphoteric oxides • Basic oxide: group 1A and 2A oxides (except BeO). BaO + H2O -> Ba(OH)2 • Amphoteric oxide: BeO and group 3A and 4A oxides. • Acid oxide: nonmetal oxide with high oxidation number, such as N2O5,SO2 etc. N2O5+ H2O --> 2 HNO3 • Nonmetal oxide with low oxidation state such as CO or NO are neutral)

  39. H F F F B F B N H F F H + OH- H N H H H H + H O H + H+ H N H N H • • • • • • • • • • • • • • H H XI. Lewis acid and base • Arrehenius acid/base: substance that increases [H+]/[OH–] in water • B. Brønstedacid/base : proton donor/acceptor C. A Lewis acid is a substance that can accept a pair of electrons (electron pair acceptor) • A Lewis base is a substance that can donate a pair of electrons (electron pair donor) H+ + acid base acid base acid base

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