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Unit 17: Acid-Base Equilibria

Unit 17: Acid-Base Equilibria

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Unit 17: Acid-Base Equilibria

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  1. CHM 1046: General Chemistry and Qualitative Analysis Unit 17:Acid-Base Equilibria Dr. Jorge L. Alonso Miami-Dade College – Kendall Campus Miami, FL • Textbook Reference: • Chapter 18 • Module # 6

  2. Acid-Base Theories Arrhenius (1883) NaOH HCl H2SO4 CO32- NH3 H2O Brønsted–Lowry (1923) BF3 AlI3 Lewis (1923-38)

  3. Acid-Base Definitions • Svante Arrhenius (1883) • Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions. HX H+ + X- {acid} • Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions. MOH M+ + OH- • Neutralization: HX + MOH MX + HOH acid base salt water

  4. Acid-Base Definitions • Brønsted–Lowry (1923) • Acid:Proton (H+) donor. … have a removable (acidic) proton which are donated to a bases. HX + H2OX- + H3O+ acidbaseconjugate b conjugate a. • Base:Proton (H+) acceptor. …must have a pair of nonbonding electrons which accepts protons. B + H2OBH+ + OH- base acid conj. a conj. b {base} • Neutralization: HX + BX- + BH+ acid base conj. b conj. a

  5. Strong Acids Are strong electrolytes and exist totally as ions in aqueous solution. HCl, HBr, HI, HNO3, H2SO4, HClO3, & HClO4 • SOLUBILITY RULES: for Ionic Compounds (Salts) • All salts containing the anions: NO3-, ClO3-, ClO4-, (C2H3O2-) are soluble. • All Cl-, Br-, and I- are solubleexcept forHg22+, Ag+, and Pb2+ salts. • All SO42- are solubleexcept forPb2+, Ba2+, and Sr2+. 100% Arrhenius: HX H+(aq) + X-(aq) HX + H2OX- + H3O+ acid base conj. bconj. a Strong acids have very weak conjugate bases 100% Brønsted–Lowry: For the monoprotic strong acids, [acid]i = [H3O+] or [H+] For strong acid: [1M HX]i = [1M H3O+] or [1M H+] For weak acid: [1M HA]i > [0.1M H3O+] or [0.1M H+]

  6. Strong Bases these substances dissociate completely (100%) in aqueous solution. SOLUBILITY RULES: for Ionic Compounds (Salts) All OH- are insolubleexcept for IA metals, (NH4+), Ca2+, Ba2+ , & Sr2+ (heavy IIA). Arrhenius: MOH  M+ + OH- Brønsted–Lowry: these substances accept protons from water to form very weak conjugate acids. Na+OH-+ H2O Na+ + HOH + OH- base acidconjugateacid conjugate base For the monobasic strong bases, [base]i = [OH-] [1M NaOH] = [1M OH-]

  7. Review of Strong Acid & Bases (according Brønsted–Lowry) Strong acidshave very weak conjugate bases HX + H2O H3O+ + X- acid base conj. a conj. b Has no affinity for H+ Example: HCl Strong baseshave very weak conjugate acids MOH  M+ + OH- baseconj. a conj. b Has no affinity for OH- Example: NaOH Later on this course: Salts that are neutral Salts of Strong acids & baseshave very weak conjugate acids & bases MX + H2O M+(aq) + X- (aq) salt conj. a conj. b Have no affinity for H+ orOH- Example: NaCl

  8. Concentrated 10M Strong Acid Strong Acid Dissociation HX HX H3O+ X- Concentrated 10M Very Weak Acid Weak Acid Dissociation HA

  9. Weak Acids Reactions betweenacidsandbasesalways yield their correspondingconjugate basesandconjugate acids. HA + H2O ↔ H3O+ + A- acid base conj. a conj. b Weak acids (HA) have very strong conjugate bases (A-) H+ H+ • The weaker the acid, the stronger their conjugate base.

  10. Weak Bases B + HOH BH+ + OH- base acid conj. a conj. b Bases react with water to produce hydroxide ion. • Weak bases have strong conjugate acids H+ H+ • The weaker the base, the stronger their conjugate acid.

  11. conjugate Acid / Base Strength HA + H2O ↔ H3O+ + A- acid base conj. a conj. b The stronger an acid, the weaker its conjugate base. B + HOH ↔ BH++OH- base acid conj. a conj. b The stronger an base, the weaker its conjugate acid.

  12. The Leveling Effect of Water + Strong acids react with water to produce H3O+, the strongest acid that can exist in an aqueous solution. Strong bases react with water to produce OH-, the strongest base that can exist in an aqueous solution. +

  13. Review of Weak Acid & Bases Weak acids(HA) have very strong conjugate bases (A-) HA + H2O↔ H3O+ + A- acidbase conj. a conj. b Has affinity for H+ Example: HF Salts: (1) of strong base & weak acid: MA  M+(aq) + A-(aq) A- + HOH  HA + OH- (hydrolysis) (2) of a weak base & strong acid: BHX  BH+(aq) + X-(aq) BH+ + HOH  B + H3O+ (hydrolysis) (3) of a weak base & weak acid: BHA ↔ BH+(aq) + A-(aq) Weak bases(B) have very strong conjugate acids (BH+) B + HOH↔ BH+ + OH- baseacidsconj. a conj. b Example: NH3 Has affinity for OH- Later on this course: Salts that are acidic or basic Example: NaF Example: NH4Cl Example: NH4F

  14. Lewis Acids and Bases(1923-38) • Lewis acidsare electron-pair acceptors (have empty valance orbitals). + + • Lewis basesare electron-pair donors. + + • Lewis acid-base neutralization reactions + {Lewis A-B} .. .. : : : : H+ + O H- H O H .. ..

  15. H2O(l) + H2O(l) H2O(l) H3O+(aq) + OH−(aq) H+(aq) + OH−(aq) Is Water an Acid or a Base? The Autoionization of Water • In purewater, a few water molecules act asacids and a few act as bases. • This is referred to as autoionization. 1 H3O+ and 1 OH- for every 10 million (107) H2O molecules {Movie}

  16. H2O(l) + H2O(l) H3O+(aq) + OH−(aq) Autoionization of Water • The equilibrium expression for this process is • [H3O+] [OH−] • Kw is referred to as the ion-product constant for water (@ 25°C). Kw = = 1.0  10−14 Kw = = 1.0  10−14 [H2O]2

  17. What proportion of H2O molecules dissociate into H3O+ & OH- ? • In pure water, Kw = [H3O+] [OH−] = 1.0  10−14 • Because in pure water [H3O+] = [OH−], Kw = [1.0  10−7 ] [1.0  10−7 ] = 1.0  10−14 1 H3O+ and 1 OH- for every 10 million (107) H2O molecules

  18. pH …..defined as the negative base-10 logarithm of the hydronium ion concentration. pH = −log [H3O+] For pure H2O: [1.0  10−7 ] = 7.0 Problem: Calculate pH when [H3O+] = 2.3 x 10-3 M 10x log log - 2.3 *10^- 3 = 2.64 base 10 log Problem: Calculate [H3O+] when pH = 2.3 ? antilog 10x log 10^ -2.3 = 5.0 * 10-3

  19. Other “p” Scales • The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions). • Some examples of other “p” scales are: • pH = −log [H3O+] • pOH = −log [OH−] • pKw = −log Kw • pKa = −log Ka • pKb = −log Kb pH = 7 pOH = 7 pKw = 14 pKa = 3.2 pKb = 4.7 [H3O+] = 1 x 10-7 [OH-] = 1 x 10-7 Kw = 1 x 10-14 Ka = 6.8 x 10-4 Kb = 1.8 x 10-5

  20. pH and pOH equilibrium in pure Water • In pure water, • [H3O+] [OH−] = Kw −log [H3O+] + −log [OH−] = −log Kw [1.0  10−7 ] [1.0  10−7 ] = 1.0  10−14 pH + pOH = pKw • Because in pure water [H3O+] = [OH−], • Kw = [1.0  10−7 ] [1.0  10−7 ] = 1.0  10−14 • + 7 = 14

  21. H2O(l) + H2O(l) H3O+(aq) + OH−(aq) pH and pOH equilibrium in Water to which Acids & Bases are Added Add acid H3O+ [H3O+] [OH−] H2O Kw = [1.0  10−7 ] [1.0  10−7 ] = 1.0  10−14 Kw = [1.0  10−6 ] [1.0  10−8 ] = 1.0  10−14 pH + pOH = pKw 6 + 8 = 14

  22. H2O(l) + H2O(l) H3O+(aq) + OH−(aq) pH and pOH equilibrium in Water to which Acids & Bases are Added Add base OH- [H3O+] [OH−] H2O Kw = [1.0  10−7 ] [1.0  10−7 ] = 1.0  10−14 Kw = [1.0  10−8 ] [1.0  10−6 ] = 1.0  10−14 pH + pOH = pKw 8 + 6 = 14

  23. [H3O+] [OH−] Notice the relationship between [H3O+] and [OH−] It is an inverse relationship! 1 [OH−] ∝ [H3O+] pH + pOH = 14

  24. pH These are the pH values for several common substances. pH + pOH = 14

  25. Ka= [H3O+] [A−] [HA] HA(aq) + H2O(l) A−(aq) + H3O+(aq) Dissociation Constants for Weak Acids • For a generalized acid dissociation, the equilibrium expression would be • This equilibrium constant is called the acid-dissociation (ionization) constant, Ka.

  26. Dissociation Constants The greater the value of Ka, the stronger the acid. pKa 3.2 3.3 4.2 4.7 7.5 9.3 9.9 pKa Salts of weak acids are strong bases

  27. HCOOH (aq)+ H2O(l) HCOO−(aq)+ H3O+(aq) [H3O+] [A−] [HA] Ka = Calculating Ka from the pH Problem: (to determine Ka, simply dissolve a known [HA]ithen measure pH) A 0.10 Msolution of formic acid, HCOOH, at 25°C has a pH = 2.38. Calculate Ka for formic acid at this temperature. Also calculate % ionization. What is the Ka expression? pH = −log [H3O+] 2.38 = −log [H3O+] −2.38 = log [H3O+] 10−2.38 = 10log [H3O+] = [H3O+] 4.2  10−3 = [H3O+] = [HCOO−]

  28. Ka= [H3O+] [A−] [HA] HCOOH (aq)+ H2O(l) HCOO−(aq)+ H3O+(aq) [4.2  10−3] [4.2  10−3] [0.10] = Calculating Ka from pH Problem: • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. Also calculate % ionization. [H3O+] = [HCOO−] = 4.2  10−3 = 1.8  10−4

  29. HCOOH(aq)+ H2O(l) HCOO−(aq)+ H3O+(aq) 4.2  10−3 0.10 [H3O+]eq [HA]initial Calculating Percent Ionization Whole (Initial) Part (@ Equi) • Percent Ionization =  100 • In this example Percent Ionization =  100 = 4.2%

  30. [H3O+] [C2H3O2−] [HC2H3O2] Ka = Calculating pH from Ka Problem: For acetic acid at 25°C is Ka =1.8  10−5 Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25°C. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) [x] [x] [0.3] = We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

  31. 1.8  10−5 = [H3O+] [C2H3O2−] [HC2H3O2] Ka = (x)2 (0.30) Calculating pH from Ka • pH = −log [H3O+] = − log x • pH = −log (2.3  10−3) • pH = 2.64 (1.8  10−5) (0.30) = x2 5.4  10−6 = x2 [H3O+] = 2.3  10−3 = x What is the pH?

  32. 2002B Q1 (b) Calculate the pH of 0.50 M Lactic Acid.

  33. Polyprotic Acids • Have more than one acidic proton. • If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

  34. [HB+] [OH−] [B] Kb = Dissociation Constants for Weak Bases Bases react with water to produce hydroxide ion. The generalized equilibrium constant expression for these reactions is where Kb is the base-dissociation constant.

  35. Weak Bases Kb can be used to find [OH−] and, through it, pH. pKb pKb 4.7 8.8 8.0 3.4 6.7 3.7 6.5 Salts of weak bases are strong acids

  36. pH & pOH for Weak Acids & Bases NH3(aq) + H2O(l) NH4+(aq) + OH−(aq) [NH4+] [OH−] [NH3] = 1.8  10−5 = Kb = (x)2 (0.15) Kb = 1.8 x 10-5 What is the pH of a 0.15 M solution of NH3?

  37. = [NH4+] [OH−] [NH3] = 1.8  10−5 Kb = (x)2 (0.15) pH of Basic Solutions (1.8  10−5) (0.15) = x2 2.7  10−6 = x2 1.6  10−3 = x2 What is pH of soln? What is x? [OH−] = 1.6  10−3M pOH = −log (1.6  10−3) pOH = 2.80 pH = 14.00 − 2.80 pH = 11.20

  38. 2002A Q1(a-b)

  39. 2005 A acid base equilibria

  40. 2005 A acid base equilibria ANSWERS

  41. c) ii

  42. 2005 B acid base equilibria

  43. 2005 B acid base equilibria ANSWERS a) b) c) d)iii

  44. d)i d)ii

  45. d)iii

  46. 2001 a/b equilibria titration

  47. 2001 answers

  48. 2002

  49. a)