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Acids and Bases

Chapter 16. Acids and Bases. Johannes N. Bronsted Thomas M. Lowry 1879-1947. 1874-1936. Both independently developed Bronsted-Lowry theory of acids and bases. Acids and Bases: A Brief Review. Classical Acids: Taste sour

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Acids and Bases

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  1. Chapter 16 Acids and Bases Johannes N. Bronsted Thomas M. Lowry 1879-1947. 1874-1936. Both independently developed Bronsted-Lowry theory of acids and bases.

  2. Acids and Bases: A Brief Review Classical Acids: Taste sour Donate H+ (called “H-plus” or “proton”) Turn litmus red Generally formed from H-Z, where Z = nonmetal Classical Bases: Taste bitter and feel soapy. Donate OH- (called “O-H-minus” or “hydroxide”) Turn litmus blue Generally formed from MOH, where M = metal Neutralization: Acid + Base  Salt + water H-Z + MOH  MZ + HOH H+ in water is actually in the form of H3O+, “hydronium”

  3. Brønsted-Lowry Acids and Bases Proton Transfer Reactions • Brønsted-Lowry acid/base definition: aciddonates H+ baseaccepts H+. • Brønsted-Lowry base does not need to contain OH-. • Consider HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq): • HCl donates a proton to H2O. Therefore, HCl is an acid. • H2O accepts a proton from HCl. Therefore, H2O is a base. • Water can behave as either an acid or a base. • Amphoteric substances can behave as acids and bases.

  4. Brønsted-Lowry Acids and Bases Conjugate Acid-Base Pairs • Whatever is left of the acid after the proton is donated is called its conjugate base. • Similarly, whatever remains of the base after it accepts a proton is called a conjugate acid. • Consider • After H2O (base) gains a proton it is converted into H3O+ (acid). Therefore, H2O and H3O+ are conjugate acid-base pairs. • After HCl (acid) loses its proton it is converted into Cl- (base). Therefore HCl and Cl- are conjugate acid-base pairs. • Conjugate acid-base pairs differ by only one proton.

  5. Brønsted-Lowry Acids and Bases Conjugate Acid-Base Pairs In each of following Bronsted-Lowry Acid/Base reactions, Which is acid? base? conjugate acid? conjugate base? HCl + H2O  H3O+ + Cl- acid base conj. acid conj. base NH3 + H2O NH4+ + OH- base acid conj. acid conj. base CH3NH2 + H2SO4 CH3NH3+ + HSO4- conj. base base acid conj. acid

  6. Brønsted-Lowry Acids and Bases Relative Strengths of Acids and Bases • The stronger the acid, the weaker the conjugate base. • The stronger the base, the weaker the conjugate acid. • H+ is the strongest acid that can exist in equilibrium in aqueous solution. • OH- is the strongest base that can exist in equilibrium in aqueous solution.

  7. Strong and Weak Acids and Bases Strong acids: completely ionized in water: HCl, HNO3, H2SO4 (also HBr, HI) Strong bases: completely ionized in water: MOH, where M = alkali M(OH)2, where M = alkaline earth Weak acids: incompletely ionized in water: any acid that is not strong - acetic acid, etc. Ka is finite. Weak bases: incompletely ionized in water: any base that is not strong – NH3, etc. Kb is finite.

  8. The Autoionization of Water The Ion Product of Water • In pure water the following equilibrium is established but [H2O]2 = constant (at 25oC) This is called the autoionization of water

  9. The Autoionization of Water In pure water at 25oC, [H3O+][OH-] = 1 x 10-14 and also, [H3O+]= [OH-] = 1 x 10-7 (From now on, for simplification, let’s use the abbreviation: [H+] = [H3O+] which means [H]+ = [OH-] = 1 x 10-7 We definepH = -log [H+] and pOH = -log [OH-] In pure water at 25oC, pH = pOH = 7.00 pH + pOH = 14 pKw = 14 Acidic solutions have pH < 7.00 Basic solutions have pH > 7.00

  10. The pH Scale Conc. Drano Battery acid

  11. The pH Scale • For [H+] = 3.4 x 10-5M, calculate pH, pOH and [OH-] • pH = 4.47, pOH = 9.53, [OH-] = 2.95 x 10-10 • (b) For [OH-] = 4.4 x 10-3M, calculate pH, pOH, and [H+] • pOH = 2.36, pH = 11.64, [H+] = 2.27 x 10-12 • (c) For pH= 8.9, calculate, pOH, [H+], [OH-] • pOH = 5.1, [H+] = 1.26 x 10-9, [OH-] = 7.94 x 10-6 • (d) For pOH= 3.2, calculate pH, [H+], [OH-] • pH = 10.8, [H+] = 1.58 x 10-11, [OH-] = 6.3 x 10-4 • The pH meter is the most accurate way to measure pH values of solutions.

  12. Use this “decision tree” to calculate pH values of solutions of specific solutions. • Is it pure water? If yes, pH = 7.00. • Is it a strong acid? If yes, pH = -log[HZ] • Is it a strong base? If yes, pOH = -log[MOH] • or pOH = -log (2 x [M(OH)2]) • Is it a weak acid? If yes, use the relationship • Ka = x2/(HZ – x), where x = [H+] • Is it a weak base? If yes, use the relationship • Kb = x2/(base – x), where x = [OH-] • Is it a salt (MZ)? If yes, then decide if it is neutral, acid, • or base; calculate its K value by the relationship • KaKb = Kw, where Ka and Kb are for a conjugate system; • then treat it as a weak acid or base. • Is it a mixture of a weak acid and its weak conjugate base? • It is a buffer; see next chapter.

  13. 2. Strong Acids Calculate the pH of 0.2 M HCl . pH = -log[H+] = -log[0.2] = 0.70 3. Strong Bases Calculate the pH of 0.2 M NaOH . pOH = -log[OH-] = -log[0.2] = 0.70 pH = 14 – 0.70 = 13.30 Calculate the pH of 0.2 M Ba(OH)2. pOH = -log[OH-] = -log[0.4] = 0.40 pH = 13.60

  14. 4. Weak Acids Weak acids are only partially ionized in solution, and are in equilibrium: (shorthand): OR: (shorthand) • Ka is the acid dissociation constant.

  15. 4. Weak Acids

  16. 4. Weak Acids Calculate pH of 0.2 M solution of acetic acid, HC2H3O2 From previous slide, Ka= 1.8 x 10-5 From now on, we’ll use the shorthand notation for HC2H3O2 = HAc Denote the acetate ion, C2H3O2- as “Ac-” Equilibrium is then: HAc  H+ + Ac- 0.2 0 0 Initial conc (M): -x +x +x change: 0.2-x x x at equilibrium: Ka= But, this is a quadratic equation!! We can assume x is small if Ka < 10-4. x = 1.90 x 10-3 = [H+] = [Ac-] pH = -log (1.90 x 10-3) = 2.72 Then, x2/0.2= 1.8 x 10-5

  17. 4. Weak Acids Polyprotic Acids • Polyprotic acids have more than one ionizable proton. • The protons are removed in steps not all at once: • It is always easier to remove the first proton in a polyprotic acid than the second. • Therefore, Ka1 > Ka2 > Ka3 etc. • Most H+(aq) at equilibrium usually comes from the first ionization (i.e. the Ka1 equilibrium).

  18. 4. Weak Acids Polyprotic Acids

  19. Carbonic acid, H2CO3, Ka1=4.3 x 10-7 H2CO3 H+ + HCO3- Ka1 = 4.3 x 10-7 HCO3- H+ + CO32- Ka2 = 5.6 x 10-11 Since Ka1>>Ka2, nearly all H+ ions come from 1st equilibrium. Therefore, the 1st equilibrium determines the pH. Calculate pH of 0.4 M H2CO3 solution. Ka = x2/HZ-x 4.3 x 10-7 = x2/0.4 x = [H+] = 4.15 x 10-4 pH = -log (4.1 x 10-4) = 3.38

  20. 5. Weak Bases • Weak bases remove protons from substances. • There is an equilibrium between the base and the resulting ions: Example: The base dissociation constant, Kb is defined as:

  21. 5. Weak Bases • The larger Kb the stronger the base.

  22. 5. Weak Bases What is pH of 0.15 M solution of methylamine, NH2CH3, a weak base? For convenience, denote it as “B” Then: B + H2O  BH+ + OH-Kb =4.4 x 10-4 Initially: 0.15 0 0 change: -x +x +x At equil: 0.15-x x x Then: - - = = 3 x 8.12 x 10 [OH ]

  23. Relationship Between Ka and Kb • For a conjugate acid-base pair KaKb = Kw (constant) • Therefore, the larger the Ka, the smaller the Kb. That is, the stronger the acid, the weaker the conjugate base. • Taking negative logarithms: -log Ka- log Kb= -log Kw pKa + pKb = pKw For HAc (acetic acid), the pKa = -log (1.8 x 10-5) = 4.74. Thus, the pKb for Ac- (acetate 0 is 14 – 4.74 = 9.26

  24. 6. Salts Salts may be acidic, basic or neutral. Salts made by a strong acid and a strong base are neutral, e.g., NaCl, KNO3. Salts made by a weak acid and a strong base are weakly basic, e.g., sodium acetate, NaAc, NaHCO3. Salts made by a strong acid and a weak base are weakly acidic, e.g., NH4Cl. Calculate the pH of a 0.35 M solution of sodium acetate. Since Ka Kb = Kw, where Ka = 1.8 x 10-5 for acetic acid, Then Kb for NaAc (sodium acetate) is Kb = Kw/Ka = 1.00 x 10-14/1.8 x 10-5 = 5.56 x 10-10. Now treat this as a weak base problem, Kb = x2/base = 5.56 x 10-10 = x2/0.35 x = [OH-] = 1.39 x 10-5 pOH = 4.85 and pH = 9.14

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