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Chapter 14 Aqueous Equilibria Acids and Bases

Chapter 14 Aqueous Equilibria Acids and Bases. Definitions of Acids and Bases. Arrhenius gave the oldest definition of acids and bases. Acids - substances that increase the H + concentration H 2 O of water: HCl H + + Cl -

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Chapter 14 Aqueous Equilibria Acids and Bases

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  1. Chapter 14 Aqueous Equilibria Acids and Bases

  2. Definitions of Acids and Bases • Arrhenius gave the oldest definition of acids and bases. • Acids - substances that increase the H+ concentration • H2O • of water: HClH+ + Cl- • Bases - substances that increase the OH- concentration • H2O • of water: NaOH Na+ + OH- • Bronsted-Lowry expanded Arrhenius’ definition to include non-aqueous solutions and substances that don’t have H+andOH-groups. • Acids - substances that can transfer a proton (H+). • Bases - substances that can accept a proton (H+). • (Acid)(Base)(Conjugate Acid)(ConjugateBase) • HA + BBH+ + A-

  3. Problem • 14.1 Write a balanced equation for the dissociation of each of the following Bronsted-Lowry acids in water. • H2SO4 (b) HSO4- (c) H3O+ (d) NH4+ • 14.2 What is the conjugate acid of each of the following Bronsted-Lowry bases? • (a) HCO3- (b) CO3-2 (c) OH- (d) H2PO4-

  4. Acid Strength and Base Strength • Generally strong acids and bases are completely dissociated in aqueous solutions. • Acids: HCl, HBr, HI, HNO3, H2SO4, HClO4 • Bases: NaOH, KOH, LiOH, Ca(OH) 2, O-2, CH3-, NH2- • Since acids and bases are chemically opposites, strong acids are very weak bases and strong bases are very weak acids. • Generally weak acids and bases are those which are NOT completely dissociated in aqueous solutions. • Acids: HF, HSO4-, H3PO4, HNO2, H2S, H2CO3, CH3COOH • Bases: F-, SO4-2, NO4-, NH4OH, H2O, CH3NH2 • The conjugate base of a weak acid is a weak base and vice-versa.

  5. Problem • 14.4 If you mix equal concentrations of reactants and products, which of the following reactions proceed to the right and which proceed to the left. • HF(aq) + NO3-(aq) ↔ HNO3(aq) + F-(aq) • (b) NH4+(aq) + CO3-2(aq) ↔ HCO3-(aq) + NH3(aq)

  6. Dissociation of Water • Water is both a weak acid and a weak base. (Acid)(Base)(Conjugate Acid)(ConjugateBase) H2O + H2OH3O+ + OH- • The overall equation is: 2H2O(l) H3O+(aq) + OH-(aq) • The equilibrium constant for the reactions is: • [H3O+][OH-] • Kc = Kw = ----------------- = [H3O+][OH-] = 1.0 x 10-14 • [H2O] • Since the concentrations of H3O+and OH- are equal for pure water, the concentration of each is • [H3O+]= [OH-] = 1.0 x 10-7

  7. Kw Calculations • In acid and basic solutions, the concentrations of H3O+and OH- are not equal. • In acidic solutions: H3O+ > OH- • In basic solutions: OH- > H3O+ • In neutral solutions: H3O+ = OH- • The product of their concentrations however would still be equal to Kw. • Problem • 14.6 The concentration of OH- in a sample of seawater is 5.0 x 10-6 M. Calculate the concentration of H3O+ ions, and classify the solution as acidic, neutral or basic. • 14.7 At 50oC the value of Kw is 5.5 x 10-14. What are the concentrations of H3O+ and OH- in a neutral solution at 50oC.

  8. The pH Scale • Expressing the [H+] and [OH-] concentrations in molarity is very inconvenient because of the very small concentrations involved. • A better method is to express the [H+] concentration as a negative logarithm (pX = -log [X]) to make the value a small, positive number. pH = - log [H+] • Conversely given the pH the [H+] concentration can be calculated: [H+] = e-pH • For acidic solutions pH < 7, for basic solutions pH > 7 and for neutral solutions pH = 7. • Other useful relationships are: • pOH = - log [OH+], pKw = - log [Kw] = pH + pOH = 14

  9. Measuring pH • pH can be measured using acid-base indicators. Indicators are weak organic acids or bases. • HI ↔ H+ + I- • color A color B • According to Le-Chatelier’s principle, addition of acids causes the equilibrium to shift to the left producing color A and addition of bases causes a right-hand shift producing color B. • A commonly used indicator is phenolphthalein which is colorless up to pH 9 but purple above pH 9. • Another is methyl orange which is red below pH 4 but orange above pH 4.

  10. Problem • 14.8 Calculate the pH of the following solutions: • A sample of seawater that has an OH- concentration of 1.58 x 10-6M • A sample of acid rain that has an H3O+ concentration of 6.0 x 10-5M. • 14.9 Calculate the concentrations of H3O+ and in each of the following solutions: • Human blood (pH 7.40) • (b) A cola beverage (pH 2.8)

  11. The pH of Strong Acids and Bases • The pH of strong acids and bases are easy to calculate since they are 100% dissociated. • This means that the [H+] and [OH-] concentrations are simply equal to the acid or base concentrations. • Ex: 0.01M HCl would have [H+] = 0.01M • pH = -log [0.01M] = 2. • 0.105M NaOH would have [OH-] = 0.105M • pOH = - log [0.105M] = 0.979, pH = 14.0 – 0.979 = 13.0 • Problem • 14.10 Calculate the pH of the following solutions: • 0.050M HClO4 (b) 0.010M Ba(OH)2 • 14.11 Calculate the pH of a solution prepared by dissolving 0.25g of BaO in enough water to make o.500L of solution.

  12. Equilibria of Weak Acids • Unlike strong acids weak acids are not 100% dissociated. • The equilibrium constant of weak acids can be calculated using the ICE method. The equilibrium constant for acids is usually designated as Ka. • HA (aq) ↔ H+(aq) + A-(aq) • [H+][A-] • Ka = ------------ • [HA] • Problem • 14.12 The pH of 0.10M HOCl is 4.23. Calculate the Ka and pKa for hypochlorous acid.

  13. Calculating Equilibrium Concentrations in Solutions of Weak Acids • For weak acids, the pH must be calculated from the equilibrium constants of the acid or base. • The ICE method is used to calculate for the concentration of the [H+] concentrations. • Since the Ka’s for weak acids are extremely small, the concentrations of the ions formed are also small compared to the original concentration of the weak acid. • The calculations may therefore be greatly simplified by making approximations.

  14. Problem • 14.14 Acetic acid, CH3COOH, is the solute that gives vinegar its characteristic odor and sour taste. Calculate the pH and the concentrations of all species present (H3O+, CH3COO-, and OH-) in: • 1.00M CH3COOH • 0.0100M CH3COOH

  15. Dissociation (%) of Weak Acids • The dissociation of weak acids can also be expressed as a percentage of the dissociated species. • Percent dissociation is defined as the concentration of the dissociated species divided by the initial concentration of the acid multiplied by 100. • [H+] or [A-] • % dissocation = ---------------- • [HA]initial • Generally, the more dilute the acid the greater the % dissociation. • Problem • 14.16 Calculate the % dissociation of HF ( Ka = 3.5 x 10-4) in: (a) 0.050M HF (b) 0.50M HF

  16. Polyprotic Acids • Polyprotic acids are acids with more than 1 replaceable H+ions. • Ex: H2SO4, H2S, H3PO4 • Polyprotic acids have more than 1 equilibrium constant Ka1 , Ka2, … corresponding to the number of replaceable H+ions. The Ka’s progressively gets smaller Ka1>Ka2>Ka3 ... • There are more species present when polyprotic acids dissociate than in a monoprotic acid. • Problem • 14.17 Calculate the pH and the concentration of all species present in 0.10M H2SO3. Ka values in Table 14.3.

  17. Equilibria in Solutions of Weak Bases • A weak base reacts reversibly with water to form the OH- ion. • B(aq) + H2O(l) ↔ BH+(aq) + OH- (aq) • [BH+][OH- ] • Kb = ---------------- • [B] • pH calculations proceed as with weak acid calculations. Instead of calculating for the [H+] however, the [OH- ] is calculated first and the pH calculated using pKw. • Problem • 14.19 Calculate the pH and the concentrations of all species present in 0.40M NH3 (Kb = 1.8 x 10-5).

  18. Relationship between Ka and Kb • The overall equilibrium constant of a series of reversible reactions is the product of the individual equilibrium constants: Koverall = K1 x K2 x K3 x … • For an conjugate acid-base pair: • Koverall = Kw = Ka x Kb • Taking the –log of all the terms above gives the relationship: pKw = pKa + pKb • Problem • 14.21 (b) Find Ka for HOCl in Appendix C, and the calculate Kb for OCl-? • (c) The value of pKa for formic acid (HCOOH) is 3.74. What is the value of pKb for the formate ion (HCOO-)?

  19. Hydrolysis of Salts • Salts of strong acids and bases are neutral. • Salts of strong acids and weak bases are acidic. • HCl + NH3 ↔ NH4+ + Cl- • acidic • NH4+hydrolyzes (reacts with water) releasing H+ making the solution acidic. • NH4+ + H2O ↔ NH4OH + H+ • Salts of strong bases and weak acids are basic. • NaOH + CH3COOH ↔ CH3COO- + Na + + H2O • basic • CH3COO-hydrolyzes releasing OH- making the resulting solution basic. • CH3COO- + H2O ↔ CH3COOH + OH-

  20. Problem 14.23 Calculate the pH of 0.20M NaNO2; Ka for HNO2 is 4.6 x 10-4. NO2- hydrolyzes to: NO2- + H2O ↔ HNO2 + OH- [H+][NO2-] Ka = -------------- ; Kw = [H+][OH-] [HNO2] [HNO2][OH-] Kw Khydr = ----------------- = ---- [NO2-] Ka

  21. Factors that Affect Acid Strength • A strong acid (HX) is one where there is complete dissociation between the cation (H+) and the anion (X-). • The stronger the bond between H – X the less likely that it will be completely dissociated. • The more electronegative the anion of the acid, the stronger the H – X bond, the weaker the acid. • For Groups VI and VII elements, the farther down the group the lesser the electronegativity, making the H – X bond weaker. The acid formed from these elements would be stronger. • For oxoacids, the more electron withdrawing (more electronegative) the atom(s) attached to the oxygen that is bonded to the hydrogen atom the more acidic the oxoacid. • H – O – X • Where O – X is an oxo anion.

  22. Problem • 14.26 Identify the stronger acid in each of the following pairs: • H2S or H2Se • HI or H2Te • HNO2 or HNO3 • H2SO3 or H2SeO3

  23. Lewis Acids and Bases • G.N. Lewis proposed an even more inclusive definition of acids and bases. • Lewis defined acids as electron-pair acceptors and bases as electron-pair donors. • According to this definition :NH3 is a base because it can donate its lone electron pair and H+ is an acid because it can accept lone pairs of electrons. • H+ + : NH3 ↔ H:NH3 • Similarly BF3 is a Lewis acid: • BF3+ : NH3 ↔ F3B:NH3

  24. Problem • 14.27 For each of the following identify the Lewis acid and Lewis base. • AlCl3 + Cl- ↔ AlCl4- • 2NH3 + Ag+ ↔ Al(NH3)2 +

  25. Recitation Problems Starting page 579 of your book 14.33 14.55a,b 14.85 14.35 14.59a 14.93c,e 14.45a-d 14.65d 14.99a,b 14.49b,d 14.79 14.107b,d

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