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Daniel L. Reger Scott R. Goode David W. Ball www.cengage.com/chemistry/reger Chapter 15 Solutions of Acids and Bases.

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Presentation on theme: "Daniel L. Reger Scott R. Goode David W. Ball www.cengage.com/chemistry/reger Chapter 15 Solutions of Acids and Bases."— Presentation transcript:

1 Daniel L. Reger Scott R. Goode David W. Ball www.cengage.com/chemistry/reger Chapter 15 Solutions of Acids and Bases

2 Acid: a substance that produces H 3 O + when dissolved in water. Base: a substance that produces OH - when dissolved in water. Arrhenius acids and bases are limited to water solutions. Arrhenius Acids and Bases

3 Acid: proton donor. Base: proton acceptor. Bronsted-Lowry acid base reaction: proton transfer from acid to base. Bronsted-Lowry Acids and Bases style

4 Conjugate Acid-Base Pairs: two species that differ by a proton. Acid : the species that contains the proton that is transferred. Conjugate base: the species formed by loss of the proton. HF(aq) + H 2 O( l ) ⇌ F - (aq) + H 3 O + (aq) HF and F - make up an acid-base conjugate pair. F - is the conjugate base of HF. Conjugate Acid-Base Pairs

5 HF(aq) + H 2 O( l ) ⇌ F - (aq) + H 3 O + (aq) HF is the acid, F - is its conjugate base (or F - is a base and HF is its conjugate acid). The reverse reaction is F - (aq) + H 3 O + (aq) ⇌ HF(aq) + H 2 O( l ) H 3 O + is the acid, H 2 O is its conjugate base. Acid-Base Conjugate Pairs

6 AcidBase HClCl - H 2 SO 4 HSO 4 - HSO 4 - SO 4 2- NH 4 + NH 3 H 3 O + H 2 O H 2 OOH - H 2 O is amphoteric - it acts as both an acid and a base. Acid-Base Conjugate Pairs

7 An Arrhenius system acid + base  water + salt is limited to reactions in water. A Bronsted-Lowry system HA + B ⇌ A - + BH + is more general - a proton is transferred from the acid (HA) to the base (B). Acid-Base Reactions

8 acid + base ⇌ conj. base + conj. acid HF + NH 3 ⇌ F - + NH 4 + HCl + H 2 O ⇌ Cl - + H 3 O + H 2 O + NH 2 - ⇌ OH - + NH 3 Bronsted-Lowry Acid-Base Reactions

9 HCN(aq) + H 2 O( l ) ⇌ H 3 O + (aq) + CN - (aq) HCN is the acid, CN - is its conjugate base. H 2 O is the base, H 3 O + is its conjugate acid. CH 3 COO - (aq)+H 2 O( l) ⇌ CH 3 COOH(aq)+OH - (aq) CH 3 COO - is the base, CH 3 COOH is its conjugate acid. H 2 O is the acid, OH - is its conjugate base. Identifying Conjugate Pairs

10 Identify the acid-base conjugate pairs HNO 2 (aq) + H 2 O( l ) ⇌ H 3 O + (aq) + NO 2 - (aq) CH 3 NH 2 (aq) + H 2 O( l ) ⇌ CH 3 NH 3 + (aq) + OH - (aq) Test Your Skill

11 H 2 O( l ) + H 2 O( l ) ⇌ H 3 O + (aq) + OH - (aq) H 3 O + and H 2 O are a conjugate acid- base pair. H 2 O and OH - are a conjugate acid- base pair. Autoionization of Water

12 H 2 O( l ) + H 2 O( l ) ⇌ H 3 O + (aq) + OH - (aq) The equilibrium constant for this reaction is called K w. K w = [H 3 O + ][OH - ] K w changes with temperature and is equal to 1.0 × 10 -14 at 25 o C. Autoionization of Water

13 H 2 O( l ) + H 2 O( l ) ⇌ H 3 O + (aq) + OH - (aq) In pure water, [H 3 O + ] = [OH - ]. K w = [H 3 O + ][OH - ] = 1.0 × 10 -14 K w = [H 3 O + ] 2 = 1.0 × 10 -14 [H 3 O + ] = [OH - ] = 1.0 × 10 -7 M Calculating Hydrogen and Hydroxide Ion Concentrations

14 Adding an acid or a base to water causes the H 3 O + and OH - concentrations to change. Calculate the hydroxide ion concentration of a solution in which the hydrogen ion concentration is 3.6 × 10 -3 M. Acidic or Basic Solutions

15 Calculate the hydrogen ion concentration in a solution in which the hydroxide ion concentration is 0.025 M. Test Your Skill

16 pH = -log 10 [H 3 O + ] Calculate the pH of a 0.0034 M solution of [H 3 O + ]. pH = -log(0.0034) = 2.47 pH values are generally given to two decimal places. A pH of 2.47 has two significant figures; the 2 serves to locate the decimal point of the original number. pH Scale

17 Calculate the hydrogen ion concentration in a solution that has a pH of 3.52. Calculating Hydrogen Ion Concentration from pH

18 The p-notation can be used for other quantities. pOH = -log[OH - ] pK w = -log K w = -log(1.0 × 10 -14 ) = 14.00 p-notation

19 [H 3 O + ][OH - ] = K w = 1.0 × 10 -14 log([H 3 O + ][OH - ]) = log K w log[H 3 O + ] + log[OH - ] = log K w -log[H 3 O + ] - log[OH - ] = -log K w pH + pOH = pK w pH + pOH = 14.00 Relating pH and pOH

20 pH + pOH = 14.00 AcidNeutralBase pH 7 pOH >7 =7 <7 Relating pH and pOH

21 HA(aq) + H 2 O( l ) H 3 O + (aq) +A - (aq) Strong acids ionize completely in solution. Memorize the six common strong acids: HCl, HBr, HI, HNO 3, HClO 4, and H 2 SO 4. 100% Strong Acids and Bases

22 Calculate the pH of a 0.050 M HCl solution. pH of a Strong Acid Solution

23 Strong bases quantitatively produce hydroxide ions in water. The most common strong bases are the group IA and soluble IIA oxides and hydroxides. Strong Bases

24 NaOH(s)  Na + (aq) + OH - (aq) Li 2 O(s) + H 2 O( l )  2Li + (aq) + 2OH - (aq) Strong Bases

25 Calculate the pH of a 0.035 M Ba(OH) 2 solution. pH of a Strong Base Solution

26 What is the concentration of a solution of HCl if the pH is 3.75? What is the concentration of a solution of KOH if the pH is 11.60? Test Your Skill

27 Weak acids and weak bases are those that do not ionize completely in water. A weak acid exists in equilibrium with its conjugate base. HCN(aq) + H 2 O( l ) ⇌ H 3 O + (aq) + CN - (aq) A weak base exists in equilibrium with its conjugate acid. CH 3 NH 2 (aq) + H 2 O( l ) ⇌ CH 3 NH 3 + (aq) + OH - (aq) Weak Acids and Bases

28 HA(aq) + H 2 O( l ) ⇌ H 3 O + (aq) + A - (aq) B(aq) + H 2 O ⇌ BH + (aq) + OH - (aq) Weak Acids and Bases

29 HA (aq) + H 2 O( l ) ⇌ H 3 O + (aq) + A - (aq) Both H 2 O and A - are bases and compete for the proton. If HA is a strong acid, A - is an extremely poor proton acceptor. HCl (aq) + H 2 O( l ) → H 3 O + (aq) + Cl - (aq) Proton bonded to Cl - Proton bonded to H 3 O + Competition for Protons

30 HA (aq) + H 2 O( l ) ⇌ H 3 O + (aq) + A - (aq) If HA is a weak acid, A - is a weak base. The stronger the weak acid, the weaker the conjugate base. HF(aq) + H 2 O ⇌ H 3 O + (aq) + F - (aq) Proton bonded to F - Proton bonded to H 3 O + Competition for Protons

31 Leveling Effect: In water, all acids stronger than H 3 O + appear to be equally strong. By using a weaker base than water as the solvent, we can differentiate between the strong acids. weakest strongest | | HNO 3 < H 2 SO 4 < HCl < HBr < HI < HClO 4 The Influence of the Solvent

32 Analytical Concentration: the total concentration of all forms of an acid; both the protonated form (the acid) and the unprotonated form (the conjugate base). The Concentration of an Acid

33 You will likely see two kinds of equilibrium calculations: 1.You measure equilibrium concentrations and calculate K or 2.you are given starting concentrations and K and calculate equilibrium concentrations. The approach is the same. 1.Write the balanced chemical equation. 2.Calculate equilibrium concentrations/iCe table. 3.Write the algebraic expression for K. 4.Substitute concentrations into expression. 5.Solve. Equilibrium Calculations

34 A 0.250 M solution of HF is determined to be 3.7% ionized. Determine K a for HF. Calculating K a for a Weak Acid

35 The pH of a 0.100 M solution of HOCl is 4.26. Calculate K a for the acid. HOCl + H 2 O ⇌ H 3 O + + OCl - Determining K a from pH

36 Calculate the pH of a 0.50 M CH 3 COOH solution. K a = 1.8 × 10 -5 CH 3 COOH + H 2 O ⇌ H 3 O + + CH 3 COO - Determining Concentrations of Species in Weak Acid Solution

37 Determine the pH of a 0.025 M solution of HCN, K a = 7.2 × 10 -10. Test Your Skill

38 Calculate the equilibrium concentrations and pH of a 0.100 M HF solution, K a = 6.3 × 10 -4. Use approximation. Calculating pH of a Weak Acid Solution

39 Is 7.9  10 -3 << 0.050? No, it is not. 5% of 0.050 is 2.5  10 -3 and the calculated value is larger. The approximation fails. We need another method to compute the root. Check Approximation

40 Successive Approximation

41 Is y 2 sufficiently close to y 1 ? Restating the question, is 7.6  10 -3 within 5% of 7.9  10 -3 ? By within 5%, we mean is 7.6  10 -3 within a range varying from 95% to 105% of 7.9  10 -3 ? The answer is yes. Since the second approximation was quite close to the first, we can accept it. Check Second Approximation

42 HA(aq) + H 2 O( l ) ⇌ H 3 O + (aq) + A - (aq) Fraction Ionized in Solution

43 Calculate fraction ionized for 0.500 M HOCl, K a = 4.0 × 10 -8. HOCl + H 2 O ⇌ H 3 O + + OCl - Fraction Ionized in Solution

44 A weak base reacts with water to form hydroxide ions. B(aq) + H 2 O( l ) ⇌ BH + (aq) + OH - (aq) Solutions of Weak Bases

45 CH 3 NH 2 + H 2 O ⇌ CH 3 NH 3 + + OH - Calculate the pH of a solution of 0.150 M methylamine, CH 3 NH 2, K b = 4.4 × 10 -4. pH of a Solution of a Weak Base

46 Calculate the pH of a 0.35 M solution of hydroxylamine, NH 2 OH, K b = 1.1 × 10 -8. Test Your Skill

47 HA(aq) + H 2 O( l ) ⇌ H 3 O + (aq) + A - (aq) A - (aq) + H 2 O( l ) ⇌ HA(aq) + OH - (aq) Relating K a and K b

48

49 The conjugate base of a weak acid is a weak base: CH 3 COOH + H 2 O ⇌ H 3 O + + CH 3 COO - K a = 1.8 × 10 -5 CH 3 COO - + H 2 O ⇌ CH 3 COOH + OH - K b = 5.6 × 10 -10 HCN + H 2 O ⇌ H 3 O + + CN - K a = 6.2 × 10 -10 CN - + H 2 O ⇌ HCN + OH - K b = 1.6 × 10 -5 Strengths of Weak Acid-Base Conjugate Pairs

50 K a for HNO 2 is 4.6 x 10 -4. Calculate K b for the NO 2 - ion. NO 2 - (aq) + H 2 O( l ) ⇌ HNO 2 (aq) + OH - (aq) Calculating K a and K b for Acid-Base Conjugate Pairs

51 K b for pyridine is 1.8 × 10 -9. Calculate K a for the pyridinium ion. Test Your Skill

52 K b = 2.8 x 10 -11 K b = 1.4 x 10 -5 K a = 7.2 x 10 -10 K a = 3.5 x 10 -4 acid Stronger acid HCN HF Weaker base Stronger Weaker CN - F - base Ranking Acid-Base Strength

53 The stronger the acid, the weaker the conjugate base. strong acid  very weak conjugate base strong base  very weak conjugate acid weak acid  weak conjugate base Conjugate Partners of Strong Acids and Bases

54 The pH of a salt solution is  =7 (neutral) if it contains the cation of a strong base (Na + ) and the anion of a strong acid (Cl - )  >7 (basic) if it contains the cation of a strong base (K + ) and the anion of a weak acid (F - )  <7 (acidic) if it contains the cation of a weak base (NH 4 + ) and the anion of a strong acid (NO 3 - ). pH of Salt Solutions

55 Determine the pH of a 0.59 M solution of CH 3 OONa. K a of CH 3 OOH = 1.8 × 10 -5. CH 3 OO - (aq) + H 2 O( l ) ⇌ CH 3 OOH(aq) + OH - (aq) The acetate ion is a base. Calculate K b for the acetate ion: The pH of a Salt Solution

56 When two or more acids are present in solution, the effects of the weaker acids may be ignored if K a is smaller by a factor of 100 or more. Mixtures of Acids

57 To calculate the pH of a mixture of HCl (strong acid) and HOCl (K a = 3.0 × 10 -8 ), ignore the HOCl; treat the solution as a solution of HCl. To calculate the pH of a mixture of HCOOH (K a = 1.8 × 10 -4 ) and HCN (K a = 7.2 × 10 -10 ), ignore the HCN; treat the solution as a solution of HCOOH. Mixtures of Acids

58 Binary hydrides contain hydrogen and one other element. Acidity depends on the HA bond strength and the stability of A -. Binary Hydrides

59 Acidity increases with decreasing H-A bond dissociation energy HF< { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/8/2296069/slides/slide_59.jpg", "name": "Acidity increases with decreasing H-A bond dissociation energy HF<

60 Acidity increases with increasing electronegativity of A. CH 4 < NH 3 < H 2 O < HF 2.5 3.0 3.5 4.0 Binary Hydrides

61 Oxyacids contain H, O, and a third element X. X–O–H X is usually a non-metal or a transition metal in a high oxidation state. Oxyacids

62 As the electronegativity of X increases, the X-O bond becomes more covalent and the O-H bond becomes more polar. X–O–H  X–O - + H + Oxyacids

63 Two factors increase acidity of the oxyacid:  Greater electronegativity of central atom. HClO 4 > HBrO 4  Greater oxidation number of central atom (more oxygen atoms). HClO 4 > HClO 3 > HClO 2 > HClO Oxyacids

64 Lewis acid: electron pair acceptor. Lewis base: electron pair donor. Lewis acid-base reaction: formation of a coordinate-covalent bond. AlCl 3 + Cl -  AlCl 4 - Coordinate-covalent bond: covalent bond in which both electrons come from one atom. Lewis Acids and Bases


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