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© 2008 Brooks/Cole 1 Chapter 16:Acids and Bases Chemistry: The Molecular Science Moore, Stanitski and Jurs.

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Presentation on theme: "© 2008 Brooks/Cole 1 Chapter 16:Acids and Bases Chemistry: The Molecular Science Moore, Stanitski and Jurs."— Presentation transcript:

1 © 2008 Brooks/Cole 1 Chapter 16:Acids and Bases Chemistry: The Molecular Science Moore, Stanitski and Jurs

2 © 2008 Brooks/Cole 2 Arrhenius Definition Arrhenius: any substance which ionizes in water to produce… Better version of the Arrhenius definition: Acid: Acid: hydronium ions (H 3 O + ) in water are acidic. Base Base: hydroxide ions (OH - ) in water are basic. acid.protons (H + ) is an acid. base.hydroxide ions (OH - ) is a base.

3 © 2008 Brooks/Cole 3 Acid-Base Reactions A strong acid is an acid that ionizes completely in water (product favored). There are six strong acids: HCl - hydrochloric acid HBr – hydrobromic acid HI – hydroiodic acid H 2 SO 4 - sulfuric acid HNO 3 – nitric acid HClO 4 – perchloric acid

4 © 2008 Brooks/Cole 4 Acid-Base Reactions Strong Acids and Bases A strong base is a base that is present entirely as ions (product favored). The hydroxides of Group IA, IIA (except Be and Mg hydroxides) are strong bases. So, w So, why is NH 3 (aq) basic?

5 © 2008 Brooks/Cole 5 Brønsted-Lowry Concept An alternative definition: NH 3 (g) + H 2 O( l ) NH 4 + (aq) + OH - (aq) indirectly produced Base: H + acceptor Acid: H + donor acid acid= proton (H + ) donor base base= proton (H + ) acceptor Works for non-aqueous solutions and explains why NH 3 is basic:

6 © 2008 Brooks/Cole 6 Weak Weak acids and bases do not fully ionize (reactant favored). Brønsted-Lowry Concept Strong Strong acids and bases almost completely ionize (product favored). HNO 3 (aq) + H 2 O( l )H 3 O + (aq) + NO 3 - (aq) Note: the products are a new acid and base pair. LeChatelier – a more dilute solution (more water), will ionize more. HF(aq) + H 2 O( l )H 3 O + (aq) + F - (aq) Base: H + acceptor Acid: H + donor Base: H + acceptor Acid: H + donor 100% 3%

7 © 2008 Brooks/Cole 7 Water’s Role as Acid or Base Water acts as a base when an acid dissolves in water: HBr(aq) + H 2 O (l) H 3 O + (aq) + Br - (aq) acid base amphiprotic Water is amphiprotic - it can donate or accept a proton (act as acid or base). But water acts as an acid for some bases: H 2 O (l) + NH 3 (aq) NH 4 + (aq) + OH - (aq) acid base acid base

8 © 2008 Brooks/Cole 8 Practice Complete the equations: (acid or base?, mass/charge balance?, single or double arrow?) HClO 4 + H 2 O CN - + H 2 O H 2 S + H 2 O NO H 2 O Ca(OH) 2 + H 2 O ClO H 3 O + HCN + HO - HS - + H 3 O + HNO 2 + HO - Ca OH -

9 © 2008 Brooks/Cole 9 Conjugate Acid-Base Pairs one Molecules or ions related by the loss/gain of one H +. Conjugate AcidConjugate Base H 3 O + H 2 O CH 3 COOH CH 3 COO - NH 4 + NH 3 H 2 SO 4 HSO 4 - HSO 4 - SO 4 2- HC l C l - donate H + accept H + not NH 4 + and NH 2 - are not conjugate (conversion requires 2 H + )

10 © 2008 Brooks/Cole 10 Identify the base conjugate to HC 2 H 3 O 2 (aq) and the acid conjugate to HCO 3 - (aq) HCO H 2 O OH - + HC 2 H 3 O 2 + H 2 O H 3 O + + Conjugate Acid Base: H + acceptor Conjugate Base Acid: H + donor Conjugate Acid-Base Pairs Show that HCO 3 - is amphoteric. C 2 H 3 O 2 - H 2 CO 3

11 © 2008 Brooks/Cole 11 Practice: Conjugate Acid-Base Pairs Write the equation for the acid or base in water and identify the acid-base pairs. HBr HSO 3 - (as an acid) HSO 3 - (as a base) PH 4 +

12 © 2008 Brooks/Cole 12 Relative Strength of Acids & Bases Strong acids are better H + donors than weak acids. Strong bases are better H + acceptors than weak bases.  Strong acids have weak conjugate bases.  Weak acids have strong conjugate bases. Strong acid+ H 2 OH 3 O + + weak conjugate base Fully ionized, reverse reaction essentially does not occur. The conjugate base is weak. Weakly ionized, reverse reaction readily occurs. The conjugate base is strong. Weak acid + H 2 OH 3 O + + strong conjugate base

13 © 2008 Brooks/Cole 13 Conjugate acid Conjugate base H 2 SO 4 HSO 4 - HBrBr - HC l C l - HNO 3 NO 3 - H 3 O + H 2 O H 2 SO 3 HSO 3 - HSO 4 - SO 4 2- H 3 PO 4 H 2 PO 4 - HFF - CH 3 COOHCH 3 COO - H 2 SHS - H 2 PO 4 - HPO 4 2- NH 4 + NH 3 HCO 3 - CO 3 2- H 2 OOH- OH - O 2- H 2 H - CH 4 CH 3 - Acid strength increasing Base strength increasing stong acids strong base extremely weak acids extremely weak bases Relative Strength of Acids & Bases

14 © 2008 Brooks/Cole 14 HF is a stronger acid than NH 4 +. (NH 3 is a stronger base than F - ) HF has greater tendency to ionize than NH 4 +. (NH 3 more readily accepts H + than F - ) Reactant Favored NH F - NH 3 + HFProblem Is the following aqueous reaction product or reactant favored? Acid strength increasing Base strength increasing Conj acid. Conj. base H 2 SO 4 HSO 4 - HBrBr - HC l C l - HFF - NH 4 + NH 3 OH - O 2- H 2 H - CH 4 CH 3 - Relative Strength of Acids & Bases

15 © 2008 Brooks/Cole 15 Carboxylic Acids C 3 H 7 COOH CH 3 COOH

16 © 2008 Brooks/Cole 16 Amines (bases) CH 3 NH 2 (CH 3 ) 2 NH(CH 3 ) 3 N proton acceptors

17 © 2008 Brooks/Cole 17 Autoionization of Water Pure water conducts a very small electrical current. Autoionization Autoionization occurs: H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) K w = ionization constant for water Heavily reactant favored. K w = [ H 3 O + ] [ OH - ] K w = 1.0 x (at 25°C) Base Acid Base

18 © 2008 Brooks/Cole 18 Ionization Constant for Water K w is a special equilibrium constant for the autoionization of water. It is T-dependent. T = 25°C (77°F) is usually used as the standard T. T (°C) K w x x x x x x K w increases with temperature. Is it endo or exothermic?

19 © 2008 Brooks/Cole 19 Autoionization of Water Hydronium and hydroxide ions are produced in equal numbers in pure water, Thus, the concentrations of H 3 O + and OH - in pure water are both 1.0 x M. K w = [ H 3 O + ] [ OH - ] 2H 2 O (l) H 3 O + (aq) + OH - (aq)

20 © 2008 Brooks/Cole 20 Autoionization of Water all H 3 O + and OH - are present in all aqueous solutions. Neutral solution. Pure water 25°C): [H 3 O + ] = M = [OH - ] Acidic solution If acid is added to water: [ H 3 O + ] is increased, disturbing the equilibrium: 2 H 2 O H 3 O + + OH - LeChatelier: equilibrium shifts to the left. [OH - ] Equilibrium is reestablished: [ H 3 O + ] > M > [OH - ] Product of ion concentrations is the same [ H 3 O + ][OH - ] = K w = 1.0 x

21 © 2008 Brooks/Cole 21 Autoionization of Water Basic solution. If base is added to water. LeChatelier: equilibrium shifts to the left, [H 3 O + ] New equilibrium: [H 3 O + ] < M < [OH - ] [ H 3 O + ][OH - ] = K w = 1.0 x at 25°C Example Calculate the hydronium and hydroxide ion concentrations at 25°C in a 6.0 M aqueous sodium hydroxide solution. K w = [H 3 O + ][OH - ] = 1.0 x NaOH (aq) is strong (100% ionized) so [OH - ] = 6.0 M. [H 3 O + ](6.0) = 1.0 x [H 3 O + ] = 1.7 x M[ OH - ] = 6.0 M 2 H 2 O H 3 O + + OH -

22 © 2008 Brooks/Cole 22 Autoionization of Water In a neutral solution, the concentrations of H 3 O + (aq) and OH - (aq) are equal. In an acidic solution, the concentration of H 3 O + (aq) is greater than that of OH - (aq). In a basic solution, the concentration of OH - (aq) is greater than that of H 3 O + (aq). The relative concentrations of H 3 O + and OH - indicate acidic, neutral or basic character of a solution:

23 © 2008 Brooks/Cole 23 At 25 o C, we observe the following conditions. Acid or Base In an acidic solution, [H 3 O + ] > 1.0 x M. In a neutral solution, [H 3 O + ] = [OH - ] = 1.0 x M. In a basic solution, [H 3 O + ] < 1.0 x M.

24 © 2008 Brooks/Cole 24 The pH scale Can describe acidity in terms of [H 3 O + ], but a logarithmic scale is more convenient: pH = −log 10 [H 3 O + ] At 25°C a neutral aqueous solution has: pH = −log 10 [1.0 x ] = −(−7.00) = 7.00 acidic solutions: [H 3 O + ]>1.0 x 10 -7, pH < 7.00 basic solutions: [H 3 O + ] 7.00

25 © 2008 Brooks/Cole 25 The pH of a Solution For a solution in which the hydronium-ion concentration is 1.0 x 10 -3, the pH is: Note that the number of decimal places in the pH equals the number of significant figures in the hydronium-ion concentration.

26 © 2008 Brooks/Cole 26 Example – pH given [H 3 O + ] ? A sample of orange juice has a hydronium-ion concentration of 2.9 x M. What is the pH?

27 © 2008 Brooks/Cole 27 Example - [H 3 O + ] given pH? The pH of human arterial blood is 7.40 (acidic or basic?). What is the hydronium-ion concentration? (Guesss….> or < 1.0x10 -7 M?) 10 logx = x

28 © 2008 Brooks/Cole 28 The pOH scale pOH = −log 10 [OH - ] A neutral solution (25°C) has: pOH = −log 10 [1.0 x ] = −(−7.00) = 7.00 Since K w = [ H 3 O + ][ OH - ] = 1.0 x −log(K W )= −log[H 3 O + ] + (−log[OH - ]) = −log(1.0 x ) pK w = pH + pOH = 14.00

29 © 2008 Brooks/Cole 29 Example An ammonia solution has a hydroxide-ion concentration of 1.9 x M. What is the pH of the solution? we first calculate the pOH: Then the pH is:

30 © 2008 Brooks/Cole 30 pH Calculations Solution A:pOH = −log[ OH - ] = 3.37 pH + pOH = pK w = pH = Solution B:pH = −log[ H 3 O + ] = 8.12 A has higher pH, B is more acidic Given two aqueous solutions (25°C). Solution A: [OH - ] = 4.3 x M, Solution B: [H 3 O + ] = 7.5 x M. Which has the higher pH? Which is more acidic?

31 © 2008 Brooks/Cole 31 Practice - WSs [H + ] K w =1x = [H + ] [OH - ] [OH - ] pH 14 = pH + pOH pOH pH= -log [H + ] [H + ] = 10 -pH pOH =-log [OH - ] [OH - ] = 10 -pOH

32 © 2008 Brooks/Cole 32 pH of Aqueous Solutions Stomach Fluids Lemon Juice Vinegar Wine Tomatoes Black Coffee Milk Pure water Blood, seawater Sodium bicarbonate Borax solution Milk of Magnesia Detergents Aqueous ammonia 1 M NaOH [H 3 O + ] [OH - ] Example Battery AcidpH Bleach

33 © 2008 Brooks/Cole 33 Measuring pH H 3 O + concentrations can be measured with an: pH meterElectronic pH meter:  fast and accurate  preferred method. IndicatorAcid-base Indicator :  substance that changes color within a narrow pH range  may have multiple color change (e.g. bromthymol blue)  one “color” may be colorless (e.g. phenolphthalein)  cheap and convenient.

34 © 2008 Brooks/Cole 34 HA(aq) + H 2 O( l ) H 3 O + (aq) + A - (aq) acid ionization constant The acid ionization constant is used to report the degree of ionization: [A - ][H 3 O + ] [HA] K a = (Water omitted, as usual) Strong acids have large K a values Weak acid have small K a values Ionization Constants of Acids and Bases When an acid ionizes in water:

35 © 2008 Brooks/Cole 35 Base Ionization Constants For a base in water: base ionization constant The base ionization constant, K b, is: K b = [BH + ][OH - ] [B] A - (aq) + H 2 O( l ) HA(aq) + OH - (aq) K b = [HA][OH - ] [A - ] B(aq) + H 2 O( l ) BH + (aq) + OH - (aq) If the base is an anion:

36 © 2008 Brooks/Cole 36 Larger K a = stronger acid Larger K b = stronger base Ionization Constants

37 © 2008 Brooks/Cole 37 Larger K a = stronger acid Larger K b = stronger base Ionization Constants

38 © 2008 Brooks/Cole 38 Acid and Base Strength Small value of K a indicates that the numerator is smaller than the denominator. Equilibrium favors reactants or only a small amount dissociates – a weak acid. Small value of K b indicates that the numerator is smaller than the denominator. Equilibrium favors reactants or only a small amount dissociates – a weak base.

39 © 2008 Brooks/Cole 39 Write the ionization equation and ionization constant expression for the following acids and bases: a.) HF b.) HBrO c.)NH 3 d.) CH 3 NH 2 Example

40 © 2008 Brooks/Cole 40 K a Values for Polyprotic Acids Some acids can donate more than one H + FormulaNameAcidic H’s H 2 SHydrosulfuric Acid2 H 3 PO 4 Phosphoric Acid 3 H 2 CO 3 Carbonic Acid2 HOOC-COOHOxalic acid2 C 3 H 5 (COOH) 3 Citric acid3 Each H + ionization has a different K a. The 1 st proton is easiest to remove The 2 nd is harder, etc.

41 © 2008 Brooks/Cole 41 K a Values for Polyprotic Acids Phosphoric acid (H 3 PO 4 ) is weak. It has three acidic protons: H 3 PO 4 (aq) + H 2 O (l) H 3 O + (aq) + H 2 PO 4 - (aq) K a = 7.5 x H 2 PO 4 - (aq) + H 2 O (l) H 3 O + (aq) + HPO 4 2- (aq) K a = 6.2 x HPO 4 2- (aq) + H 2 O (l) H 3 O + (aq) + PO 4 3- (aq) K a = 3.6 x K a decreasing H + is harder to remove

42 © 2008 Brooks/Cole 42 Practice – Relative Strength (K a and K b ) The following reactions all have K eq >1. NO HFHNO 2 + F - CH 3 COO - + HFCH 3 COOH + F - HNO 2 + CH 3 COO HNO CH 3 COOH Arrange the substances based on their relative base strength. F - CH 3 COOH HF HNO 2 - CH 3 COO - HNO 2 1 – strongest base 2 – intermediate base 3 – weakest base 4 – not a B-L base

43 © 2008 Brooks/Cole 43 Molecular Structure and Acid Strength What makes a strong acid? A weak H–A bond, so H + can be easily removed! smaller bond energy larger acid strength HX Bond Energy (kJ) K a HF weak acid 5667 x HC l 4311 x 10 7 HBr3661 x 10 8 HI2991 x strong acids binary acids Consider the binary acids HF, HC l, HBr, HI. H-A, bond energy decreases down a group. Acid strength increases.

44 © 2008 Brooks/Cole 44 Molecular Structure and Acid Strength What makes a strong acid? A weak H–A bond, so H + can be easily removed! binary acids Consider the binary acids SiH 4, PH 3, H 2 S, HCl. The higher electronegativity, across a period, makes the bond more polar and hence more ionic – a stronger acid.

45 © 2008 Brooks/Cole 45 Oxoacids  hydrogen, oxygen and one other element H O-E  higher the electronegativity on E, stronger the acid as this weakens the bond between the O and H Molecular Structure and Acid Strength HOCl HOBr HOI Electronegativity: Cl = 3.0, Br = 2.8, I = 2.5 K a : 3.5 × × ×

46 © 2008 Brooks/Cole 46 The acid strength increases as the number of oxygen atoms bonded to E increases. Strong oxoacid has at least two oxygen atoms per acidic H atoms. Molecular Structure and Acid Strength HClO HClO 2 HClO 3 * HClO 4 K a : 3.5 × × ≈ 10 3 ≈ 10 8

47 © 2008 Brooks/Cole 47 Problem Solving Using K a and K b Example: K a from pH Lactic acid is monoprotic. The pH of a M solution was 2.43 at 25°C. Determine K a for this acid. HA(aq) + H 2 O( l ) H 3 O + (aq) + A - (aq) Using the pH information: - log [H 3 O + ] = 2.43[H 3 O + ] = = M These hydronium ions are produced by: H 2 O( l ) + H 2 O( l ) H 3 O + (aq) + OH - (aq) and from the autoionization of water:

48 © 2008 Brooks/Cole 48 K a = = 1.4 x (0.0037)(0.0037) (0.100 – ) HA(aq) + H 2 O( l ) H 3 O + (aq) + A - (aq ) [ ] initial x (water autoionization) change [ ] equil – K a = [H 3 O + ][A - ] [HA] Problem Solving Using K a and K b

49 © 2008 Brooks/Cole 49 Problem Solving Using K a and K b Example: pH from K a Determine the pH of a M propanoic acid solution at 25°C. K a = 1.4 x What % of acid is ionized? C 2 H 5 COOH + H 2 O( l ) H 3 O + (aq) + C 2 H 5 COO - (aq) K a = = 1.4 x [H 3 O + ][C 2 H 5 COO - ] [C 2 H 5 COOH]

50 © 2008 Brooks/Cole 50 HA(aq) + H 2 O( l ) H 3 O + (aq) + A - (aq) [ ] initial * 0 change -x +x +x [ ] equil – x x* x *Ignore the water contribution 1.4 x = x 2 (0.100 – x) [H 3 O + ][A - ] [HA] K a = 1.4 x = Problem Solving Using K a and K b Determine the pH of M propanoic acid soln. at 25°C. K a = 1.4 x What % of acid is ionized?

51 © 2008 Brooks/Cole 51 Problem Solving Using K a and K b Because K a is small, assume x < [HA] K a = 1.4 x = ≈ x 2 (0.100 – x) x x = (1.4 x )(0.100) = M ( << M; the approximation is good) pH = -log( ) = 2.93 %-ionized = x100 = 1.18 % <5% rule can assume [HA] eq ~[HA] 0

52 © 2008 Brooks/Cole 52 What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x Example 3 – calculating pH from known K b Initial Change -x +x Equilibrium 0.20-x xx

53 © 2008 Brooks/Cole 53 Example 3 – calculating pH from known K b What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x

54 © 2008 Brooks/Cole 54 Example 3 – calculating pH from known K b What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x Solving for x we get

55 © 2008 Brooks/Cole 55 Example 3 – calculating pH from known K b What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x Solving for pOH

56 © 2008 Brooks/Cole 56 Practice K a from pH Para-aminobenzoic acid (PABA), HC 7 H 6 NO 2, is used in some sunscreen agents. A solution is made by dissolving mol of PABA in enough water to make mL of solution. The solution’s pH = What is the K a for PABA? (1.9x10 -5 ) pH from K a Barbituric acid (Ka=1.1x10-4) is used in the manufacture of some sedatives. What is the pH for a M solution of barbituric acid? (2.07)

57 © 2008 Brooks/Cole 57 Relationship between K a and K b values For an acid-base conjugate pair: HA and A - [HA][OH - ] [A - ] K a x K b = [H 3 O + ][A - ] [HA] = [H 3 O + ][OH - ]= K w KaKa KbKb When two reactions are added, their equilibrium constants are multiplied (chapter 14).

58 © 2008 Brooks/Cole 58 Relationship between K a and K b values Phenol, C 6 H 5 OH, is a weak acid, K a = 1.3 x at 25°C. Calculate K b for the phenolate ion C 6 H 5 O - K a x K b = 1.0 x K b == 7.7 x x x

59 © 2008 Brooks/Cole 59 Acid Base Neutralization Reactions Salt Salt = ionic compound formed in acid + base reaction Strong acid + strong base Strong acid + strong base form neutral salts. HC l (aq) + NaOH(aq) NaC l (aq) + H 2 O( l ) H + (aq) + C l - (aq) + Na + (aq) + OH - (aq) Na + (aq) + C l - (aq) + H 2 O( l ) HX(aq) + MOH(aq)MX(aq) + H 2 O( l ) acidbasesalt Net ionic equation: H + (aq) + OH - (aq) H 2 O( l ) H 3 O + (aq) + OH - (aq) 2 H 2 O( l )

60 © 2008 Brooks/Cole 60 H 3 O + (aq) + OH - (aq) H 2 O (l) + H 2 O (l) K eq = 1/K w = 1x10 14 strong base strong acid weak base weak acid Na + and C l - are spectator ions. Na + is the v. weak conjugate acid of a v. strong base. Cl - is the v. weak conjugate base of a v. strong acid. Final solution has pH = 7. Salts of Strong Bases and Strong Acids H 3 O + (aq) + C l - (aq) + Na + (aq) + OH - (aq) Na + (aq) + C l - (aq) + 2 H 2 O( l ) 100%

61 © 2008 Brooks/Cole 61 Salts of Strong Bases and Weak Acids Weak acid + strong base Weak acid + strong base form basic salts. CH 3 COOH(aq) + NaOH(aq) CH 3 COONa(aq) + H 2 O (l) Net ionic: CH 3 COOH(aq) + OH - (aq) CH 3 COO - + H 2 O (l) CH 3 COOH(aq) + Na + (aq) + OH - (aq) Na + (aq) + CH 3 COO - (aq) + H 2 O 100% final solution

62 © 2008 Brooks/Cole 62 acetate is a weak base (much stronger than water) solution is basic (pH > 7.0). CH 3 COO - (aq) + H 2 O (l) CH 3 COOH(aq) + OH - (aq) Calculate pH of solution from K b of CH 3 COO -. hydrolysis A hydrolysis reaction – water is broken apart. CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) baseacidbase acid Salts of Strong Bases and Weak Acids 100%

63 © 2008 Brooks/Cole 63 pH of a Salt Solution For a weak acid + strong base, pH depends on K b Larger K b = stronger base Example What is the pH of a 1.50 M aqueous solution of Na 2 CO 3 ? K b (CO 3 2- ) = 2.1 x (or equal volumes of 1.5M H 2 CO 3 and 3.0 M NaOH are mixed, what is the pH of the mixture?)

64 © 2008 Brooks/Cole 64 Na + is the conjugate acid of NaOH (strong base).  It remains 100% ionized  Na + has no effect on pH. CO 3 2- is the conjugate base of HCO 3 - (weak acid)  Most CO 3 2- ions undergo hydrolysis  Generates HCO 3 - and OH -  Changes the pH CO 3 2- (aq) + H 2 O (l) HCO 3 - (aq) + OH - (aq) What is the pH of a 1.50 M aqueous solution of Na 2 CO 3 ? K b = 2.1 x pH of a Salt Solution

65 © 2008 Brooks/Cole 65 CO 3 2- (aq) + H 2 O (l) HCO 3 - (aq) + OH - (aq) [ ] initial change - x +x +x [ ] equil x xx K b = 2.1 x = = ≈ [HCO 3 - ][OH - ] [CO 3 2- ] x 2 (1.50 – x) x x = 1.77 x pOH = −log(1.77 x ) = 1.75 pH = = What is the pH of a 1.50 M aqueous solution of Na 2 CO 3 ? K b = 2.1 x pH of a Salt Solution

66 © 2008 Brooks/Cole 66 Salts of Weak Bases and Strong Acids NH 3 (aq) + HC l (aq) NH 4 + (aq) + C l - (aq) The products of the neutralization reaction form an acidic solution: NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H 3 O + (aq) NH 3 (aq) + H 3 O + (aq) + C l - (aq) NH 4 + (aq) + C l - (aq) + H 2 O (l) Net ionic: NH 3 (aq) + H 3 O + (aq)NH 4 + (aq) + H 2 O (l) The final pH depends on the K a, larger K a = more acidic

67 © 2008 Brooks/Cole 67 pH of a Salt Solution Find the pH of a M NH 4 Br. Is this solution acidic, basic or neutral? K a (NH 4 + ) = 5.6 x The solution contains NH 4 + ions and Br - ions. NH 4 + is the conjugate acid of NH 3 (a weak base).  It will partially ionize in solution giving NH 3 and H 3 O +  Forms an acidic solution. Br - is the conjugate base of HBr (a strong acid)  Br - ions stay fully ionized  Does not change the pH.

68 © 2008 Brooks/Cole 68 NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H 3 O + (aq) [ ] initial change - x +x +x [ ] equil x xx x = 8.57 x pH = −log(8.57 x ) = 5.07 K a = 5.6 x = = ≈ [NH 3 ][H 3 O + ] [NH 4 + ] x 2 (0.132 – x) x pH of a Salt Solution pH of a M aq. solution of NH 4 Br? Is this acidic, basic or neutral? K a (NH 4 + ) = 5.6 x Acidic!

69 © 2008 Brooks/Cole 69 Salts of Weak Bases and Weak Acids The most difficult. Consider qualitative results. e.g. F - (aq) + H 2 O (l) HF(aq) + OH - (aq) NH 3 (aq) + HF (aq) NH 4 + (aq) + F - (aq) NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H 3 O + (aq) K a (NH 4 + ) = 5.6 x ; K b (F - ) = 1.4 x K a (weak acid) > K b (weak base) The ammonium reaction is more favorable. The acid wins! The solution will be acidic!

70 © 2008 Brooks/Cole 70 Acids, Bases and Salts Strong acid strong basepH = 7 Strong acid + strong base→ salt solution, pH = 7 Strong acid weak base pH < 7 Strong acid + weak base → salt solution, pH < 7 Weak acid strong base pH > 7 Weak acid + strong base → salt solution, pH > 7 Weak acid weak base pH = ? Weak acid + weak base → salt solution, pH = ? Need K’s “strongest wins”

71 © 2008 Brooks/Cole 71 Neutral Ions in Water (Table 16.5) AnionsCl - NO 3 - Br - ClO 4 - I - CationsLi + Ca +2 Na + Sr +2 K + Ba +2

72 © 2008 Brooks/Cole 72 What is the pH of a 0.10 M NaCN solution at 25 o C? The K b for CN - is 2.5 x Example Sodium cyanide gives Na + ions and CN - ions in solution. Only the CN - ion hydrolyzes. pH =11.2

73 © 2008 Brooks/Cole 73 What is the pH of a 0.15 M NH 4 NO 3 solution at 25 o C? The K a for NH 4 + is 5.6 x Example NH 4 NO 3 gives NH 4 + ions and NO 3 - ions in solution. Only the NH 4 + ion hydrolyzes.

74 © 2008 Brooks/Cole 74 The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium. The Common Ion Effect Consider a solution of acetic acid (HC 2 H 3 O 2 ), in which you have the following equilibrium.

75 © 2008 Brooks/Cole 75 The Common Ion Effect The equilibrium composition would shift to the left (LeChatelier Principle) and the degree of ionization of the acetic acid would decrease. If we were to add NaC 2 H 3 O 2 to this solution, it would provide C 2 H 3 O 2 - ions which are present on the right side of the equilibrium. This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect.

76 © 2008 Brooks/Cole 76 What is the pH of a solution that is 0.025M in formic acid? The K a for formic acid is 1.7 x What is the pH of the same solution after M in sodium formate, NaCH 2 O is added? Example Initial Change Equilibrium x +x +x x x x From NaCH 2 O 0.018

77 © 2008 Brooks/Cole 77 Initial Change-x +x Equilibrium0.025-x x0.018+x Example x [H 3 O + ]= 2.4 × pH = -log(H 3 O + ) = -log(2.4 x ) = 3.62


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