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Applications of Equilibrium Acids and Bases Chapter 14 Acids and Bases 14.1 the Nature of Acids and Bases Acids

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Applications of Equilibrium Acids and Bases Chapter 14 Acids and Bases 14.1 the Nature of Acids and Bases Acids Applications of Equilibrium Acids and Bases Chapter 14 Acids and Bases 14.1 The Nature of Acids and Bases Acids - Bases Acids and Bases 14.1 The Nature of Acids and Bases Acids - first recognized as class of substances that taste sour. Bases (alkalis are soluble bases) - characterized by bitter taste + slippery feel (drain cleaners are highly basic mixtures) Arrhenius “Acids produce hydrogen ions in aqueous solutions, while bases produce hydroxide ions” Limited application Bronsted-Lowry “An acid is a proton donor and a base is a proton acceptor.” ⇋ + - HA(aq) + H2O(l) H3O (aq) + A (aq) Acid Base Conjugate Acid conjugate base N.B. The conjugate base is all that is left of the acid molecule after a proton is lost. N.B. The conjugate base is all that is left of the acid molecule after a proton is lost. The conjugate acid is formed when the proton is transferred to the base. N.B. The conjugate base is all that is left of the acid molecule after a proton is lost. The conjugate acid is formed when the proton is transferred to the base. A conjugate acid-base pair consists of two substances related to each other by the donating and accepting of a single proton. 2 conjugate acid-base pairs: HA and A- H2O and H3O+ (hydronium ion)* (*hydronium ion is actually more complicated) The equilibrium system is a competition for the proton between the two bases: H2O and A- If H2O is a much stronger base than A- (if H2O has a greater affinity for H+ than A- does) the equilibrium position will be far to the right (favoring products); most of the acid dissolved will be in the ionized form. (For strong acids like nitric acid, hydrochloric acid, sulfuric acid, up to now we have assumed the acids are fully ionized in aqueous solution.) ⇋ + - HA(aq) + H2O(l) H3O (aq) + A (aq) Equilibrium expression for system above: Ka = [H3O+][A-] = [H+] [A-] [HA] [HA] Ka is called the acid dissociation constant Ka has the same form as that for the simple dissociation into ions HA(aq) ⇋ H+ (aq) + A-(aq) But, don’t forget the important role the water plays in causing the acid to ionize in the first place. Write the simple dissociation (ionization) reactions (omitting water) for each of the following acids. a) Hydrochloric acid HCl(aq) b) Acetic acid HC2H3O2(aq) Ethanoic acid + c) The ammonium ion NH4 + d) The anilinium ion C6H5NH3 (aq) 3+ e) The hydrated aluminum(III) ion [Al(H2O)6] Write the simple dissociation (ionization) reactions (omitting water) for each of the following acids. a) Hydrochloric acid HCl(aq) HCl(aq) ⇋ H+ (aq) + Cl-(aq) Write the simple dissociation (ionization) reactions (omitting water) for each of the following acids. b) Acetic acid HC2H3O2(aq) Ethanoic acid ⇋ + - HC2H3O2(aq) H (aq) + C2H3O2 (aq) Write the simple dissociation (ionization) reactions (omitting water) for each of the following acids. + c) The ammonium ion NH4 + ⇋ + NH4 (aq) H (aq) + NH3(aq) Write the simple dissociation (ionization) reactions (omitting water) for each of the following acids. + d) The anilinium ion C6H5NH3 (aq) + ⇋ + C6H5NH3 (aq) H (aq) + C6H5NH2(aq) Write the simple dissociation (ionization) reactions (omitting water) for each of the following acids. 3+ e) The hydrated aluminum(III) ion [Al(H2O)6] 3+ ⇋ + 2+ [Al(H2O)6] H (aq) + [Al(H2O)5OH] (aq) BrØnsted-Lowry model extends to reactions in gas phase. ⇋ HCl(g) + NH3(g) NH4Cl (s) 14.2 Acid Strength Defined by equilibrium position of its dissociation (ionization) reaction: ⇋ + - HA(aq) + H2O(l) H3O (aq) + A (aq) Strong Acids: Strong Acids Ka large Equilibrium lies far to the right[H+] approx. equal to [HA]0 Almost all the original HA is dissociated (ionized) at equilibrium A strong acid yields a weak conjugate base (which means the water molecules win the competition for the H+ ions) Weak Acids Ka small Equilibrium lies far to the left [H+]<< [HA]0 Almost all of the original HA is still present at equilibrium A weak acid yields a conjugate base, A-, that is a stronger base than water. Weak Acids Strong Acids Phosphoric acid H3PO4 Nitrous acid HNO2 Hypochlorous HOCl Organic acids - have a carboxyl group Acetic (ethanoic acid) Benzoic acid Weak Acids Strong Acids Phosphoric acid H3PO4 Hydrochloric acid Nitrous acid HNO2 HCl(aq) Hypochlorous HOCl Nitric acid HNO3 (aq) Organic acids - have a Perchloric acid carboxyl group HClO4(aq) Acetic (ethanoic acid) Benzoic acid Sulfuric Acid - diprotic acid (2 acidic protons) + - H2SO4(aq) → H (aq) + HSO4 (aq) Strong acid, virtually 100% ionized in water Sulfuric Acid - diprotic acid (2 acidic protons) + - H2SO4(aq) → H (aq) + HSO4 (aq) Strong acid, virtually 100% ionized in water - However, the HSO4 ion is only a weak acid ⇋ 2- HSO4-(aq) H+(aq) + SO4 (aq) Notice most acids are oxyacids - Sulfuric Acid - diprotic acid (2 acidic protons) + - H2SO4(aq) → H (aq) + HSO4 (aq) Strong acid, virtually 100% ionized in water - However, the HSO4 ion is only a weak acid ⇋ 2- HSO4-(aq) H+(aq) + SO4 (aq) Notice most acids are oxyacids - acidic proton is attached to an oxygen atom. Page 628 Table 14.2 Values of Ka for some common monoprotic acids Note that Ka values for strong acids are not listed - why? Problems finding Ka for strong acids Ka = [H+] [A-] [HA] very small and hence very uncertain So far to right, prevents accurate calculation of Ka. Relative Base Strength Using table 14.2 arrange the following species according to their strengths as BASES. (proton acceptors) H2O, F-, Cl-, NO2- and CN- Relative Base Strength H2O, F-, Cl-, NO2- and CN- Water is a stronger base than the conjugate base of a strong acid, but a weaker base than the conjugate base of a weak acid. Cl- < H2O < conjugate bases of weak acids Weakest base strongest bases From table: Ka HF > Ka for HNO2 > Ka for HCN Relative Base Strength H2O, F-, Cl-, NO2- and CN- Ka HF > Ka for HNO2 > Ka for HCN So combined order of increasing base strength Cl- < H2O < F- < NO2- < CN- Note HF is a weak acid - why might that be? Water - as an acid and a base Water - as an acid and a base Water is Amphoteric - can behave as acid and base Autoionization of water Kw ion product or dissociation constant for water Kw = [H3O+] [OH-] = [H+][OH-] For water at 25oC experiments show that [H+] = [OH-] = 1.0 x 10-7 M At 25oC Kw = [H+][OH-] = (1.0 x 10-7)(1.0 x 10-7) = 1 x 10-14 Important to remember Kw is temperature dependent. When we find pH - will also be temperature dependent. Kw cont. In any aqueous solution at 25oC, the product of [H+] and [OH-] must always equal 1.0 x 10-14. Neutral solution, [H+] = [OH-] Acidic solution [H+] > [OH-] Basic solution [H+] < [OH-] But in each case, at 25oC, Kw = [H+][OH-] = 1 x 10-14 Calculating [H+] and [OH-] Example 14.3 Calculate [H+] or [OH-] as required for each of the following solutions at 25oC, and state whether the solution is neutral, acidic, or basic. a) 1.0 x 10-5 M OH- -14 Kw = [H+][OH-] = 1 x 10 [H+] = Calculating [H+] and [OH-] Example 14.3 Calculate [H+] or [OH-] as required for each of the following solutions at 25oC, and state whether the solution is neutral, acidic, or basic. a) 1.0 x 10-5 M OH- -14 Kw = [H+][OH-] = 1 x 10 [H+] = 1 x 10 -14 / 1.0 x 10-5 = 1.0 x 10-9M [OH-] > [H+] basic b) 1.0 x 10-7 M OH- -14 Kw = [H+][OH-] = 1 x 10 [H+] = 1 x 10 -14 / 1.0 x 10-7 = 1.0 x 10-7M [OH-]= [H+] neutral c) 10.0 M H+ [OH-] = 1 x 10 -14 / 10 = 1.0 x 10-15 M [OH-] < [H+] acidic Effect of Temperature on Kw Example 14.4 o -13 At 60 C the value of Kw is 1 x 10 . a) Using Le Chatelier’s principle, predict whether the reaction ⇋ + - 2H2O (l) H3O (aq) + OH (aq) Is exothermic or endothermic. Effect of Temperature on Kw ⇋ + - 2H2O (l) H3O (aq) + OH (aq) Kw increases from 1 x 10-14 to 1 x 10-13 as temperature increases from 25 to 60oC. Effect of Temperature on Kw ⇋ + - 2H2O (l) H3O (aq) + OH (aq) Le Chatelier’s principle states that a system at equilibrium will respond to stress on the equilibrium in such a way as to oppose the stress imposed. \ Effect of Temperature on Kw ⇋ + - 2H2O (l) H3O (aq) + OH (aq) Here, we heated the system so the system would have been trying to overcome the increase in temperature by favoring the endothermic direction for the equilibrium system (since this would consume energy). The reaction as written must be endothermic from left to right. Effect of Temperature on Kw b) Calculate [H+] and [OH-] in a neutral solution at 60oC. For a neutral solution [H+] =[OH-] -13 and since Kw = 1 x 10 = [H+] [OH-] [H+] = √1 x 10 -13 = 3 x 10-7 M 14.3 The pH Scale pH = -log[H+] For a solution where [H+] = 1.0 x 10-7 M pH = - (-7.00) = 7.00 14.3 The pH Scale pH = -log[H+] For a solution where [H+] = 1.0 x 10-7 M pH = - (-7.00) = 7.00 Significant figures for logarithms: The number of decimal places in the log is equal to the number of significant figures in the original number. 14.3 The pH Scale pH changes by 1 for every power of 10 change in [H+] hence pH decreases as [H+] increases Kw = [H+][OH-] log Kw = log [H+] + log [OH-] -log Kw = -log[H+] - log [OH-] Thus pKw = pH + pOH -14 and since Kw = 1.0 x 10 -14 pKw = -log (1.0 x 10 ) = 14.00 -14 pKw = -log (1.0 x 10 ) = 14.00 Thus for any aqueous solution at 25oC, pH and pOH add up to 14.00 pH + pOH = 14.00 Write out Sample Exercise 14.5 Calculating pH and pOH Write out Sample Exercise 14.6 Calculating pH 14.4 Calculating the pH of Strong Acid Solutions Page 634 - 635 - read and make own notes.
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