Presentation on theme: "Acid and Base Equilibrium"— Presentation transcript:
1 Acid and Base Equilibrium Chapter 16 Brown LeMay
2 Basic ConceptsAcids – sour or tart taste, electrolytes, described by Arrhenius as sub that inc. the H+ conc.Bases – bitter to the taste, slippery, electrolytes, describes by Arrhenius as sub that inc. the OH- conc.
3 Dissociation of WaterPure water exists almost entirely of water molecules. It is essentially a non-electrolyte.Water ionizes to a small extent – auto-ionizationThe equilibrium expression isH20(l) <-> H+(aq) + OH-(aq)Kw = [H+] [OH-]
4 Since the H+ ion does not exist alone in water Kw is often expressed Kw = [H3O+] [OH-]because water conc is constant it does not appear in the expressionThe proton in water
5 Values for Kw and [H+] and [OH-] Kw = [H3O+] [OH-] = 1.0 x 10-141.0 x = [X] [X]1.0 x = X2X = 1.0 X 10-7 = [H3O+] = [OH-]
6 The Bronsted – Lowry definition holds true for situations not involving waterAcids donate protonsBases accept protons
7 HCl(g) + NH3(g) NH4+(g) + Cl-(aq)\ donates H+ accepts H+ conj acid conj baseBL-Acid BL-Basenotice that the reaction doesn’t happen in water and that the OH- concentration has not increased
8 Conjugate Acids & Bases and Amphoteric Substances HNO2(aq) + H2O(l) NO2-(aq)+H3O+(aq)conjugateacid base base aciddonates - accepts protonsnote HNO2 is considered a Arrenius acid H+ inc in H20 also a BL because donates a Protonall Arrenius acids and bases are also BL acids and bases – however all BL acids and bases may or may not be arrenius acids or bases water is not an Arrenius base in this example
9 every acid has a conjugate base formed from the removal of a proton from that acid every base has a conjugate acid formed from the addition of a proton to that base
10 Amphotheric Substances NH3(aq)+H20(l) NH4+(aq) + OH-(aq)Base Acid Conj Acid Conj Basep-acc p-donarnote water is acting as an acid in this reaction and a base in the previous one that makes it a amphortic substance
11 Strengths of Acid, Bases The stronger the acid is the weaker its conjugate base (weaker acid stronger conj base)The stronger the base the weaker its conjugate acid (weaker base stronger conj acid)Stronger acids and bases ionize to a greater extent than do weak acids and bases.
12 Strong Acids – dissociate completely into ions HNO3(aq) H+(aq) + NO3-(aq)the production of H+ ions from the acid dominates – ignore the H+ donated from the water it is insignificantThe equilibrium lies so far to the right because HNO3 doesn’t reformThe neg log of the H+ from the acid determines pH.Strong bases also dissociate completely and the conc of OH- from the base is the only factor considered when calculating pH.
14 The pH scalepH – is defined as the neg log (base-10) of the H+ ion concentrationpH = -log [H+]What is the pH of a neutral solution[H+] = 1.0 x 10^-7pH = -log [1.0 x 10^-7pH = 7
15 Strong acids and the pH scale An acidic solution must have a [H+] conc greater than 1.0 X ex 1.0 X 10-6-log [1.0 X 10-6] = pH 6What is the pH of a basic solution?A basic solution is one in which the [OH-]is greater than 1X10-7.
16 Calc. pH of strong basic solutions Calculate the pH of a sol that has a [OH-] con. Of 1.0 X10-5Kw = 1x10-14 = [H+] [OH-]Kw = 1x10-14 = [H+] [1.0X10-5][H+] = 1x = 1.0 X 10-91.0X10-5pH = -log 1.0 x pH = 9
17 The “p” Scale The negative log of a quantity is labeled p (quantity) Ex: we could reference the quantity ofOH- directly: pOH = -log[OH-]From the definition of Kw-log Kw =(-log [H+]) (-log [OH-])= -log 1x10-4Kw = pH + pOH = 14
18 Calc. pH using the p scale Ex. OH- conc = 1.0X10-5-log 1.0X10-5 = 5 = pOHpH + pOH = 14pH + 5 = 14pH = 9
19 Weak Acids partially ionize in aqueous solution mixture of ions and un-ionized acid in sol.WA are in equilibrium (H20 is left out because it a pure liquid)HA(aq) + H20(l) H30+(aq) + A-(aq)Ka is the acid dissociation constantKa = [H30] [A-] = [H+] [A-][HA] [HA]
20 Acid Dissociation Constant The larger the Ka value the stronger the acid is – more product is in solution
21 Weak BasesWeak bases in water react to release a hydroxide (OH-) ion and their conjugate acid:Weak Base(aq) + H2O(l) Conjugate Acid(aq) + OH-(aq)
22 A common weak base is ammonia NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)Since H2O is a pure liquid it is not expressed in the equilibrium Kb expressionKb = [NH4+][OH-] (base dissociation[NH3] constant)Kb always refers to the equilibrium in which a base reacts with H2O to form the conjugate acid and OH-
23 Lewis Acids and BasesReview - An Arrhenius acid reacts in water to release a proton - base reacts in water to release a hydroxide ionIn the Bronstead-Lowry description of acids and bases: acid reacts to donate a proton - a base accepts a proton
24 G.N.Lewis defination -Lewis acid is defined as an electron-pair acceptorLewis base is defined as an electron-pair donor
25 In the example with ammonia, the ammonia is acting as a Lewis base (donates a pair of electrons), and the proton is a Lewis acid (accepts a pair of electrons)
26 Lewis is consistent with the description by Arrhenius, and with the definition by Bronstead-Lowry. However, the Lewis description, a base is not restricted in donating its electrons to a proton, it can donate them to any molecule that can accept them.
27 Calculating the pH of a Weak Acid What is the pH of an aq sol that is M pyruvic acid HC2H3P3? Ka = 1.4x10-4 at 25oCHC2H3P3 H+ + C2H3P3-IC -X X XE X +X X
28 Ka = [H+] [C2H3P3-] pluggin in the values [HC2H3P3] from the table1.4x10-4 = X2( X)1.4x10-4 (0.0030) = X24.2x x10-4x = X2X x10-4x-4.2x10-7 = 0 a quadraticignore the neg sol x = 5.82 x10-4pH = -log 5.82 x10-4 pH = 3.24
29 Learning CheckWhat is the pH at 25oC of a solution made by dissolving a 5.00 grain tablet of aspirin (acetylsalicylic acid) in liters of water? The tablet contains 0.325g of the acid HC9H7O4. Ka = 3.3x10-4 mm = 180.2g/lH+ = 9.4x10-4 pH = 3.03
30 BuffersA solution that resists changes in pH when a limited amount of an acid or base is added to it.Buffers contain either a weak acid and it’s conj. base or a weak base and it’s conj. acid.
31 Examples Ex. Weak acid and conj. base equal molar amounts Strong Acid addedH+ + A- HAconj. base weak acidthe conj base interacts with the H+ ionsfrom the strong acid changing them to aweak acid
32 Strong base addedOH + HA HOH + A-weak acid conj basethe weak acid interacts with the OH- ion from the base to form water and the conj. baseIf the concentration of A- and HA are large and the amount of H+ or OH- is small thesolution will be buffered or the change in pH will be minimized.
33 Buffering capacity – the amount of acid or base a buffer can react with before a significant change in pH occursRatio of acid to conj base – unless the ratio is close to 1 ( between 1:10 and 10:1) will be too low to be useful.
34 Calculating the pH of a buffer Note: a solution of 0.10 M acedic acid and its conj base 0.20 M acetate from sodium acetate is a buffer solution pH = 5.07Ex. Calc. the pH of a buffer by mixing 60.0 ml of M NH3 with 40.0ml of M NH4Cl.
35 1St cal the conc. of each species M = moles/litersmol of NH3 0.10M = X/0.060 l = molmol of NH4 0.10M = X/0.040 l = mol[NH3] = mol/ l = M[NH4] = mol/ l = MNH3 + H2O NH4+ + OH-I M OC -X X XE x x X
37 What is the pH of a buffer prepared by adding 30. 0ml of What is the pH of a buffer prepared by adding 30.0ml of .15M HC2H3O2 to 70ml of .2M NaC2H3O2?[HC2H3O2] = .15M = x/.03 = .0045/.01 = .045[C2H3O2] = .20M =x/.07 = .0140/.01 =ka = 1.7x10-5HHeq pH = pKa + log [conj base][acid]pH = log(.140 = 5.3.045)
38 Adding an acid or a base to a buffer Calc the ph of 75ml of the buffer solution of(0.1M HC2H3O2 and 0.2M NaC2H3O2)to which 9.5 ml of 0.10M HCl has been added. Compare the change to that of adding HCl to pure water.H+ + C2H3O2 - HC2H3O2H+ = 0.10M = n/.0095l = molesC2H3O2 - = 0.2M = n/.075 = molesHC2H3O2 = 0.10M = n/.075 = moles