Acid and Base Equilibrium Chapter 16 Brown LeMay

Basic Concepts Acids – 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.

Dissociation of Water Pure water exists almost entirely of water molecules. It is essentially a non-electrolyte. Water ionizes to a small extent – auto-ionization The equilibrium expression is H20(l) <-> H+(aq) + OH-(aq) Kw = [H+] [OH-]

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 expression The proton in water

Values for Kw and [H+] and [OH-] Kw = [H3O+] [OH-] = 1.0 x 10-14 1.0 x 10-14 = [X] [X] 1.0 x 10-14 = X2 X = 1.0 X 10-7 = [H3O+] = [OH-]

The Bronsted – Lowry definition holds true for situations not involving water Acids donate protons Bases accept protons

HCl(g) + NH3(g)  NH4+(g) + Cl-(aq)\ donates H+ accepts H+ conj acid conj base BL-Acid BL-Base notice that the reaction doesn’t happen in water and that the OH- concentration has not increased

Conjugate Acids & Bases and Amphoteric Substances HNO2(aq) + H2O(l)  NO2-(aq)+H3O+(aq) conjugate acid base base acid donates - accepts protons note HNO2 is considered a Arrenius acid H+ inc in H20 also a BL because donates a Proton all 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

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

Amphotheric Substances NH3(aq)+H20(l)  NH4+(aq) + OH-(aq) Base Acid Conj Acid Conj Base p-acc p-donar note water is acting as an acid in this reaction and a base in the previous one that makes it a amphortic substance

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.

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 insignificant The  equilibrium lies so far to the right because HNO3 doesn’t reform The 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.

Acids and Bases Strong Acids HClO4 prechloric H2SO4 sulfuric HI hydroiodic HCl hydochloric HBr hydrobromic HNO3 nitric Strong Bases GI hydroxides ex. NaOH,KOH G2 Hydroxides Sr(OH)2 GI Oxides ex. Na2O, K2O GI,II Amides ex. KNH2,Ca(NH2)2

The pH scale pH – is defined as the neg log (base-10) of the H+ ion concentration pH = -log [H+] What is the pH of a neutral solution [H+] = 1.0 x 10^-7 pH = -log [1.0 x 10^-7 pH = 7

Strong acids and the pH scale An acidic solution must have a [H+] conc greater than 1.0 X 10-7 ex 1.0 X 10-6 -log [1.0 X 10-6] = pH 6 What is the pH of a basic solution? A basic solution is one in which the [OH-] is greater than 1X10-7.

Calc. pH of strong basic solutions Calculate the pH of a sol that has a [OH-] con. Of 1.0 X10-5 Kw = 1x10-14 = [H+] [OH-] Kw = 1x10-14 = [H+] [1.0X10-5] [H+] = 1x10-14 = 1.0 X 10-9 1.0X10-5 pH = -log 1.0 x 10-9 pH = 9

The “p” Scale The negative log of a quantity is labeled p (quantity) Ex: we could reference the quantity of OH- directly: pOH = -log[OH-] From the definition of Kw -log Kw =(-log [H+]) (-log [OH-])= -log 1x10-4 Kw = pH + pOH = 14

Calc. pH using the p scale Ex. OH- conc = 1.0X10-5 -log 1.0X10-5 = 5 = pOH pH + pOH = 14 pH + 5 = 14 pH = 9

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 constant Ka = [H30] [A-] = [H+] [A-] [HA] [HA]

Acid Dissociation Constant The larger the Ka value the stronger the acid is – more product is in solution

Weak Bases Weak bases in water react to release a hydroxide (OH-) ion and their conjugate acid: Weak Base(aq) + H2O(l) Conjugate Acid(aq) + OH-(aq)

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 expression Kb = [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-

Lewis Acids and Bases Review - An Arrhenius acid reacts in water to release a proton - base reacts in water to release a hydroxide ion In the Bronstead-Lowry description of acids and bases: acid reacts to donate a proton - a base accepts a proton

G.N.Lewis defination - Lewis acid is defined as an electron-pair acceptor Lewis base is defined as an electron-pair donor

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)

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.

Calculating the pH of a Weak Acid What is the pH of an aq sol that is 0.0030M pyruvic acid HC2H3P3? Ka = 1.4x10-4 at 25oC HC2H3P3  H+ + C2H3P3- I 0.0030 0 0 C -X X X E 0.0030-X +X +X

Ka = [H+] [C2H3P3-] pluggin in the values [HC2H3P3] from the table 1.4x10-4 = X2 (0.0030-X) 1.4x10-4 (0.0030) = X2 4.2x10-7-1.4x10-4x = X2 X2 + 1.4x10-4x-4.2x10-7 = 0 a quadratic ignore the neg sol x = 5.82 x10-4 pH = -log 5.82 x10-4 pH = 3.24

Learning Check What is the pH at 25oC of a solution made by dissolving a 5.00 grain tablet of aspirin (acetylsalicylic acid) in 0.500 liters of water? The tablet contains 0.325g of the acid HC9H7O4. Ka = 3.3x10-4 mm = 180.2g/l H+ = 9.4x10-4 pH = 3.03

Buffers A 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.

Examples Ex. Weak acid and conj. base equal molar amounts Strong Acid added H+ + A-  HA conj. base weak acid the conj base interacts with the H+ ions from the strong acid changing them to a weak acid

Strong base added OH + HA  HOH + A- weak acid conj base the weak acid interacts with the OH- ion from the base to form water and the conj. base If the concentration of A- and HA are large and the amount of H+ or OH- is small the solution will be buffered or the change in pH will be minimized.

Buffering capacity – the amount of acid or base a buffer can react with before a significant change in pH occurs Ratio 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.

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 = 5.07 Ex. Calc. the pH of a buffer by mixing 60.0 ml of 0.100 M NH3 with 40.0ml of 0.100 M NH4Cl.

1St cal the conc. of each species M = moles/liters mol of NH3 0.10M = X/0.060 l = 0.0060mol mol of NH4 0.10M = X/0.040 l = 0.0040mol [NH3] = 0.0060 mol/ 0.100 l = 0.060 M [NH4] = 0.0040 mol/ 0.100 l = 0.040 M NH3 + H2O  NH4+ + OH- I 0.060M O.040 0 C -X +X +X E 0.060-x 0.040+x X

Kb = [NH4+] [OH-] ( 0.040+X)X 1.8X10-5 [NH3] (0.060-X) Ignore X 1.8X10-5 = 0.040X X = 2.7X10-5 0.060 -log (2.7x10-5) = 4.6 pOH pH = 9.4 Or using Henderson - Hasselbalch equation pOH = pKb + log [conj acid] = 4.74 + log ( [0.04] [B] [0.06]) = 4.6 pOH pH= 9.4

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 = .140 ka = 1.7x10-5 HHeq pH = pKa + log [conj base] [acid] pH = 4.77 + log(.140 = 5.3 .045)

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 -  HC2H3O2 H+ = 0.10M = n/.0095l = 0.00095 moles C2H3O2 - = 0.2M = n/.075 = 0.0150 moles HC2H3O2 = 0.10M = n/.075 = 0.0075 moles

Neutralization Reaction C2H3O2 = C2H3O2 moles – H+ moles 0.0150n - .00095n = 0.014 HC2H3O2 = Orginial Conc. + Conc Contributed by reaction 0.075 moles + 0.00095 = 0.0085mol [C2H3O2] = 0.014mol/0.085l = 0.16M [HC2H3O2] = 0.0085/0.085 = 0.10M pH = 4.76 + log (.16/.10) = 4.96