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Chapter 16 - Acid-Base Equilibria Homework: 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 27, 29, 30, 31, 34, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 52, 53, 55, 56, 61, 62, 65, 71, 72, 73, 74, 75, 81, 82, 83, 85, 86, 87, 88, 92, 16.2 - Bronsted-Lowry Acids and Bases In chapter 4, introduced the Arrhenius concept of acids and bases An Arrhenius acid was something that produced a H+ ion in an aqueous solution An Arrhenius base was something that produced a OH- ion in an aqueous solution Unfortunately, Arrhenius concept limits acids and bases to aqueous solutions. Bronsted and Lowry expanded and made a more general definition of acids and bases. The H+ Ion in Water. HCl ( g ) H (aq) Cl (aq) H 2O Given this equation, we see that hydrogen chloride ionizes in water to form H+(aq). An H+ ion is a proton with no valence electrons This small, positively charged particle interacts strongly with the nonbonding electrons pairs of the water molecules This forms hydrated hydrogen ions The hydronium ion When H+ interacts with water, it forms the hydronium ion H3O+ The hydronium ion will then form bonds with other water molecules, making large clusters of hydrated hydrogen ions H5O2+ or H9O4+ Chemists will use H+(aq) and H3O+(aq) interchangeably. It is the hydrated proton that is responsible for the properties of aqueous solutions of acids Proton-Transfer Reactions When we closely look at the reaction when HCl dissolves in water, we fid the HCl molecule actual transfers an H+ ion (proton) to the water molecule So we can represent the reaction between an HCl molecule and a water molecule to form hydronium and chloride ions HCl(g) + H2O(l) H3O+(aq) + Cl-(aq) Defining Acids and Bases Bronsted and Lowry defined acids and bases in terms of their ability to transfer protons An acid is a substance (molecule or ion) that can donate a proton to another substance A base in a substance that can accept a proton So when HCl dissolves in water, HCl acts as a Bronsted-Lowry acid It donates a proton to H2O And H2O would act as a Bronsted-Lowry base because it accepts a proton Because the emphasis is on proton transfer the concept also applies to reactions that do not involve solutions HCl + NH3 Cl- + NH4+ HCl is the Bronsted-Lowry acid NH3 is the Bronsted-Lowry base Another Look NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Ammonia is an Arrhenius base because adding it to water produces OHIt is a Bronsted-Lowry base because it accepts a proton from H2O The H2O is the Bronsted-Lowry acid because it donates a proton to the NH3 molecule Acids and bases always work together to transfer a proton A substance can function as an acid ONLY if there is another substance to simultaneously act like a base To be a Bronsted-Lowry acid, a molecule or ion MUST have a hydrogen atom it can lose as an H+ ion To be a Bronsted-Lowry base, a molecule or ion MUST have a nonbinding pair of electrons to use to bind the H+ ion Both? Some substances can act like an acid in one reaction and a base in another H2O is a Bronsted-Lowry base when reacting with HCl H2O is a Bronsted-Lowry acid when reacting with NH3 HCl(g) + H2O(l) H3O+(aq) + Cl-(aq) NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) A substance that is capable of being both an acid and a base is called amphiprotic Acts a base when combined with something more acidic than itself Acts as an acid when combined with something more basic than itself Conjugate Acid-Base Pairs In any acid-base equilibrium both the forward and reverse reaction involve proton transfers Example Consider the reaction of an acid (HX) with water HX(aq) + H2O(l) X-(aq) + H3O+(aq) In the forward reaction HX donates a proton to H2O So HX is the acid, and H2O is the base In the reverse reaction, H3O+ donates a proton to the X- ion An acid and a base such as HX and Xthat differ only because of the presence of a proton are called a conjugate acidbase pair Every acid has a conjugate base Formed from the removal of a proton from the acid OH- is the conjugate base of H2O X- is the conjugate base of HX Likewise Every base has a conjugate acid Formed by adding a proton to the base H3O+ is the conjugate acid of H2O HX is the conjugate acid of X- Examples with Equations In any acid-base reaction, we can identify two sets of conjugate acid-base pairs. Nitrous acid and water remove H+ HNO2(aq) + H2O(l) Acid Base NO2-(aq) + H3O+(aq) Conjugate base add H+ Conjugate acid NH3and H2O reaction add H+ NH3(aq) + H2O(l) Base Acid NH4+(aq) + OH-(aq) Conjugate acid remove H+ Conjugate base Example What is the conjugate base of each of the following acids HClO4 H2S HS- PH4+ ClO4- PH3 HCO3 CO3-2 Example 2 What is the conjugate acid for each of the following bases? CN SO42 HSO4- H2O HCN H3O+ HCO3 H2CO3 Example 3 The hydrogen sulfite ion (HSO3-) is amphiprotic Write an equation for the reaction of HSO3with water, in which the ion acts as an acid Show the conjugate acid-base pairs Then write an equation for the reaction of HSO3- with water, in which the ion acts as an base Also identify the conjugate acid-base pairs HSO3- as acid then base HSO3-(aq) + H2O(l) H3O+(aq) SO32- (aq) + Conjugate pairs HSO3-(aq) and SO32- (aq) are acid and conjugate base, respectively + H2O(l) and H3O (aq) are base and conjugate acid, respectively HSO3-(aq) + H2O(l) H2SO3(aq) + OH (aq) Conjugate pairs HSO3-(aq) and H2SO3(aq) are base and conjugate acid, respectively H O(l) and OH-(aq) are acid and conjugate base, Relative Strengths of Acids and Bases Some acids are better proton donors than others Likewise, some bases are better proton acceptors than others If we arrange acids in order of their ability to donate a proton... We find the easier they give up a proton, the less easily its conjugate base accepts a proton Similarly, the more easily a base accepts a proton, the less easily the conjugate acid gives up a proton Generally The stronger an acid (more easily it gives up a proton), the weaker it s conjugate base. If we know something about the strength of an acid, we also know something about the strength of its conjugate base. Example table shown on pg. 674 We can break up acids and bases into 3 broad categories, based on their behavior in water Table Three Broad Categories 1. Strong Completely transfer their protons to water No undissociated molecules in solution Conjugate bases have a negligible tendency to attract protons Three Broad Categories 2. Weak acids/Weak Base Only partially dissociate in water Exist in solution as a mixture of acid molecules and their constituent ions Conjugate bases show slight ability to attract protons Three Broad Categories 3. Negligible acids/Strong Base Do not act like acids in water Conjugate bases are very strong Equilibrium? In every acid-base reaction, the position of the equilibrium favors the transfer of the proton to the stronger base. Which means the equilibrium mixture has more of the weaker acid and weaker base and less of the stronger acid and stronger base Because strong acid has a weak conjugate base (and vice versa) Which really means the reaction will favor the side with the stronger base Example For the following proton-transfer reaction (i.e. acid-base reaction), predict whether the equilibrium lies predominantly to the left (Kc < 1) or to the right (Kc > 1) HSO4-(aq) + CO32-(aq) SO42-(aq) + HCO3(aq) Use table on pg. 674 to determine the stronger acid/base HSO4-(aq) + CO32-(aq) (aq) SO42-(aq) + HCO3- Equilibrium will favor the proton going to the stronger of the two bases The two bases in this equation are CO32- (in the forward reaction) and SO42- (in the reverse reaction) Looking at the chart on pg. 674, I see CO32- is lower on the chart for bases than SO42This means CO32- is a stronger base Reaction favors side with stronger base, so this reaction favors the products Kc < 1 16.3 - The Autoionization of Water One of the most important properties of water is its ability to act as wither an acid or base. In the presence of an acid, water acts like a base (proton acceptor) In the presence of a base, water acts like an acid (proton donor) In fact, one water molecule can donate a proton to another water molecule This is called the autoionization of water H2O + H2O H3O+ + OHWhen this happens, no individual no individual molecule remains ionized molecule remains ionized for long ionized for long Very rapid reaction At room temperature, about 2 out of every about 2 out of every 109 molecule are The Ion Product of Water Because the ionization of water is an equilibrium process, we can write the following equilibrium-constant expression for it H2O(l) + H2O(l) H3O+(aq) + OH-(aq) Kc = [H3O+][OH-] Where did the H2O go? Not in expression, because is a pure liquid Kc = [H3O+][OH-] Because this equilibrium-costat expressio refers specifically to the autoioizatio of water, we use the symbol Kw to represet this Kw is the io-product costat for water At 25ºC, Kw = 1.0x10-14 Playing with the equation Since H+(aq) and H3O+(aq) both mean the hydrated proton, we can also say H2O(l) H+(aq) + OH-(aq) Likewise, Kw can also be expressed as Kw = [H+][OH-] = 1.0x10-14 This equilibrium-constant expression and Kw are VERY IMPORTANT Remember these! Why is this important? It is not only applicable to pure water, but to any aqueous solution Although equilibrium between H+ and OH(along with other ionic equilibria) are affected by the presence of other ions in solution, we tend to ignore those ions except in work requiring a LOT of accuracy So the equilibrium-constant expression we just made can be used to find either [H+] or [OH-] for a dilute aqueous solution A solution where [H+] = [OH-] is called neutral Most solutions are not neutral. In most solutions, as one of the above increases, the other decreases This keeps the product of their concentrations equal to 1.0x10-14 When [H+] > [OH-], the solution is acidic When [H+] < [OH-], the solution is basic Example Determine the values of [H+] and [OH-] in a neutral solution at 25ºC Remember, that in a neutral solution [H+] = [OH-] Kw= [H+][OH-]=(x)(x) = 1.0x10-14 x2 = 1.0x10-14 x = 1.0x10-7 M = [H+] = [OH-] In an acidic solution [H+] Example 2 Calculate the concentration of H+(aq) in a solution in which [OH-] is 0.010 M a solution in which [OH-] is 1.8x10-9 M Unless told otherwise, assume temperature is 25ºC Why is this important? A solution in which [OH-] is 0.010 M [H+][OH-] = 1.0x10-14 [H+] = 1.0x10-14/ [OH-] [H+] = 1.0x10-14/ 0.010 [H+] = 1.0x10-12 Is this solution acidic or basic? Basic [OH-] > [H+] A solution in which [OH-] is 1.8x10-9 M [H+][OH-] = 1.0x10-14 [H+] = 1.0x10-14/ [OH-] [H+] = 1.0x10-14/ 1.8x10-9 [H+] = 5.6x10-6 Is this solution acidic or basic? Acidic [OH-] < [H+] The pH scale The concentration of H+(aq) is usually very small For convenience, we usually express this concentration in terms of pH pH is the negative logarithm (in base 10) pH = -log[H+] Solution pH Neutral solution [H+] = [OH-] = 1.0x10-7 pH = -log(1.0x10-7) pH = -(-7.00) = 7.00 So the pH of a neutral solution is 7.00 at 25ºC Acidic solution Acidic solution has [H+] > 1.0x10-7 Let’s have a pH of 1.0x10-3 pH = -log(1.0x10-3) pH = -(-3.00) = 3.00 So the pH for an acidic solution is less than 7.00 The more acidic the solution, the lower the pH Basic solution [H+] < [OH-] Suppose [OH-] = 2x10-3 M First we find [H+] (because we need it for pH) [H+] = Kw / [OH-] = 1.0x10-14 / 2.0x10-3 = 5.0x10-12M pH = -log(5.0x10-12) pH = 11.30 So the pH of a basic solution is greater than 7.00 The more basic the solution, the higher the pH Notes About pH A change in [H+] by a factor of 10 causes the pH to change by 1 So a solution with a pH of 6 has 10x the concentration of H+ than a solution of pH 7 Also, a solution with a pH of 1 has 1000x the concentration of [H+] than a solution with a pH of 4. pH Table Other Scales The negative log is also useful for expressing the amounts of other small numbers p So really, A common p p just means –log number is pOH pOH = -log[OH-] Why is this useful? -log[H+] + (-log[OH-]) = -logKw or pH + pOH = 14.00 Measuring pH The pH of a solution can be measured quickly and accurately with a pH meter We can also use chemicals called indicators An indicator is a colored substance that itself can exist in either an acid or a base form The two forms have different colors If you know the pH at which the indicator turns from one color to the other, you can determine whether a solution has a higher or lower pH value than this Indicators 16.5 - Strong Acids and Bases The chemistry of aqueous solutions often critically depends on the pH of the solution Therefore, it is important to look at how the pH of the solution relates to strong acids and bases Strong acids as bases are called strong electrolytes This means they exist in aqueous solutions entirely as ions Strong Acids The seven most common strong acids include six monoprotic acids and one diprotic acid HCl, HBr, Hi, HNO3, HClO3 and HClO4 H2SO4 For these strong acids, when in solution with these, the solutions exists entirely of H3O+ and its conjugate base Example Nitric acid example HNO3(aq) + H2O(l) H3O+(aq) + NO3(aq) No equilibrium arrows Reaction goes to completion to products This reaction is also seen as HNO3(aq) H+(aq) + NO3-(aq) In an aqueous solution of a strong acid, the acid is normally the only significant source of H+ ions If pH is more than 6, then we also need to consider the H+ ions from the autoionization of H2O Finding the pH of a strong monoprotic acid is simple The [H+] equals the original concentration of the acid Example If I have a 0.20 M solution of HNO3(aq) Then the [H+] also equals 0.20 M Diprotic acids (like H2SO4) are a bit more complex We will deal with these later Example 2 What is the pH of a 0.040 M solution of HClO4? If HClO4 has a concentration of 0.040, then [H+] also has a concentration of 0.040 pH = -log[H+] = -log(0.040) = 1.40 Strong Bases There are relatively few common strong bases The most common soluble strong bases are ionic hydroxides of the alkali metals and the alkaline earth metals NaOH, LiOH, Ca(OH)2, Mg(OH2), etc. These compounds completely dissociate into ions in an aqueous solution So 0.30 M NaOH is really 0.30 M Na+(aq) and 0.30 M OH(aq) In other words, we do not find undissociated NaOH in a solution Because strong bases dissociate entirely (in an aqueous solution), finding their pH is also relatively straightforward. Example What is the pH of a 0.028 M solution of NaOH? Note: Two ways to do this 1. 2. Find the [H+] Find pOH, then convert to pH I will do both Find the [H+] [OH-][H+] = 1.0x10-14 [H+] = 1.0x10-14 / 0.028 = 3.57x10-13M pH = -log(3.57x10-13) = 12.45 Find pOH, then convert to pH pOH = -log(0.028) = 1.55 pH + pOH = 14.00 14.00 - pOH = pH 14.00-1.55 = 12.45 Example 2 Ca(OH)2 is a strong base that dissociates in water to give TWO OHions per formula unit This means that the concentration of OHwill be twice that of the concentration of Ca(OH)2 What is the pH of a 0.0011 M solution of Ca(OH)2? [OH-] = 2x0.0011 M = 0.0022M pOH = -log(0.0022) = 2.66 pH = 14.00 - pOH = 14.00 - 2.66 = 11.34 Other Strong Bases If a substance reacts with water to form OH(aq), it will be strongly basic Most commonly these will be compounds with the oxide (O2-) ion Ionic metal oxides (Na2O and CaO) are often used in these cases With the metal oxides, each mole of the O2- reacts to form 2 moles of OH- Ionic hydrides (H-) and nitrides (N3-) also will act as bases in water 16.6 - Weak Acids Most acidic substances are weak acids Will only partially ionize in aqueous solution We use the equilibrium constant for the ionization reaction to find out how much a weak acid ionizes For example purposes, we will represent a weak acid as HA Therefore We can write a reaction with a weak acid as such HA (aq) + H2O(l) H3O+(aq) + A-(aq) or HA H+(aq) + A-(aq) Each of these means the same thing Equilibrium Constant Expression Kc Because [H2O] is a liquid (and solvent), we leave it out of the equilibrium-constant expression So, depending on which reaction equation we used H O A 3 HA Kc H A HA Slight changes As we did for the ion-product constant for the autoionization of water, we change the subscript constant to tell the type of equation to which it belongs Weak acids, so change the Kc to Ka Ka is called the acid-dissociation constant Many Ka values given in Appendix D of book The magnitude of Ka tells us the tendency of the acid to ionize in water The larger the value of Ka, the stronger the acid Ka will usually be less than 10-3 Finding Ka from pH To find one or the other of these things, we treat these problems like an equilibrium problem (from the last chapter) Example We have a 0.10 M solution of formic acid (HCHO2) and measured the pH using a pH meter. pH was measured to be 2.38 Find the Ka for formic acid at this temperature Step 1 - Write the Equation HCHO2(aq) H+(aq) + CHO2-(aq) Once we have the equation, we can write the equilibrium-constant expression Step 2 - Determine concentration of H+ pH = -log[H+] = 2.38 log[H+] = -2.38 [H+] = 10-2.38 = 4.2x10-3M Step 3 - Ice HCHO2(aq) Initial l Change e H+(aq) + CHO2-(aq) HCHO2 H+ CHO2- 0.10 M M 0 0 -4.2x10-3 M (0.10-4.2x10- +4.2x10-3 M +4.2x10-3 M -3 -3 Step 4 - Neat Assumption Remember, in a weak acid, we will end up with only a small amount of H+ Equilibrium concentration of formic acid (0.10 - 4.2x10-3) .10 - .0042 = .096 Here s the deal: 0.096 is so close to .10 (only off by 4 thousands of M), that we just assume the concentration of formic acid to be 0.10 What does this assumption mean for us? It means generally we can assume the change in the concentration of the weak acid is so small, its equilibrium concentration will be the same as its initial! We can now plug our values into our equilibrium-constant expression Step 5 - Plug Everything Into Equilibrium-Constant Expression HCHO2(aq) H CHO 4.2 10 4.2 10 1.8 10 Ka H+(aq) + CHO2-(aq) 3 3 2 HCHO2 0.10 This is a reasonable answer Most weak acids have a Ka between 10-3 and 1010 4 Using Ka to Calculate pH Knowing the Ka AND the initial concentration of the weak acid, we can find [H+] Which then lets us find the pH of the weak acid solution Example 0.30 M solution of acetic acid, HC2H3O2 HC2H3O2(aq) H+(aq) + C2H3O2+(aq) Ka at 25ºC is 1.8x10-5 According to the formula for acetic acid, the hydrogen that ionizes is the one separate from the rest (it is attached to the oxygen) We write the formula in this way to emphasize that only that one hydrogen is ionized Step 1 - Write the EquilibriumConstant Expression HC2H3O2(aq) H+(aq) + C2H3O2-(aq) Ka at 25ºC is 1.8x10-5 Step 2 - ICE HC2H3O2(aq) H+(aq) + C2H3OHC 2 2(aq) H3O2( H+(aq) C2H3O2aq) (aq) Iitia l 0.30 M Chag e -x M Equili (0.30 brium -x) M 0 +x 0 M M +x x M M EquilibriumCostat Expressio H C H O 1.8 x10 Ka 2 3 2 HC 2 H 3O2 ( x)( x) 5 Ka 1.8 x10 0.30 x 5 ( x)( x) Ka 1.8 x10 5 Simplify 0.30 x This will tur ito the quadratic But we ca simplify this, by remember that Ka is VERY SMALL Because Ka is so small, we recogize equilibrium will be FAR to the left, ad that x will be VERY small compared to the iitial cocetratio of acetic acid Thus, we ca say that x New(better) equatio! Ka = x2/0.30 = 1.8x10-5 x2 = (0.30)(1.8x10-5) = 5.4x10-6 x = (5.4x10-6).5 = 2.3x10-3 [H+] = x = 2.3x10-3 M Was Our Assumptio Valid? We ca check to see if our assumptio that the iitial cocetratio - x was approximately equal to the iitial cocetratio Fid the percet of the acid that ioized To do this, divide the [H+] by the iitial cocetratio of the Assumptio Check [H+]/[HC2H3O2] x 100% 0.0023 M/ 0.30 M x 100% =0.77% Our assumptio was valid Weak Acid Properties We fid the cocetratio of H+ is oly a very small percetage of the cocetratio of the acid Which meas oly a small amout of the acid actually doates a proto Because there is Properties iclude Electrical coductivity Rate of reactio a active metal with Polyprotic Acids May acids have more tha oe ioizable H atom These acids are called polyprotic acids These ted to ioize i steps Sulfurous Acid example (H SO ) Ka1 ? The umbers o the costats refer to the particular proto that ioizig ad Ka2 So ka1 refers reactio that first proto Ad ka2 refers reactio that secod proto Ka1 > Ka2 to the ioizes is the to the ioizes the After the first proto left, we have a egatively charged compoud. Usually the Ka values for the first ad secod disassociatio costat differs by at least a factor of 103 Note: Sulfuric acid (H2SO4) has a Ka1 as large Makig Life Easy Because Ka1 is so much larger tha ay of the other K values for a give acid, almost all of the [H+] comes from the first ioizatio As log as the other Ka values differ by a factor of 103 or Example The solubility of CO2 i pure water at 25ºC at 0.1 atm is 0.0037 M. The common practice is to assume that all of the dissolved CO2 is in the form of the carbonic acid (H2CO3), which is produced by the reaction between the CO2 and H2O. CO2(aq) + H2O(l) H2CO3(aq) What is the pH of a 0.0037 M solutio of H2CO3? Ka1 = 4.3x10-7 Ka2 = 5.6x10-11 Note: Sice Ka2 differs from Ka1 by more tha 103, we treat the acid H2CO3(aq) + H (aq) + HCO3 (aq) H2CO3 Iitia 0.003 l 7 M Chag e -x M Equili (0.00 brium 37 x)M H+ HCO3- 0 0 +x M M +x x M M equilibrium costatexpressio H HCO K a1 3 H 2CO3 x x .0037 x 4.3x10 7 We make same assumptio x x .0037 x 4.3 x10 7 x x 4.3x107 .0037 5 x 4.0 x10 [ H ] 5 pH log[ H ] log( 4.0 x10 ) 4.40 16.7 - Weak Bases May substaces behave as weak bases i water They react with water, ad pull protos from H2O Formig the cojugate acid of the base ad OH- ios B(aq) + H2O HB+ + OH- The most commoly ecoutered weak base is ammoia NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) So it would have the NH 4 OH equilibrium-costat Kb expressio of NH 3 Kb is the special equilibrium costat used Usig Kb to calculate [OH-] Frakly, use the same procedure you would have used to fid [H+] usig Ka Use ICE Treat the equilibrium cocetratio of the weak base as the same as the iitial cocetratio of the weak base Types of Bases Weak It ca be a little tricky to determie a weak base from a chemical formula However, weak bases fall ito two geeral categories Category 1 The first category icludes eutral substaces with a atom with a obodig pair of electros These obodig pairs of electros are where the base accepts the proto Category 2 This category icludes the aios of weak acids I other words: The cojugate base of a weak acid Determie the Cocetratio of a Salt A solutio is made by addig solid sodium hypochlorite (NaClO) to eough water to make 2.00 L of solutio has a pH of 10.50. How may moles of NaClO were added to the water? Basic Necessary Data pH of solutio is 10.50 Kb for NaClO is 3.33x10-7 (foud data table) i Step 1: Fid [OH-] pOH = 14.00 - pH = 14.00 - 10.50 = 3.50 -log[OH-] = 3.50 [OH-] = 10-3.50 = 3.16x10-4 M So this is the equilibrium cocetratio of OH- Step 2 - ICE ClO- HClO OH- M 0 0 Chag e 3.16x 10-4M +3.16 x104M +3.16 x104M Equili (x- 3.16x 3.16x Iitia l Step 3 - Fid [ClO-] Kb HClO OH ClO 4 4 (3.16 x10 )(3.16 x10 ) 7 Kb 3 . 3 x 10 ( x 3.16 x104 ) Mathmathmathmathmath 4 2 (3.16 x10 ) 4 x 3.16 x10 0.30M 7 3.3x10 Step 4: moles Fid Molarity 0.30M NaClO of is 0.30 moles 2.00 L 0.60 mol 1L Relatioship Betwee Kaad Kb Ka x Kb = Kw or pKa + pKb = pKw= 14.00 What does this mea? If we kow the K value for a cojugate acid/base of what we are Example Fid the basedissociatio costat for the fluoride io. Kb for F- is ot listed i ay tables However, Ka for HF is give Ka for HF is 6.8x10-4 K = K / K = Base Properties of Salt Solutios We ca assume that whe salts dissolve i water, they are completely dissociated Most salts are strog electrolytes Cosequetly, the acid-base properties of salt solutios are Ability to React with Water I geeral, a aio (-) i solutio ca be cosidered the cojugate base of a acid Cl- is the cojugate base of HCl ad C2H3O2is the cojugate base of HC2H3O2 Whether a aio Origial Acid from Cojugate Base To idetify the acid ad assess its stregth, we simply add a proto to the aios formula - plus a gives H proto (H+) If the acid is a strog acid, the the aio i questio will be a egligible base Therefore, the aio will ot affect the pH of However, if the H is NOT oe of the seve strog acids, the it is a weak acid I this case, the ao is a weak base Which meas the aio will react (to a small extet) with Aios that have ioizable protos (such as HSO3-) are amphiprotic They will act as either a acid or a base Behavior depeds o the relative sizes of Ka ad Kb for the io Ability to React with Water Polyatomic catios whose formulas cotai oe or more protos ca be cosidered the cojugate acid of weak basis NH4+ is the cojugate acid of the weak base NH3 So NH4+ will react with water to lower the pH However, ios of of Catio ad Aio i Solutio If a aqueous salt solutio cotais a aio that does NOT react with water a catio that does NOT react with water we expect the pH to be eutral If the solutio cotais a aio that reacts If solutio cotais A catio that reacts with water to form a hydroium io A aio that does ot react with water We expect the pH to be acidic If the the solutio cotais A catio that reacts with water to form a hydroium io A aio that reacts with water to produce a I 1. 2. 3. Summary A aio that is the cojugate base of a strog acid will ot affect the pH of a solutio A aio that is the cojugate base of a weak acid will cause a icrease i pH A catio that is the cojugate acid of a weak base will cause a decrease i pH Other metal ios will cause a decrease i pH 6. Whe a solutio cotais both the cojugate base of a weak acid ad the cojugate acid of a weak base, 5. Example List the followig i order of icreasig pH 0.1 M Ba(C2H3O2)2 ii.0.1 M NH4Cl iii.0.1 M NH3CH3Br iv.0.1 M KNO3 i. Aalysis i. ii. 0.1 M Ba(C2H3O2)2 Ba2+ is the io of oe of the heavy group 2 elemets, will ot affect pH C2H3O2- is the cojugate base of a weak acid. Will make solutio basic 0.1 M NH4Cl Cl- is the cojugate base of a strog acid. Will ot affect pH NH4+ is the cojugate iii. 0.1 NH3CH3Br NH3CH3+ is the cojugate acid of a weak base. Will make solutio acid Br- is the cojugate base of a strog acid. Will ot affect pH M Kb of NH3 is 1.8x10-5 Kb of NH2CH3 is 4.4x104 Sice NH3 has smaller Kb, ad is weaker 0.1 M iv. KNO3 K+ io is the of the strog base KOH NO3- is the cojugate base of the strog acid HNO3 Neither will react with water, makig solutio eutro Overall icreasig pH Example 2 Predict whether the salt Na2HPO4 will form a acidic or basic solutio o dissolvig with water Na2HPO4 is amphiprotic 2 Possible reactios HPO42-(aq) H+ (aq) + PO43-(aq) HPO42-(aq) + H2O H PO -(aq) + OH-(aq) HPO42-(aq) PO43-(aq) H+ (aq) + This is oe of the give equatios withi the chapter. Ka = 4.2x10-13 HPO42-(aq) H2PO4-(aq) + + H2O OH-(aq) We ca fid the Kb for this by its cojugate acid, H2PO4Ka = 6.2x10-8 Ka x Kb = Kw Therefore, Kb= 1.6x10- Base Behavior ad Chemical Structure Whe a substace is dissolved i water, it may behave as a acid, base, or be either. We will look at how the chemical structure of a substace determies Factors that Affect Acid Stregth A molecule cotaiig H will trasfer a proto ONLY if the H- bod is polarized Meaig that the has a relatively large electroegativity Which is why HCl is so acidic (Cl has a large electroegativity) Why NaH is basic Because the H atom possesses a egative charge ad will But that all! is ot A secod factor that helps determie whether a molecule that has a H- bod will doate a proto is the stregth of the bod Very strog bods are less easily dissociated Ad still ot all! A third factor is the stability of the cojugate base, I geeral, the greater the stability of the cojugate base, the stroger the acid So the overall stregth of a acid is depedet upo these three factors Biary Acids I geeral, the H- bod stregth is the most importat factor i fidig the acid stregth amog biary acids i which is i the same group i the periodic table A biary acid is oe cotaiig just hydroge ad oe other elemet The stregth of a acid teds to decrease as the Across the Periodic Table Bod stregth chage is less movig across a row i the periodic table tha dow the table Bod polarity becomes the major factor determiig the acidity of a biary acids i the same row. So acidity geerally icreases as the electroegativity of Oxyacids May commo acids, such as H2SO4 cotai oe or more O-H bods Acids i which OH groups ad possibly additioal oxyge atoms are boud to a cetral atom are called oxyacids. We will look at what factors determie Cosider a OH group boud to some atom, Y, which might i tur have other groups attached to it - Y - O - H At oe extreme, Y might be a metal, such as Na, K or Mg Because of low electroegativity, electros are pulled to the oxyge, ad ioic compoud cotaiig OH- is formed These would be sources of OH-, ad would be a base Y as a ometal cotiued Agai, as Y becomes more electroegative, compoud becomes more acidic Two 1. 2. reasos for this As electro desity is draw toward Y, the OH bod becomes weaker ad more polar, favorig the loss of H+ Because the cojugate Eve More Oxyge! May oxyacids cotai more oxyge atoms boded to the cetral atom Y The additioal oxyges (all electroegative) pull electro desity from the O-H bod This further icreases its polarity Summary For oxyacids that have the same umber of OH groups ad the same umber of O atoms Acid stregth icreases with icreasig electroegativity the cetral atom of For oxyacids that Additioally Because oxidatio umber of the cetral atom icreases as the umber of attached O atoms icreases I a series of oxyacids, the acidity icrease as the oxidatio umber Example Arrage the compouds i the followig series i order of icreasig acidity AsH3, All biary acids So we look at electroegativity HI, NaH, H2O More electroegative acidic NaH has lowest electroegativity, so basic = more most Example 2 Arrage the compouds i the followig series i order of icreasig acidity H2SeO3, H2SeO4, H2O Number of oxyges boded to cetral atom is the biggest determier here Acidity of oxyacids icreases with # of oxyge atoms boded So H2SeO4 is a stroger acid 16.11 - Lewis Acids ad Bases For a substace to be a proto acceptor (a Bosted-Lowry base) it must have a ushared pair of electros to bid the proto. G.N. Lewis revised the defiitio of acidbase reactios to focus o these pairs of electros A Lewis acid is a So Every base weve discussed so far is a electro pair door far... Which meas that everythig that is a Brosted-Lowry base is also a Lewis base However, i the Lewis theory, a base ca doate its electro pair to somethig NH3 has us bee a base for But cosider reactio betwee NH3 ad BF3 BF3 acts as a electro pair acceptor Makig it a Lewis acid Solvets So far, weve emphasized water as our solvet ad the proto as our source of acidic properties Brosted-Lowry acid-base defiitios most useful whe doig this Beefit of the Lewis acid-base defiitios is it allows us to treat Commoalities of Lewis Acids May Lewis acids have a icomplete octet of electros Like BF3 Also, may simple catios ca act as Lewis acids Fe3+ will take electros from cyaide ios to form ferricyaide io, Fe(CN)63- Hydrolysis of Metal Ios As we have see, most metal ios act like acids i a aqueous solutio. Lewis acid-base theory explais this Because metal ios are + charged, they attract the Patters Acid-dissociatio costats for hydrolysis reactios usually icrease with icreasig charge ad decreasig radius of the atom Cu2+ io has smaller charge ad larger radius tha Fe3+ is