Chapter 20 The Properties of Acids and Bases

Chapter 20 The Properties of Acids and BasesAcids and bases play a key role in many areas of chemistry. Often they are part ofequilibrium systems, so now that you understand equilibrium better we can begin todiscuss the intricacies of acid-base systemsAgain, I and going to skip around a little bit in the order I present things20-1 Acids and BasesArrhenius DefAcids make H + also taste sourBases make OH - also taste bitter and make hands slipperyWorks only in aqueous solutions+In reality H does not exist+Actually H3O the hydronium ionBrønsted- LowryAcids = Proton DonorsBases = Proton AcceptorsFocus on proton-transfer reactionsWorks in non-aqueous solutionsWill be focus for rest of chapterLewis Theoryacid - electron pair acceptorElectrophilebase - electron pair donorNucleophileEspecially good in understanding organic reactions like NH +BF20-2 WaterPure neutral water has some hydronium and hydroxide ions+ -2 H2O H3O + OHacts as both donor and acceptorKey Definition:K = [H O ][OH ] = 1x10w 3+ - -14+ -X = [H3O ] = [OH ]-14 21x10 =X-14X=SQRT(1X10 )3 3X = 1x10 -7 + -Neutral solution when [H3O ] = [OH ]Basic solution when [H3O ]< [OH ] or [OH ] >1x10 & H3O > 1x10Acidic solution when [H O ]> [OH ] or [OH ] 1x10+ - - -7 + -7+ - - -7 + -73 3

220-3 Strong Acids and BasesKey Definition:Strong acids are acids that undergo 100% dissociation in waterStrong bases are bases that undergo 100% dissociation in waterThe common strong acids and bases are listed in table 20.1YOU NEED TO MEMORIZE THESE COMPOUNDSSample calculation+ - -What are the concentrations of H3O , Cl and OH in .03M HCl100% dissociation so+ -[H3O ]= [Cl ] = .03M-[OH ] comes from K W-14 + -K W = 1x10 = [H3O ][OH ]-14 -1x10 = .03[OH ]- -14 -13[OH ] = 1x10 /.03 = 3.33x10What are the concentrations of H O , Ca and OH in .03M Ca(OH)100 % dissociation so-[OH ] = .03M Ca(OH) 2 x 2 OH/1Ca(OH) 2 = .06M2+ 2+[Ca ] = .03M Ca(OH) 2 x 1 Ca /1Ca(OH) 2 = .03M+[H3O ] comes from KW-14 + -K W = 1x10 = [H3O ][OH ]-14 +1x10 = .06[H3O ]+ -14 -13[H O ] = 1x10 /.06 = 1.67x103+ 2+ -3 2Neutral IonsNow let’s look at the back reaction- +If HCl + H2O Cl + H3O go to 100% completion-How much of a tendency does Cl have to do the back reaction:- -To Cl + H2O HCl + OHZip- Zippo - Zilch-Nada-So Cl is a neutral ionAnd so on for all the anions of the strong acidsSimilarly for Na +So the cations of strong bases are also neutralThe neutral ions are summarized in table 20.3YOU NEED TO MEMORIZE THESE IONS20-4 Carboxylic Acids Skip temporarily

20-5 pH and Acidity+ -You will find that the [H ] and [OH ] concentrations are are important in many+ -reactions. Even if [H ] and [OH ] are not products or reactants in a reaction, they+often serve as catalysts. In addition, all living organisms carefully try to keep [H ]+ -carefully regulated because small fluctuations in [H ] or [OH ] can easily kill acell+ - -14Generally the concentrations of [H ] and [OH ] in water range from 1M to 1x10M. With such a broad range of concentrations it is convenient to switch to alogarithmic scale. This scale is attributed to Søren Sørenson. The p of pH isthose to mean the ‘power’ or exponent of the hydrogen ion concentrationKey Equation+ +pH = -log[H O ] = -log[H ]3Example problems+Just a page ago we calculated that .03M HCl had a [H ] of .03MWhat is the pH of this solution?+pH = -log [H ]= -log(.03)= -(-1.52=1.52We also determined that a.03M Ca(OH) 2 had an [H ] of 1.67x10What is the pH of this solution?+pH = -log [H ]-13= -log(1.67x10 )= -(-12.778)=12.778+ -133You can do the same thing with [OH - ]Key EquationpOH = -log[OH - ]Sample Problem-Continuing with the [OH ] side of the previous problems, the .03M Ca(OH) 2 had-an [OH ] .06MpOH = -log(.06)= - (-1.222)

As we tackle the OH side of the .03M HCl solution I want to show you somethingIn the original problem we had+[H ] = .03Mthen we used-14 + -K W = 1x10 = [H3O ][OH ]to find that- -14 -13[OH ] = 1x10 /.03 = 3.33x10-13so pOH = -log (3.33x10 ) = -(-12.48) = 12.48Bu there is a shortcutKey EquationpH + pOH = 14(I can derive this for you, but you probably don’t want to sit through the math)Continuing examplewe already said that the pH of this solution was 1.52using 14 = 1.52 + pOHpOH = 14-1.52 = 12.48and you get exactly the same numberMore on pH scaleEarlier I said that+ -Neutral solution when [H3O ] = [OH ]Basic solution when [H3O ]< [OH ] or [OH ] >1x10 & H3O > 1x10Acidic solution when [H O ]> [OH ] or [OH ] 1x10+ - - -7 + -7+ - - -7 + -73 3Making the equivalent statements using the pH scale we haveNeutral pH = pOH = 7Basic solution is when pH >7 or pOH 1!-log(5) = -.7-.7 !!14 = pH + pOH; pOH = 14.7!4

One last trick problem I may throw at you-8What is the pH of 1x10 HCl?Well, that should be easy-8pH = -log(1x10 )= - (-8)=8But what is wrong with that answer? HCl is an acid, but you just got a basic pH!-7 +The problem is that you forgot about water. Water is furnishing 1x10 H+ -8 -7so H = 1x10 from the HCl and 1x10 from the water for total of1.1x10 -7-7pH = -log(1.1x10 ) = 6.96 an acid pHAdding the water contribution on top of your original acid or base is only-7necessary when the acid (or base) concentration is < or = 1x10 . It is also a bitof a quick and dirty approximation, when you get to analytical I will show you theproper way to handle this problem.+pH [H ]By now you know that I want you to be able to do calculations forward and+backwards. So how do you calculate [H ] from pH?Example CalculationIf an HNO solution has a pH of 4.83, what is the concentration of [HNO ]?3 3+pH = -log [H ]+4.53 = -log[H ]+-4.53 = log [H ]How do you get rid of the log function? Do 10^ on both sides of the equation-4.53On the left side 10^-4.53 = 10On the right side 10^log cancels out so-4.53 +10 = [H ]= 2.95x10 -5Since HNO 3 is a strong acid with 100% ionization+ -5[H ] = [HNO 3] = 2.95x10Summarizing into a single equationKey Equation+ -pH[H ] = 105

620-6 Weak Acids and BasesIf strong acids and bases have 100% dissociation, what do you think happenswith weak acids and bases? You got it,

3. list initial concentrations and start your ICE tableHA+ -H + AInitial .006 0 04. Calculate Q to determine what direction(if any) the reaction will shiftWith 0 products we don’t need a Q, you know it will go to products5. using a single variable(X), define change for each concentration needed toreach equilibrium using X.+Here, since we want to calculate pH which comes from [H ] we make+[H ]=X-[A ] = X[HA] = -X6. Substitute modified concentration terms into the ICE table and the equilibriumexpressionHA+ -H + AInitial .006 0 0Change -X X XEq .006-X X XK a = XX/(.006-X)-5K a = 1.8x10 (from table 20.4)-51.8x10 = XX/(.006-X)7. Solve the equilibrium expression for the unknownThere are two ways to solve this, long and shortStart with the long-5 2(.006-X)1.8x10 =X-7 -5 21.08x10 -1.8x10 X = X2 -5 -7X + 1.8x10 X -1.08x10 =02Using the quadratic -b +/- sqrt(b -4AC) / 2A-5 -10 -7-1.8x10 +/- sqrt(3.24x10 + 4.32x10 ) / 2-5 -7-1.8x10 +/- sqrt(4.323x10 ) / 2-5 -4-1.8x10 +/- 6.575x10 /2Ignoring the - root-4 -5(6.575x10 -1.8x10 )/2-46.395x10 /23.18x10 -4-4pH = -log(3.18x10 ) = 3.49Now the short, quick and dirtyIf X si small, the reaction does not go far to the right so.006-X ~ .006-5 21.8x10 ~ X /.006-5 -4X = sqrt(.0061.8x10 ) = 3.29x10 ; pH = 3.488

9A little off, but not too bad-4The quick and dirty method usually works if the K a is 10 or less-3if you see a K of 10 or greater you should use the quadratica20-8 Successive ApproximationThe Q&D answer we got above was a lot faster than the quadratic, but us wasoff a bit-4 -4 -43.28x10 - 3.18x10 = .10x10-4 -4.10x10 /3.18x10 x 100% = ~3% errorThe method of successive approximation is a way to get closer to the realanswer without the quadraticIn the Q&D we did an approximation that .006-X ~ .006But once we solved the equation and found that X = 3.28x10 -4we can improve our answer-4Out improved answer is that .006-3.28x10 =.005672And not solve the problem with is better approximation-5 21.8x10 ~ X /.005672-5 -4X = sqrt(.0056721.8x10 ) = 3.29x10 ; pH = 3.48X = 3.19x10 -4And now our answer is < 1% off.You can even set up a spread sheet to do this over and over. But most of thetime a single iteration will be just fine for most weak acid problemsNow that you know how to handle weak acids, and you have seen the table withlots of different weak acids it is time to look at the most common weak acid seen inorganic chemistry the carboxylic acid20-4 The carboxylic acid (out of order)The most common acid you see in organic chemistry is the carboxylic acid oftenrepresented with the atoms COOHThe structure is shown above, where R represents H or any C containing group

Two of the acids you may be familiar with are acetic acid and formic acid. Aceticacid is the acid (and the smell) found in vinegar, Formic acid is the irritant foundin and bites ( it was first isolated in 1600's by distilling ants!)10The acidic proton on these structures is the H on the COOH group.If you think about it you might wonder why. After all H-O-H with OH bonds is notacidic, and alcohols C-O-H groups are not acidic either. Th answer lies is thefact that the anion that is formed when the proton leaves is stabilized byspreading out the negative charge over 2 atoms as shown below:Is is interesting that you can actually see this in x-ray structures. The C=O bondis 123 pm, the C-O-H bond is 136 pm and the C-O bond in then resonance stablizedanion is 127 pm20-9 K B and base strengthNow let’s turn our attention to weak basesThe generic base reaction is:B + H O BH + OH -2-So in this case the OH that you associate with something being basic is theproduct of the reaction of a base with waterNotice in the reaction B the base was a proton acceptor which matches ourBronstead Lowry defintion

11Or H HH-N: + H-O-H H-N-H + OH -H HThe pair of electrons are donated to water which agrees with the Lewis model ofa baseAgain this is a equilibrium reaction with a K, but we usually use the symbol K B forthe base ionization or base protonation constant. In a similar manner as theacidsKey ConceptThe larger the K B, the more the reaction goes to the right, so the stronger thebaseTable 20.5Key EquationpK ‘s for -log(K )BBClicker questionRank the bases ammonia, aniline, and ethylamine from weakest to strongestbaseAll of the weak bases found in this table are amines, or nitrogen containingorganic molecules. The lone pair of electrons on the nitrogen that can eitheraccept protons or donate electrons (depends on the definition of base you use)is what these bases all have in common+Calculating the [H ] or pH due to a base follows the same kind of logic ascalculation involving acids, you just have to add one more stepSample CalculationLet’s calculate the pH of a .04M ethylamineAgain we attack this problem just like any other equilibrium problem1. Write the balanced reaction CH3CH2NH 2 + H2O CH3CH2NH 3 + OH -+ -I will shortcut this to B + H2O BH + OH2. Write the equilibrium expression that corresponds to the balanced reaction+ -K B =[BH ][OH ]/[B]3. list initial concentrations and start your ICE table+ -B H2O BH + OHInitial .04 Constant 0 04. Calculate Q to determine what direction(if any) the reaction will shiftWith 0 products we don’t need a Q, you know it will go to products5. using a single variable(X), define change for each concentration needed to

each equilibrium using X.Here, since we want to calculate pH which can be calculated form pOH- -which comes from [OH ] we make [OH ]=X- +[OH ] = X = [BH ][B] = -X6. Substitute modified concentration terms into the ICE table and the equilibriumexpression+ -B H2O BH + OHInitial .04 Const 0 0Change -X X XEq .04-X X XK B = XX/(.04-X)-4K B = 4.5x10 (from table 20.5)-44.5x10 = XX/(.04-X)7. Solve the equilibrium expression for the unknownThere are three ways to solve this, long and short, and successive approximationLet’s just do the successive approximationst1 approximation.04-X ~ X-4 24.5x10 =X /.04-4X = sqrt (4.5x10 .04)x ~ .00424ND2 approximation.04 -X ~.04 - .00424 ~ .03576-4 24.5x10 =X /.03576-4X = sqrt (4.5x10 .03576)x ~ .0040-But X = [OH ] and we want pH, so-[OH ] =.004, pOH = -log(.004)= 2.40pH = 14-2.40 = 11.6020-10 Conjugate Acid-Base PairsSo far we have concentrated on the forward reaction HA + H2O A + H3O+ -Or B + H O BH + OH2But now let’s look at the reverse reactions- + + -A + H O HA + H O Or BH + OH B + H O3 2 2Note that re reverse reactions are also acid base reactions- + + -A + H3O HA + H2O Or BH + OH B + H2OH Acceptor H Donor H donor H acceptorAcid Base Acid Base- +12

Key DefinitionSo the acid’s anion is a base in the reverse reaction - we call this a conjugatebase and a base’s cation is an acid in the reverse reaction - we call this aconjugate acidSample problemsWhat are the conjugates bases of the following acids?- +HClO 3, H2PO 4 , and CH3NH3Acids are proton donators, so what ion do you have left after they have donated- 2-their proton - ClO , HPO and CH NH3 4 3 2What are the conjugate acids of the following bases?-2 -CH NHCH , HPO , F3 3 4Bases are proton acceptors so what do you have after you have accepted a+ -proton? - CH NH CH , H PO , HF3 2 3 2 4So the conjugate form of the weak acids are bases, and will make the solutionbasic. Also the conjugate form of the weak bases are acids, and will make thesolution acidic. Now a reminder. What is the conjugate form of HCl, and is it anacids or a base. Similarly, what is the conjugate form of NaOH, and will it makea solution acid or base? (The anions of strong acids are neutral, the cations ofstrong bases are neutral !)K b’s of weak acids-So now you know that the acetate anion, CH3COO should be basic because it isa conjugate base. But how strong a base is it?Well considerCH COOH + H O CH COO + H O , K = 1.75x10- + -53 2 3 3 aFor the reverse reaction- +CH COO + H O CH COOH + H O Reverse so K = 1/K3 3 3 2 aH O + H O H O + OH+ -2 2 3 wKNow if we add the above two equations together, what do we do with the K’s?(Multiply them)The result is- -CH3COO + H2O CH3COOH + OH and this is our base reaction for the acetateion.13

14-14 -5 -10K B = K w/K A = 1x10 /1.75x10 = 5.71x10Key EquationThe K b for the conjugate base of a weak acid = K w/KASimilarly for basesThe K for the conjugate acid of a weak base = K /KOr, more simply, K K =KA W BA B WSample calculation+What is the K for ammonia (NH )A 4-5K B for ammonia = 1.8x10 (table 20.5)K = K /K = 5.7x10 -10A w B20-11 Salt solutionsWe started this chapter looking at strong acids and bases, then we worked onweak acids and bases, and just now we did the conjugate forms of the acids andbases. Now let’s put all these different pieces together into a compound called asalt.Back in chapter 10 you learned that a salt is the product of an acid baseneutralization reaction, so let’s start there.What do you get when you neutralize an acid like HCl with NaOH?HCl + NaOH NaCl(aq) + H2O+ -Early on in this chapter you learned that Na and Cl were neutral ions, so the pHof the neutralized solution is indeed, neutral.Now what about CH3COOH and NaOH?CH COOH + NaOH CH COONa(aq) + H O3 3 2But as an ionic compound CH COONa Na + CH COO+ -Na neutral CH COO basic (C.B of a weak acid)3+ -3 3+ -How about HCl and NH 3 NH 4 (acidic) and Cl (neutral)So neutralization reactions can leave you neutral, or acidic or basic!What is a person to do? On top of that, sometimes I won’t even give you theneutralization reaction, I will just give you the salt, say NaHSO 3. Are you up theproverbial creek without a paddle?Key Procedure:1. Break each salts into its component anion and cation2. Ignore the neutral cations and anions3. Identify the acid/base properties of the remaining cation or anion

Table 20.7 is a nice table that summarizes the acid/base properties of many ionsbut you shouldn’t have to memorize this table, you should be able to use logicrememberthe cations of strong bases are neutral, so this means basically the firsttwo column of the periodic tableThe anions of strong acids are neutral, and you shoul dknow your straongacids by nowSo you can ignore all of the aboveLooking at what remainsThe basic anions are the conjugate bases of weak acidsThe acidic cations are the conjugate acids of weak bases (and Ithink all have a N in them!)Table 20.7 does have a few that don’t fit above. Let’s examine them moreclosely1. There are no basic cations!2. Cations with >+2 charge tend to be acidicLast semester we talked about waters of hydration, that is watersthat are bound to + cationsWhen you have a highly charge cation, the cation actually grabs- +the water so strongly that it splits the water into OH and H . The- +OH stays attached to the metal and the H is releasedExample:3+ 3+ 2+ +Al (aq) is actually [Al(H O) ] (aq) [Al(OH)(H O) ] (aq) + H (aq)2 6 2 515Added material NOT in your textAcid-Base properties of OxidesThere is another class of compounds that have acid/base properties that we canidentify fairly easily and that is the oxides.In nature you will find oxides of virtually every element. Oxides can be eitheracidic or basic. You now have enough chemistry behind you that you can beginto understand these properties and to predict them of unknown compounds.Let’s start with the acidic oxides. When a covalent oxide dissolves in water is willform an acidic solution. Examples:SO 3 (g) + H2O(l) H2SO 4(aq)SO 2 (g) + H2O(l) H2SO 3(aq)CO 2 (g) + H2O(l) H2CO 3(aq)2NO (g) + H O(l) HNO (aq) + HNO (aq)2 2 3 2

16On the other hand when ionic oxides dissolve in water we get basic solutionsCaO(s) + H2O(l) Ca(OH) 2(aq)Na2O(s) + H2O(l) 2NaOH(aq)K2O(s) + H2O(l) 2KOH(aq)and these are called basic oxidesIn our covalent oxides the oxygen has a covalent bond to the cental atom andthe H, and the bond between the central atom and the O is stronger than thebond between the O and the H so the H may be easily removedIn the ionic compound there is no covalent bond between the metal and the O 2-anion. The oxide anion, in fact, has a high affinity for protons (a strong base)and will pull the off of water2- -O (aq) + H2O(l) 2OH (aq)Making most ionic oxides basicBut hold it... In the lab you should have found that CrO 3, a metal oxide had a pH+6of 2 or less. What gives? (In this compound you have Cr We just talked abouthow metal ions with strong + Charge can be acids, here the acidity of the +6metal ion has overpowered the basicity of the metal-oxide)Practice problems Maybe Clicker questionsIdentify the following compounds as acids or basesNaCl (N)CH3COOK (B)RbF (B)NH4NO 3 (A)CaO (B)SO 2 (A)CH COONH (Trick question - N)3 420-12 Polyprotic AcidsLet’s talk a bit about H2SO 4 and H3PO4These are examples of polyproitic acids, acid that have more than one proton togive. There ae several polyprotic acids listed in table 20.9 of your text.If you look at the K’s of the polyprotics, you will see that form most of them thefirst K is relatively strong, but the second is much weaker. Can you figure outwhy?