Note for CHM 116 with Professor Berger at IPFW
Note for CHM 116 with Professor Berger at IPFW
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Date Created: 02/06/15
174 This points out the general observation that as the concentration of the weak acid decreases the degree of ionization increases This can be rationalized in terms of LeChatelier s Principle As the solution is diluted the side of the equilibrium with more particles is favored This also raises an important question When can x be neglected relative to the quantity from which it is being subtracted We ve already said that if a quantity is less than 5 of the quantity from which it is being subtracted the original quantity is negligible However until you solve the equation you can t compare x to the quantity quotom which it is being subtracted a and st05Ca a X 0050 2 00025 c 2 a A Ca 095 c 095 c C So if fa 2 380xwillbe g 5 ofCa Your book says 100 This doesn t follow the 5 rule Example What is the dissociation of a 010 M aqueous solution of the weak base methylamine CH3NH2 at 25 Cif its Kb is 44 x 10 A What is the pH of the solution 175 The followmg two table 1151 dissom hon constants for a vanay of weak elds and bases Iespechvely 3129c Acid Dissociaiion Consiams a125 C Substance Formula Ka Acetic acid HCZHBO2 17 gtlt1Ci395 Benzoic acid H127H502 63 x10395 Boric acid HSEIOa 59 gtlt1039m Carbonic acid HZCOa 43 x 10397 HCOB39 43 gtlt1Ci39M Cyanic acid HOCN 35 x10quot Formic acid HCHQ2 17 x10quot Hydrocyanic acid HCN 49 x 1039m Hydro uoric acid HF 68 x 1Dquot Hydrogen sulfate ion HSOA39 11 x 10392 Hydrogen sul de H23 89 x 10398 HS 12 X 103913 Hypochlorous acid HCIO 35 x 10398 Nitrous acid HNo2 45 x10quot Oxalic acid HZCZO4 56 gtlt1Ci392 HcZO 51 x10395 Phosphoric acid HSPOA 69 x 10393 H2120 62 x108 HF39OAZ 48 gtlt1Ci3913 Phosphorous acid HZPHOB39 16 x 10392 HPHOS 7 x 10397 Propionic acid HCBHEO2 13 gtlt1Ci395 Pyruvic acid HCBHBOa 14 x 10quot Slllflll39DLIS acid HZSOa 13 x 10392 HSOB39 63 gtlt1Ci39E 175 Base Dissocia on Cons1ams a125 C Substance Formula K Ammoma NH3 18 x105 Anmne CBHSNH2 4 2 x 101D D1methy1am1ne CHawH 51x10quot Ewamme CZH NHZ 4 7 x10quot Hydrazme NM 17 x105 Hydroxy amme NHZOH 1 1 x108 Methy amme CH3sz 44 x10quot Py dme CSHSN 14 x109 Urea NHZCONHZ 15 x10 Polypmhc Amds Polypxohc and Helhose winch can dissom le mme than one pmlon p8 molecule HAaq1lt110 a 1430 aqHA g lg 1quot Hgo llAl39 HA aqujo a 1130 1qu g Kn HAj Fox anypolypmhc and K gt 1g gtK 3 177 Sul lric acid illustrates this phenomenon well stOAaq H200 e H30aq HSOA aq strong 0 s02quot HSOA aqH20l H3Oaq 042 aq Kn2 11 x10392 HSO4 as does phosphoric acid 0 POquot H3P04aq H200 H3Oaq HzPOA aq Ka1 W 69 x10393 H3P04 0 HPOZ39 H21gt04 aq H200 H3Oaq HPOf aq K32 62 x10398 H2P04 0 PO37 HPOf aq H200 H3Oaq P043 aq Ka3 M 48 x1013 HPOZ For most polyprotic acids successive values of Ka differ by 45 orders of magnitude Example Calculate the concentrations of all species present in a 010 M solution of tartaric acid HzCAHAOS a weak diprotic acid We can abbreviate tartaric acid as HzTar K 92x104andKa2is43x105 a1 178 The results of these calculations point out additional generalities regarding the ionization of polyprotic acids If KB gt 20 Kai the hydronium ion from the rst ionization effectively suppresses the second ionization and the yield of hydronium ion is essentially all due to the rst ionization The concentration of the doubly deprotonated acid A2 is numerically equal to Kaz Ions as Acids and Bases Salts may be regarded as ionic compounds which result from the neutralization of an acid by a base In some instances the salt will exhibit no acidbase behavior but in many cases a solution of the salt will be either acidic or basic Consider the reaction of acetic acid CH3COOH with sodium hydroxide NaOH CH3COOH aq NaOH aq CH3COONa aq H20 1 Since acetic acid is a weak acid while NaOH and sodium acetate are a strong base and strong electrolyte respectively we can write the net ionic equation as CH3COOH aq OH aq a CH3COO aq H20 1 If we react 02 M CH3COOH with an equal volume of 02 M NaOH the resulting solution is 01 M in Na and CH3COO and has a pH of about 89 This is clearly a basic solution Similarly if we react 02 M NH3 a weak base with an equal volume of 02 M HCl a strong acid NH3 aq H30 aq NHX aq H20 1 the resulting solution is 01 M in NH and Cl and has a pH of about 51 This is clearly an acidic solution Neither of these neutralized solutions is in fact neutral This is due to the fact that certain ions will undergo a reaction with water called a hydrolysis reaction In the rst case the acetate ion CH3COO undergoes hydrolysis CH3COO aq H20 1 CH3COOH aq OH aq and in the second case the ammonium ion NHX undergoes hydrolysis NHX aq H20 1 NH3 aq H30 aq and you can clearly see why the resulting solutions are basic and acidic respectively 179 In general the anion of a weak acid will have appreciable basic character and will undergo hydrolysis to yield the weak acid itself and OH ions resulting in a solution with pH gt 7 The anions of strong acids on the other hand have no tendency to act in this manner The cation of a weak base will undergo hydrolysis to yield the weak base itself and the hydronium ion resulting in a solution with pH lt 7 The mtions of strong bases have no such tendencies In general there are four different possible combinations of weak and strong acids and bases The salt obtained by the reaction of a strong acid and weak base will be acidic due to hydrolysis of the cation The salt obtained by the reaction of a weak acid and strong base will be basic due to hydrolysis of the anion The salt obtained by the reaction of a strong acid and strong base will be neutral since neither the cation nor the anion undergo hydrolysis The salt obtained by the reaction of a weak acid and weak base could be acidic basic or neutral since both the cation and the anion undergo hydrolysis If Ka gt Kb the solution will be acidic If Ka lt Kb the solution will be basic If Ka Kb the solution will be neutral How do we calculate the pH of a salt solution First it is necessary to identify the ion or ions that are subject to hydrolysis Consider the 01 M solution of NaCH3COO we mentioned above The anion undergoes hydrolysis but the cation does not CH3COO aqH20 1 CH3COOHaqOH aq We are not likely to find a value for Kb of CH3COO in any table However we can easily write the expression for Kb and find the value of Ka for CH3COOH Ka 17 x 10 5 at 25 C CH3COOHOH b CHSCOO 30 If we multiply this by and rearrange we get 3 1710 CH3COOHOH H3O CI13COOH X CHSCOO 39 HSO CIISCOO 39HSO b X HSO 0H 39 You should recognize the rst part of this rearranged expression as the reciprocal of the expression for Ka of acetic acid and the second part as the expression for KW Therefore This is a general result for any conjugate acid base pair K Kb Kw Therefore for CH3COO 14 Kb E 533 x 1010 K 17 x 10 5 Another method to arrive at this result is illustrated in your text Now back to the question What is the pH ofa 010 M solution ofNaCH3COO Here we made an implicit assumption that the autoionization of water could be neglected as a source of OH Since OH b gt 45 x 10 7 this is a reasonable assumption Note that this is a very small value so the extent of hydrolysis is not very large lt0008 but does result in a pH of about 889 1711 Salts of a weak base and strong acid where cation hydrolysis occurs may be treated in a similar manner In this case the mtion acts as an acid and 1 Kafbew Calculations for the salt of a weak acid and weak base where both cation and anion hydrolysis occurs become quite involved Consider a solution of NHAF NH aq P120 1 NH3 aq H3O aq Ka 56 x 10 10 F aq H20 1 HF aq OH aq Kb 14 x 10 11 2 H20 1 H301 aq OH aq KW 10 x 10 14 These three equilibria must be satisfied simultaneously With 6 unknowns NHf NH3 F HF H3O and OH we must be able to write and simultaneously solve 6 independent equations in order to describe the chemical system Qualitatively however we can say that since Ka gt Kb cation hydrolysis will be slightly favored and an acidic solution will result Salts that contain small highly charged cations may also undergo hydrolysis to yield acidic solutions The following equations illustrate the point Fe 6 H20 a FeHQO63 FeH2063 H20 FeH205OH2 H301 Here Fe is acting as a Lewis acid and H20 as a Lewis base The presence of a concentrated center of positive charge Fe3 tends to draw electron density away quotom the H atoms in the coordinated water making them susceptible to dissociation as H as shown The Common Ion Effect Suppose we have a solution of the weak acid HF HF aq H20 1 H3O aq F aq LeChatelier s Principle tells us that if we dissolve some solid NaF a strong electrolyte and source of F ion the equilibrium will be shifted to the left suppressing the ionization of HF Similarly if we add HCl a strong acid and source of H3O ions the equilibrium will also shift to the left suppressing the ionization of HF 1712 These are both illustrations of the common ion effect which is a shift in the position of an ionic equilibrium caused by the addition of a solute that provides an ion which is part of the equilibrium Earlier in the chapter we calculated that the ionization of a 050 M solution of HF at 25 C was about 37 We would expect that the ionization will be somewhat smaller if sufficient NaF was added to the solution to make the solution initially 01 M in F HF aq H20 1 F aq H3O aq I 050 010 0 A 7x x x eq 050 7 x 010x x 3 HF 050 x We ll make an assumption which might simplify the calculation Assume x ltlt 010 K drumMW 050 3 x 34x103 x100068 050 x H301 34x 10 3M HF 050 7 00034 s 050M pH flog H3O flog 34x 10 3 247 This points out a general observation that when the anion of a weak acid is added to a solution of the weak acid the dissociation of the acid is suppressed and the pH increases Similarly if the cation of a weak base is added to a solution of the weak base the ionization of the weak base is suppressed and the pH decreases Solutions containing a weak acid and its conjugate base or a weak base and its conjugate acid play an important role in acidbase chemistry These solutions are referred to as buffers and the next section will deal in great detail with these important solutions 1713 Buffers A buffer is a solution containing either a weak acid and its conjugate base or a weak base and its conjugate base which is able to resist large changes in pH when acids or bases are added to it within certain limits referred to as the buffer capacity The utility of a buffer is illustrated by comparing the effect of the addition of either strong acid or base to a buffered solution to the effect of adding the same acid or base to an unbuffered solution Consider 100 L ofpure water at 25 C which has a pH of700 Now we add 040 g 0010 mol of NaOH This increases the concentration of OH to 001 M and raises the pH to 12 The addition of 036 g HCl 0010 mol would increase the hydronium ion concentration to 0010 M and reduce the pH to 200 Now consider a buffer solution containing 01 M HClO and 0035 M NaClO The pH of this buffer solution is 700 Ka for HClO is 35 x 10 8 HClO aq H20 1 ClO aq H3O aq This solution will resist changes to its pH when limited amounts of strong acid or base are added to it HClO will react with OH and OCl will react with H3Oto keep the pH change in check HClO aq OH aq a C10 aq H20 1 C10 aq H3O aq a HClO aq H20 1 Adding 0010 mol HCl to 100 L of the buffer results in the following changes HC10aq H20 1 C10 aq H3Oaq 010 0035 10x 10 7 0010 011 0025 0 Lq 0117 x 0025 x x z H30CIO39 0025 xx 3 5 X 104 HC10 011 x Assume x ltlt 0025 2 H30 ClO 39 0025x a 35 x 108 HC10 011 x 154x107 H301 pH flogl54x107 681 HOW did we calculate the initial pH of the buffer 1714 HClO aq H20 1 ClO aq H3O aq 010 0035 0 7x x x 010 7 x 0035 x x We know by now that a HC10 010 x and our assumption that x is negligible relative to 0035 and 010 will hold H30C1039 HC10 rearranging x HC10 Hp 1 K3 0C1 taking the 710g of each side 1715 HClO C10 39 log 0 logK log logK log Ha 0C1 HOCl 01quot pH pKa log m 746 log 746 046 700 HOCl 39 010 The last equation is known as the HendersonHasselbalch Equation which in its most general form is base H K 10 P P a 8 acid Therefore the rst important characteristic of a buffer its pH is given by the Henderson Hasselbalch Equation and is determined by the pK of the weak acid or base and the ratio of the concentrations of the conjugate acidbase pair For a buffer that is prepared from a weak base and its conjugate acid the pK in the HendersonHasselbalch Equation is that of the conjugate acid The other important characteristic of a bu er is its buffer capacity The buffer capacity corresponds to the maximum amount of either acid or base it can neutralize before undergoing a large change in pH The buffer capacity is determined by the concentrations of the conjugate acid and base pair A 1 M HFF buffer has 10 times the buffer capacity of a 01 M HFF buffer In general the pH of a buffer will not change by more than 05 pH units as long as the amount of strong acid or base added is less than half the amount of conjugate base or weak acid present in the buffer The use il pH range of a buffer is pH pKa 1 1 This means that the ratio of acid to base must be between 1 to 10 and 10 to 1 for reasonable buffer actiVity Example What is the pH of a buffer made by mixing 120 mL of 020 M NH3 with 100 mL of 015 MNHACl at 25 C Kb 18 x 10 5 for NH3 17 16 Distribution Curves For any acidconjugate base or basconjugate acid system we can derive a relationship between pH and the fraction of molecules that are ionized and unionized Fixing the pH xes the ratio of acid to base for a given conjugate acidbase pair as shown in the equation below base H Klo p p gacid Since pKa is a constant for a given acid we an easily calculate the ratio of base to acid at any given pH Consider a hypothetical weak acid HA whose Ka is 10 x 10 4 pKa is therefore 400 The following table gives us the values of log baseacid and base acid for several values of pH Knowing the ratios as a mction of pH allows us to plot a distribution curve for the acid pH wastde baseacid pH bassgmd baseacid 000 7400 000010 500 100 10 050 7350 000032 550 150 32 100 7300 00010 600 200 100 150 7250 00032 650 250 320 200 7200 0010 700 300 1000 250 7150 0032 750 350 3200 300 7100 010 800 400 10000 350 7050 032 850 450 32000 400 000 050 900 500 100000 450 050 32 950 550 320000 1717 fraction AcidBase Titrations An acidbase titration is a procedure for determining the amount of acid or base in a sample by determining the volume of base or acid solution of known concentration required to completely react with it In the accompanying gure for instance a sample of acid is contained in the Erlenmeyer ask along with a few drops of the indicator phenolphthalein The buret contains a solution of NaOH whose concentration is known to a high degree of precision and accuracy The buret allows us to determine the volume of the NaOH solution required to completely react with the acid in the ask The indicator turns from colorless to pink at the rst excess of OH The equivalence point of a titration is the point at which the amount of titrant added is the exact amount necessary for a stoichiometric reaction with the analyte Unfortunately we cannot directly observe the equivalence point What we observe is the endpoint of the titration The endpoint is the point of the titration at which the indicator just changes color Judicious selection of an indicator is required so that the endpoint of the titration closely approximates the equivalence pomt 1718 The slrnplesl case 15 a stmhg amdshong baselllmhon Consldex the Illmhon 0f2500 mL of 0100 MHCl Wlth 0100 MNaOH H3039 OH a 2 H10 Inlllally H301 1000 100x10 MandOH 100x10 BM andsopH Now we add 500 mLNaOH The total Volume new 3000 mL 003000 L 1130 OH lt 2 H10 lmh l 1110185 250 x10 3 0 moles added 500 x10 1 nal 1110185 200 x10 3 0 11301 W 667 x 101M pH 110 To mach the endpmnl we must add 2500 mL 0fthe NaOH soluhon At lhls pmnl we have added exactly smug OH to mac completely Wllh the H302 shhce nelthex Na 01 C1 has any Iendmcy tc undeng hydmlysls the soluhon w l be nelmal at the eqlnvalence palm Aftex 3000 mL cfthe NaOH solullon 15 added we have an excess ofOH 500 x 10 mal OI139 833 x 10quot M pH 1192 006000 1 The changes m pH 4 dunng the course ofa vm NmagH mamn are lllushaled 2 D by a mmhon curve a 5 plot 0pr ofa scluhch m M 1 as a f mc m m Vhena tha1em m volume oflllmnl S cm i added g 25 a Blue 73 ammalm 3a 4 m an 15 2 h 46 m u 5 m 15 2s 25 Cm 15 40 46 m Vulums quaoH added mu w mu 1 1a m an 195 2m 22 us we 7m 1 m 1159 a 75 1137 a as 222 ms ms 252 1719 Now let us consider the case of the titration of a weak base by a strong acid Everything we say regarding this titration can be said regarding the titration of a weak acid by a strong base The titration of a weak base by a strong acid is somewhat more complicated than the use of a strong acidstrong base Consider the titration of 250 mL of 0100 M NH3 with 0100 M HCl We must consider four distinct stages of the titration 1 Prior to addition of any acid Initially we have a solution of a weak base We already know how to calculate the pH of such a solution NH3 aq H20 1 OH aq NH aq I 0100 0 0 A 7x x x eq 0100 7 x x x b 18x105 NH3 0100 X Assumexltlt 0100 2 18 x10395 x g13x 10 3 0100 03900 x 100 13 0100 H301 39 77 x10 12 pH 1111 2 After addition of acid has begun but prior to equivalence point Here we have a solution that contains both the weak base and its conjugate acid You should recognize a solution of this composition as a buffer We can use the Henderson Hasselbalch equation to calculate the pH of the solution We will calculate the pH after 1500 mL of HCl have been added 1720 NH3 31 H30 3 1 H20 1 NH g initial moles 000250 0 moles added 150 x 10 3 nal moles 000100 000150 pH pKa log M 926 log 0001 926 018 908 acid 000150 3 At the equivalence point At the equivalence point we have added exactly enough acid to react completely with the base so we have a solution that contains the conjugate acid of a weak base we already know how to deal with this situation NH aq H20 1 NH3 aq H30 aq K 5 39 556 x103910 and NH 00500M NH aq H20 1 NH3 aq H30 aq H 0050 0 20 A l l X x x eq 0050 7 x x x K H30 1 NH3 XX 556 x 1010 quot NH4 00500 x XSH3O 53x106M pH 528 14 12 10 4 A ex the Eqmvalence Pond A ex the eqmvalence p01nlthexe 15 an excess csto and the pH 15 delenmned by the concemmhon oflhe excess H302 Calculate the pH a ex 4000 mL of Hcl have been added AI thls pmnl we have an excess emulsoo L x a volume 0f5500 mL mm 5 1n 000150 0101 006500 1 15 23 x 10quot M Msmyl 120 MW 2U 25 3D 35 vdldme uf H01 added mL Edwslsnss palm AD 0100 mnl pH 154 1721 vdlums H01 sddsd mL 0 1 2 3 4 5 10 15 20 21 22 24 25 2E 27 2E 29 30 35 AD 45 50 000150 mall130 1n PH 1113 1054 1032 1013 BBB BBB 944 and EBB 854 839 788 52B 270 240 222 210 200 170 152 140 130 1722 AaldrBase Indlcalms We bne y menhoned andrb se lndlcamxs m the pxevmlls chapten Aaldrbase lndlcalms are watex soluble cxgamc dyes winch are ellhex weak ansted and 01 bases winch have dlffexmt cclcxs m then mmzed and unmmzed fcnns HIn a9 1410 a In a9 H30 ac h1aq1lt1101 Hln aq OH ac The pH cfthe soluhon detemnnes the xatm cfthe ccncentxaucn cfthe mmzed to unmmzed fcxms and hence the calm cfthe scluucn Genemlly when m 2 lo the color ofthe baslc form w l predomlnale and when HID HTI In 1 z 10 the calm cfthe amdlc form w l pxedcnnnate The cclcx cfthe mdlcalm w l change Iapldly when 01 g s 10 and lb leads to the rule cfthumb that the useful pH mnge ofan mdlcalm 1s pH pK 1 1 1n selechng an mdlcalm for a hlmhon u 1s lrnpo anl to select an mdlcalm whose calm uansmcn mnge falls Within the sleeplynsmg po mn ofthe unaucn curve to avmd hlmhon snot 1n the example below phenolphlhalem 1s a smiable mdlcalm bntbmmcxesol geen us not because It mdexgoes us calm changewell before the equwalence palm Wm mm N am quot30 w quotk pmapmum x calmness In smum Palm s sue MW smmswm 2 nsmlsznzsanasmwsn Mume m m nu ma ml 1723 Solutions of Very weak Acids and Bases In any aqueous solution of a weak acid we have two sources of H3Oz 1 Ionization of the weak acid HA aq H20 1 A aq H3O aq 2 Autoionization of water 2 H20 1 OH aq H3O aq So the total H3O concentration H3OL is given by the sum of the concentrations of H3O from the ionization of the weak acid H3Oa and the autoionization of water H3OW H30L H301a H30w When H3OIa lt 45 x 10 7 M H3OIW is greater than 5 of ngOIL and the autoionization of water cannot be ignored as a source of H3O Similarly in any aqueous solution of a weak base we have two sources of OH 1 Ionization of the weak acid B aq H20 1 BH aq OH aq 2 Autoionization of water 2 H20 1 OH aq H3O aq So the total OH concentration OH L is given by the sum of the concentrations of OH from the ionization of the weak base OH a and the autoionization of water OH W OH I OH la 0H 1w When OH a lt 45 x 10 7 M OH W is greater than 5 of OH L and the autoionization of water cannot be ignored as a source of OH We will proceed by looking at acids but everything we say regarding acids and H3O can also be said regarding bases and OH 1724 In any acidbase equilibrium H3OL is the appropriate hydroniurn ion concentration to use H30 1 A 1 So Ka HA and KW H30 l OH According to our equilibria H3Oa A and H3OW OH This leads to an equation known as a charge balance equation H3OL A OH This equation simply tells us that the total concentration of positive charge must equal the total concentration of negative charge Hence the name charge balance equation Rearranging A H30L 0H So H301t H30t OH 2 H30 112 H30 t0H HA HA H3013 Kw quot HA If the degree of ionization of the weak acid is so small that we cannot ignore the autoionization of water as a source of H301 it stands to reason that A ltlt HA and the equilibrium concentration of HA HA is essentially equal to the initial concentration of HA HA Ka HA H3013 Kw H3013 Ka HA Kw So H30 11 KaHAi 1g This equation should only be used in the case of a very weak acid or a very dilute solution where KW is not negligible compared to the product Ka HA I
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