GEN CHEM & QUAL ANALY
GEN CHEM & QUAL ANALY CHM 2046
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Date Created: 09/18/15
CHAPTER 15 ACIDS AND BASES Table 152 of the text contains a list of important Brznsted acids and bases 21 both why b base c acid d base e acid 1 base g base h base i acid j acid Recall that the conjugate base of a Brznsted acid is the species that remains when one proton has been removed from the acid a nitrite ion N027 b hydrogen sulfate ion also called bisulfate ion H5047 c hydrogen sulfide ion also called bisulfide ion HST d cyanide ion CNT e formate ion HCOOT In general the components of the conjugate acidibase pair are on opposite sides of the reaction arrow The base always has one fewer proton than the acid a The conjugate acidibase pairs are l HCN acid and CNT base and 2 CH3COOT base and CH3COOH acid b l HCO37 acid and C0327 base and 2 HCO37 base and HzCO3 acid c d e V l HzPO47 acid and HPO427 base and 2 NH3 base and NH4Jr acid V l HClO acid and C107 base and 2 CH3NH2 base and CH3N39H3Jr acid V 1 H20 acid and OH7 base and 2 C0327 base and HCO37 acid V The conjugate acid of any base is just the base with a proton added a st t HPO4Z b HZCO3 g H2504 c Hco3 h Hso4 d H3PO4 i Hso3 e H2PO4T a The Lewis structures are M if 39o c c O H and 39o c c o39 b Hir and CszO4 can act only as acids C2HO47 can act as both an acid and a base and C204 can act only as a base The conjugate base of any acid is simply the acid minus one proton a CHzCICOO b 104 c H2P04 d HPO42 e P0437 1 H504 g 5042 h 103 i 5032 i NH3 k Hs 1 sz m oc1 CHAPTER 15 ACIDS AND BASES 417 1515 H l4gtlt1073M 714 KW 7 I39OXIO 7 71x10 12M H l4gtlt10 3 OH l 1516 OH 062M 714 11 KW 1390 X10 16 x 103914 M 0H5 062 1517 a HCl is a strong acid so the concentration of hydrogen ion is also 00010 M What is the concentration of chloride ion We use the definition of pH pH flogH 710g00010 300 b KOH is an ionic compound and completely dissociates into ions We first find the concentration of hydrogen ion K I 714 H W 13X 10514101 QH 076 The pH is then found from its defining equation pH flogH 710g13 X 1014 1389 1518 a BaOHz is ionic and fully ionized in water The concentration of the hydroxide ion is 56 X 1074 M Why What is the concentration of Bah We find the hydrogen ion concentration KW 7 10 gtlt10714 OH 56gtlt10 4 H 18x10 11M The pH is then pH flogH flogl8 X 10711 1074 b Nitric acid is a strong acid so the concentration of hydrogen ion is also 52 X 1074 M The pH is pH flogH 710g52 X 10 4 328 1519 Since pH 7 logH we write H linH a H 10 23942 38x 10 3M c 11quot 10quot 6 11 x 10 7M b 11quot 10 21 62x 10le d 11quot 10 153900 10x 10 15M Strategy Here we are given the pH of a solution and asked to calculate H Because pH is defined as 1520 pH logH we can solve for H by taking the antilog of the pH that is H lOTPH Solution From Equation 154 of the text a pH 710g H 520 logH 7520 418 1521 1522 1523 CHAPTER 15 ACIDS AND BASES To calculate H we need to take the antilog of 7520 7520 H 10 63x 10 6M Check Because the pH is between 5 and 6 we can expect H to be between l X 1075 M and 1 X 1076 M Therefore the answer is reasonable b pH 710g Ht 1600 logH 71600 71600 H 10 10x 10 16M C Strategy We are given the concentration of 0H7 ions and asked to calculate H The relationship between H and 0H7 in water or an aqueous solution is given by the ionproduct of water KW Equation 153 of the text Solution The ion product of water is applicable to all aqueous solutions At 25 C KW 10gtlt 10 14 HOH Rearranging the equation to solve for H we write 7 10x10 14 7 10x10 14 27 x 10396 M OH 37 gtlt10 9 HF Check Since the 0H7 ltl gtlt 1077Mwe expect the H to be greater than l X 1077 M pH H Solution is lt7 gtl0gtlt1077M acid gt7 ltl0gtlt 1077M basic 7 10 X 10 7M neutral a acidic b neutral c basic The pH can be found by using Equation 158 of the text pH 1400 7 pOH 1400 7 940 460 The hydrogen ion concentration can be found as in Example 154 of the text 460 flogH Taking the antilog of both sides H 25 x 10 5 M 1524 1525 1526 1531 1532 CHAPTER 15 ACIDS AND BASES 419 l L 0360 mol X 198 x 10393 mol KOH 1000 mL 1L 550 mL gtlt KOH is a strong base and therefore ionizes completely The OHT concentration equals the KOH concentration because there is a 11 mole ratio between KOH and OHT OH 0360M pOH flogOH7 0444 We can calculate the 0H7 concentration from the pOH pOH 1400 7 pH 1400 71000 400 OrpOH OH 1 l0gtlt1074M Since NaOH is a strong base it ionizes completely The 0H7 concentration equals the initial concentration of NaOH NaOH 10 gtlt1074 molL So we need to prepare 546 mL of 10 X 1074 MNaOH This is a dimensional analysis problem We need to perform the following unit conversions molL gt mol NaOH gt grams NaOH 546 mL 0546L 74 gNaOH 546 mL X 1390 X10 01 NaOH X 403900 g NaOH 22 gtlt 10393 g NaOH 1000 mL soln 1 mol NaOH 184 g HCl gtlt 316 1HHC11 Molarity of the HCl solution is 393 g 0762 M 662 X 107 L pH flog0762 0118 A strong acid such as HCl will be completely ionized choice b A weak acid will only ionize to a lesser extent compared to a strong acid choice c A very weak acid will remain almost exclusively as the acid molecule in solution Choice d is the best choice 1 The two steps in the ionization of a weak diprotic acid are H2Aaq H200 HaoYaq HAKaq HAYaq H200 HaoYaq NM The diagram that represents a weak diprotic acid is c In this diagram we only see the first step of the ionization because HA7 is a much weaker acid than HzA 420 1533 1534 1535 1536 1537 1538 1543 CHAPTER 15 ACIDS AND BASES E V Both b and d are chemically implausible situations Because HA7 is a much weaker acid than HzA you would not see a higher concentration of A 7 compared to HA7 a strong acid weak acid g strong acid b weak acid c strong acid first stage of ionization e weak acid 1 weak acid h weak acid i weak acid A O V a strong base b weak base c weak base d weak base e strong base The maximum possible concentration ofhydrogen ion in a 010 M solution ofHA is 010 M This is the case if HA is a strong acid lfHA is a weak acid the hydrogen ion concentration is less than 010 M The pH corresponding to 010M H is 100 Why three digits For a smaller H the pH is larger than 100 why a false the pH is greater than 100 b false they are equal c true d false 21 false they are equal c true b true find the value of logl00 on your calculator d false if the acid is strong HA 000M The direction should favor formation of F7aq and HzOl Hydroxide ion is a stronger base than fluoride ion and hydro uoric acid is a stronger acid than water C17 is the conjugate base of the strong acid HCl It is a negligibly weak base and has no affinity for protons Therefore the reaction will not proceed from left to right to any measurable extent Another way to think about this problem is to consider the possible products of the reaction CH3COOHaq Cliaq gt HClaq CH3COOiaq The favored reaction is the one that proceeds from right to left HCl is a strong acid and will ionize completely donating all its protons to the base CH3COOT We set up a table for the dissociation C5H5COOHaq Haq csHscooxaq Initial M 010 000 000 Change 1V1 7x x x Equilibrium 1V1 010 7 x x x K HC6H5COO a C6H5COOH 2 65 gtlt1075 x o1o 7 x x2 65 X1075x765 X 10 6 o CHAPTER 15 ACIDS AND BASES 421 Solving the quadratic equation x 25gtlt1073M Hl pH ilog25 gtlt 10 3 260 This problem could be solved more easily if we could assume that 010 7 x m 010 If the assumption is mathematically valid then it would not be necessary to solve a quadratic equation as we did above Resolve the problem above making the assumption Was the assumption valid What is our criterion for deciding 1544 Strategy Recall that a weak acid only partially ionizes in water We are given the initial quantity of a weak acid CH3COOH and asked to calculate the concentrations of H CH3COOi and CH3COOH at equilibrium First we need to calculate the initial concentration of CH3 COOH In determining the H4r concentration we ignore the ionization of H20 as a source of H so the major source of Hir ions is the acid We follow the procedure outlined in Section 155 of the text Solution Step 1 Calculate the concentration of acetic acid before ionization 1 mol acetic acid 4 00560 g acet1c ac1d X 933 X 10 mol acet1c ac1d 6005 g acet1c acid 933 gtlt1074 mol 00500 L soln 00187 M acetic acid Step 2 We ignore water39s contribution to H We consider CH3COOH as the only source of H4r ions Step 3 Letting x be the equilibrium concentration of H4r and CH3COOi ions in molL we summarize CH3COOHaq Haq CH3COO aq Initial 1V1 00187 0 Change 1V1 7x x x Equilibrium 1V1 00187 7 x x x Step 4 Write the ionization constant expression in terms of the equilibrium concentrations Knowing the value of the equilibrium constant K3 solve for x You can look up the Ka value in Table 153 of the text K HCH3COO a CH3COOH 1 8 x 10 5 39 00187 7 x At this point we can make an assumption thatx is very small compared to 00187 Hence 001877x m 00187 18x10 5 w 39 00187 x 58x 10 4M 11 CH3COO CH3COOH 001877 58 X 10 4M 00181 M 422 1545 1546 1547 CHAPTER 15 ACIDS AND BASES Check Testing the validity of the assumption 58 gtlt10 4 00187 gtlt 100 31 lt 5 The assumption is valid First we find the hydrogen ion concentration 7620 H io PH 10 63gtlt1077M If the concentration of H is 63 X 1077 M that means that 63 X 1077 M of the weak acid HA ionized because of the 11 mole ratio between HA and Hf Setting up a table HAW HM AM Initial 111 0010 0 0 Change M 63 X 10 7 63 X 10 7 63 X 10 7 Equilibrium M 6 0010 63 X 10 7 63 X 10 7 Substituting into the acid ionization constant expression HA 7 63gtlt10 763 x1077 HA 7 0010 Ka 40x 103911 We have omitted the contribution to H due to water A pH of 326 corresponds to a H of 55 X 1074 M Let the original concentration of formic acid be I If the concentration of H is 55 X 1074 111 that means that 55 X 1074 M of HCOOH ionized because of the 11 mole ratio between HCOOH and Hi HCOOHaq Haq Hcoo aq lnitial 111 0 0 Change M 55 gtlt1074 55 gtlt1074 55 gtlt1074 Equilibrium M 1 55 X 10 4 55 gtlt1074 55 gtlt1074 Substitute Ka and the equilibrium concentrations into the ionization constant expression to solve for I H HCOO a HCOOH l7gtlt10 4 5395X10742 x 55 x1074 I HCOOH 23 x 10 3M a Set up a table showing initial and equilibrium concentrations C5H5COOHaq Haq c6H5c00 aq Initial M 0 20 000 000 Change 111 7x x x Equilibrium 111 020 7 x x x CHAPTER 15 ACIDS AND BASES 423 Using the value of Ka from Table 153 of the text Htiic6H5coo Ka C6H5COOH 65 gtlt1075 x o2o 7 x We assume thatx is small so 020 7 x m 020 2 65 gtlt10 5 x 020 x 36gtlt10 3M Ht csHscoo 73 Percent ionization Wx 100 18 020 M b Set up a table as above C5H5COOHaq Haq c6H5c00 aq Initial M 000020 000000 000000 Change 111 7x x x Equilibrium 111 000020 7 x x x Using the value of Ka from Table 153 of the text K Ht C6H5COO a C6H5COOH 2 65 gtlt10 5 x 000020 7 x In this case we cannot apply the approximation that 000020 7 x m 000020 see the discussion in Example 158 of the text We obtain the quadratic equation x2 65 X 1075x7 13 x1078 o The positive root of the equation is x 86 X 1075 M Is this less than 5 of the original concentration 000020 M That is is the acid more than 5 ionized The percent ionization is then 86 X 10 5 M Percent ionization gtlt100 43 000020 M Note that the extent to which a weak acid ionizes depends on the initial concentration of the acid The more dilute the solution the greater the percent ionization see Figure 154 of the text 1548 Percent ionization is defined as ionized acid concentration at equilibrium 0 percent ionization X 100 initial concentration of acid 424 CHAPTER 15 ACIDS AND BASES For a monoprotic acid HA the concentration of acid that undergoes ionization is equal to the concentration of H4r ions or the concentration of A7 ions at equilibrium Thus we can write H percent ionization X 100 HAlo a First recognize that hydro uoric acid is a weak acid It is not one of the six strong acids so it must be a wea aci Step 7 Express the equilibrium concentrations of all species in terms of initial concentrations and a single unknown x that represents the change in concentration Let 7x be the depletion in concentration in olL of HF From the stoichiom etry of the reaction it follows that the increase in concentration for both Hir and F7 must be x Complete a table that lists the initial concentrations the change in concentrations and the equilibrium concentrations HFaq HM Flag lnitial 1V1 060 0 0 Change ItI 7x x 76c Equilibrium 1V1 060 7 x x x Step 2 Write the ionization constant expression in terms of the equilibrium concentrations Knowing the value of the equilibrium constant K3 solve for x i Ka H F 1 HF You can look up the Ka value for hydro uoric acid in Table 153 of your text 7 1 x 10 4 39 060 7 x At this point we can make an assumption thatx is very small compared to 060 Hence 060 7 x m 060 Oftentimes assumptions such as these are valid if K is very small A very small value of K means that a very small amount of reactants go to products Hence x is small If we did not make this assumption we would have to solve a quadratic equation 71 X 10 4 w 06 Solving for x x 0021M Ht Step 3 Having solved for the H calculate the percent ionization H percent ionization X 100 HFlo Mx100 35 060 M 1549 1550 CHAPTER 15 ACIDS AND BASES 425 b 7 c are worked in a similar manner to part a However as the initial concentration of HF becomes smaller the assumption that x is very small compared to this concentration Will no longer be valid You must solve a quadratic equation Hum x2 HF 00046 7 x b Ka 71x10 4 x2 71 gtlt 1074x7 33 x1076 0 x 15x10 3M 73 Percent ionization MX 100 33 00046 M 7 2 0 K2 7 H F l x 7lgtlt10 4 HF 000028 7 x x27lgtlt1074x720gtlt1077 0 x 22gtlt1074M 22gtlt10 4M gtlt 100 79 000028 M Percent ionization As the solution becomes more dilute the percent ionization increases Given 14 ionization the concentrations must be H A7 014 gtlt 0040M 00056M HA 00407 00056M 0034M The value of Ka can be found by substitution 000562 0034 K m a 92 X 10 4 HA The equilibrium is C9H804aq HaqC9H7O47aq Initial 1V1 020 0 0 Change 1V1 7x x x Equilibrium 1V1 020 7 x x x a K H llC9H7O4l C9H804l 2 30 gtlt1074 x 020 7 x 426 CHAPTER 15 ACIDS AND BASES Assuming 0207x m 020 x H 77gtlt10 3M x 77gtlt10 3M Percent ionization X 100 gtlt100 39 020 020 M b At pH 100 the concentration of hydrogen ion is 010 M H 107PH The extra hydrogen ions will tend to suppress the ionization of the weak acid LeChatelier39s principle Section 145 of the text The position of equilibrium is shifted in the direction of the unionized acid Let39s set up a table of concentrations with the initial concentration of H4r equal to 010M chsomq 7 HM C9H704 aq Initial 020 010 0 Change ItI 7x x x Equilibrium It1 020 7 x 010 x x K a C9H804l 3 Ogtlt104 x010 x 39 020 7 x Assuming 020 7 x m 020 and 010 x m 010 x 60x 1074M 74 x gtlt100 MXIOO 030 020 020M Percent ionization The high acidity of the gastric juices appears to enhance the rate of absorption of unionized aspirin molecules through the stomach lining In some cases this can irritate these tissues and cause bleeding 1553 a We construct the usual table NH3aq H200 7 NH4aq OHM Initial M 010 000 000 Change 7x x x Equilibrium It1 010 7 x x x Kb NHJHOH NH31 2 18 gtlt10 5 x 010 7 x Assuming 010 7 x m 010 we have 2 18 gtlt10 5 x 010 1554 CHAPTER 15 ACIDS AND BASES 427 x 13x10 3M OH pOH ilogl3gtlt 10 3 289 pH 1400489 1111 By following the identical procedure we can show b pH 896 Strategy Weak bases only partially ionize in water Baq HzOl BHaqOH aq Note that the concentration of the weak base given refers to the initial concentration before ionization has started The pH of the solution on the other hand refers to the situation at equilibrium To calculate Kb we need to know the concentrations of all three species B BH and 0H7 at equilibrium We ignore the ionization of water as a source of 0H7 ions Solution We proceed as follows Step 1 The major species in solution are B OHS and the conjugate acid BH Step 2 First we need to calculate the hydroxide ion concentration from the pH value Calculate the pOH from the pH Then calculate the OH7 concentration from the pOH pOH 14007pH 14007 1066 334 pOH ilogOH7 ipOH logOH7 Taking the antilog of both sides of the equation 10 pOH OH 73 34 OH 10 46gtlt1074M Step 3 If the concentration of OH7 is 46 X 1074 M at equilibrium that must mean that 46 X 1074 M of the base ionized We summarize the changes Baq 4 H200 BHaq 4 0H aq Initial 1V1 030 0 0 Change M 746 X 10 4 46 X 10 4 46 X 10 4 Equilibrium M 030 7 46 X 10 4 46 X 10 4 46 X 10 4 Step 4 Substitute the equilibrium concentrations into the ionization constant expression to solve for Kb Kb BHOH B 74 2 Kb m 71gtlt 10397 030 428 1555 1556 1561 CHAPTER 15 ACIDS AND BASES A pH of 1122 corresponds to a H of 603 gtlt 10 12 Mand a OH of 166 X 10 3 M Setting up a table N39H3aq H200 N39H4aq 0H aq Initial M 1 000 0 00 Change M 7166 X 10 3 166 X 10 3 166 X 10 3 Equilibrium M 17 166 X 103 166 gtlt1073 166 gtlt1073 Kb w NH3l 18 X105 7 166 X 10 3166 x1073 17166gtlt10 3 73 Assuming 166 X 10 is small relative to x then x 015M NH3 The reaction is NH3aq HzOl NH4aq 0H aq Initial A4 0080 0 0 Change A4 7x x x Equilibrium A4 0080 7 x x x At equilibrium we have K NHMOH a NH3l 2 18x10 5 x s x 0080 7 x 0080 x l2gtlt1073M 73 Percent NH3 present as NH gtlt 100 15 If Ka1 gtgt Ka2 we can assume that the equilibrium concentration of hydrogen ion results only from the first stage of ionization In the second stage this always leads to an expression of the type c yy K c 7 y 32 where 0 represents the equilibrium hydrogen ion concentration found in the first stage If 0 gtgt K32 we can assume Ci y m c and consequently y K32 Is this conclusion also true for the second stage ionization of a triprotic acid like H3PO4 1562 CHAPTER 15 ACIDS AND BASES 429 The pH of a 0040MHCl solution strong acid is pH flog0040 140 Follow the procedure for calculating the pH of a diprotic acid to calculate the pH of the sulfuric acid solution Strategy Determining the pH of a diprotic acid in aqueous solution is more involved than for a monoprotic acid The first stage of ionization for HzSO4 goes to completion We follow the procedure for determining the pH of a strong acid for this stage The conjugate base produced in the first ionization H5047 is a weak acid We follow the procedure for determining the pH of a weak acid for this stage Solution We proceed according to the following steps Step 1 H2504 is a strong acid The first ionization stage goes to completion The ionization of HzSO4 is HzSO4aq HaqHSO4 aq The concentrations of all the species H2504 H and HSO47 before and after ionization can be represented as follows HzSO4aq HM Hso aq 0 Initial 1V1 0040 0 Change M 70040 0040 0040 Final M 0 0040 0040 Step 2 Now consider the second stage of ionization H5047 is a weak acid Set up a table showing the concentrations for the second ionization stage Letx be the change in concentration Note that the initial concentration of H4r is 0040 M from the first ionization Hsomq HYaq so xaq Initial M 0040 0040 0 Change ItI 7x x x Equilibrium 1V1 0040 7 x 0040 x x Write the ionization constant expression for K3 Then solve for x You can find the Ka value in Table 155 of the text HSOi Hson a 13 X104 0040 xx 0040 7 x Since Ka is quite large we cannot make the assumptions that 0040 7 x m 0040 and 0040 x m 0040 Therefore we must solve a quadratic equation x2 0053x7 52 x1074 0 7 70053 t o0532 7 41752 x1074 21 70053 t 0070 2 x x 85x10 3M or x 70062M 430 1563 1564 CHAPTER 15 ACIDS AND BASES The second solution is physically impossible because you cannot have a negative concentration The first solution is the correct answer Step 3 Having solved for x we can calculate the H4r concentration at equilibrium We can then calculate the pH from the Hir concentration H 0040Mx 004085 gtlt 1073A4 0049M pH flog0049 131 Without doing any calculations could you have known that the pH of the sulfuric acid would be lower more acidic than that of the hydrochloric acid There is no H2504 in the solution because HSO47 has no tendency to accept a proton to produce HzSO4 W hy We are only concerned with the ionization Hsomaq HYaq so aq Initial M 020 000 000 Change A4 7x x x Equilibrium A4 020 7 x x x K a Hso 13 x 10 2 m 020 7 x Solving the quadratic equation x H 5042 0045M Hsof 020 7 0045M 016 M For the first stage of ionization Hzc03aq HYaq HCO aq Initial M 0025 0 00 0 00 Change A4 7x x x Equilibrium A4 0025 7 x x x K HHCO a1 7 H2C03l Z 2 42 X 10 7 7 x N x 0025 7 x 0025 x 10x10 4M 1567 1568 1569 1570 CHAPTER 15 ACIDS AND BASES 431 For the second ionization HCOgiaq Haq CO3ziaq Initial M 10 X 10 4 10 X 10 4 000 Change 1V1 7x x x Equilibrium M 10 X 104 7 x 10 X 104 x x Huico 32 Hcog 4 8 X1041 10 X1074 xx 10 x1074x l0gtltlO 47x 10gtlt10 4 x 48 X 10 11M Since HCO37 is a very weak acid there is little ionization at this stage Therefore we have H HCO3 10 x 10 4 M and C032 x 48 x 10 1114 The strength of the HiX bond is the dominant factor in determining the strengths of binary acids As with the hydrogen halides see Section 159 of the text the HiX bond strength decreases going down the column in Group 6A The compound with the weakest HiX bond will be the strongest binary acid H2Se gt H2S gt H20 All the listed pairs are oxoacids that contain different central atoms whose elements are in the same group of the periodic table and have the same oxidation number In this situation the acid with the most electronegative central atom will be the strongest a H2504 gt steO4 b H3PO4 gt H3ASO4 The CHClzCOOH is a stronger acid than CHZClCOOH Having two electronegative chlorine atoms compared to one will draw more electron density toward itself making the 07H bond more polar The hydrogen atom in CHClzCOOH is more easily ionized compared to the hydrogen atom in CHZClCOOH The conjugate bases are C5H507 from phenol and CH3OT from methanol The C5H5OT is stabilized by resonance 00150 The CH3OT ion has no such resonance stabilization A more stable conjugate base means an increase in the strength of the acid 432 1575 1576 1577 1578 CHAPTER 15 ACIDS AND BASES a The Kit cation does not hydrolyze The Bri anion is the conjugate base of the strong acid HBr Therefore Bri will not hydrolyze either and the solution is neutral pH m 7 b Al3 is a small metal cation with a high charge which hydrolyzes to produce H4r ions The N037 anion does not hydrolyze It is the conjugate base of the strong acid HNO3 The solution will be acidic pH lt 7 c The Ba2 cation does not hydrolyze The C17 anion is the conjugate base of the strong acid HCl Therefore Cli will not hydrolyze either and the solution is neutral pH m 7 d Bi3 is a small metal cation with a high charge which hydrolyzes to produce H1r ions The N037 anion does not hydrolyze It is the conjugate base of the strong acid HNO3 The solution will be acidic pH lt 7 Strategy In deciding whether a salt will undergo hydrolysis ask yourself the following questions Is the cation a highly charged metal ion or an ammonium ion Is the anion the conjugate base of a weak acid If yes to either question then hydrolysis will occur In cases where both the cation and the anion react with water the pH of the solution will depend on the relative magnitudes of Ka for the cation and Kb for the anion see Table 157 of the text Solution We first break up the salt into its cation and anion components and then examine the possible reaction of each ion with water a The NaJr cation does not hydrolyze The Bri anion is the conjugate base of the strong acid HBr Therefore Bri will not hydrolyze either and the solution is neutral b The Kit cation does not hydrolyze The 0327 anion is the conjugate base of the weak acid H5037 and will hydrolyze to give H8037 and OHT The solution will be basic Both the N39H41r and N027 ions will hydrolyze N39H4Jr is the conjugate acid of the weak base NH3 and N027 is the conjugate base of the weak acid HNOz From Tables 153 and 154 of the text we see that the Ka ofN39H4Jr 56 X 10710 is greater than the Kb ofNOzi 22 X 10711 be acidic c V Therefore the solution will d Cr3 is a small metal cation with a high charge which hydrolyzes to produce H4r ions The N037 anion does not hydrolyze It is the conjugate base of the strong acid HNO3 The solution will be acidic There are two possibilities i MX is the salt of a strong acid and a strong base so that neither the cation nor the anion react with water to alter the pH and ii MX is the salt of a weak acid and a weak base with Ka for the acid equal to Kb for the base The hydrolysis of one would be exactly offset by the hydrolysis of the other There is an inverse relationship between acid strength and conjugate base strength As acid strength decreases the proton accepting power of the conjugate base increases In general the weaker the acid the stronger the conjugate base All three of the potassium salts ionize completely to form the conjugate base of the respective acid The greater the pH the stronger the conjugate base and therefore the weaker the acid The order of increasing acid strength is HZ lt HY lt HX 1579 1580 CHAPTER 15 ACIDS AND BASES 433 The salt sodium acetate completely dissociates upon dissolution producing 036 M Na and 036 M CH3COOT ions The CH3COOT ions will undergo hydrolysis because they are a weak base CH3COOiaq HzOl CH3COOHaq OH7aq Initial M 036 000 000 Change A4 7x x x Equilibrium A4 036 7 x x x Kb CH3COOHOH CH3COO 56 gtlt10 10 x 036 7 x Assuming 036 7 x m 036 then x OHS l4gtlt1075 pOH 7logl4gtlt1075 485 pH 14007485 915 The salt ammonium chloride completely ionizes upon dissolution producing 042 M N39H4 and 042M C17 ions N39H4Jr will undergo hydrolysis because it is a weak acid N39HIr is the conjugate acid of the weak base NH3 Step 1 Express the equilibrium concentrations of all species in terms of initial concentrations and a single unknown x that represents the change in concentration Let 7x be the depletion in concentration m olL of NH4 From the stoichiometry of the reaction it follows that the increase in concentration for both H30 and NH3 must be x Complete a table that lists the initial concentrations the change in concentrations and the equilibrium concentrations NH4aqHzOl 7 NH3aq H30aq Initial M 042 000 000 Change A4 7x x x Equilibrium A4 042 7 x x x Step 2 You can calculate the Ka value for N39H4Jr from the Kb value of NH3 The relationship is Ka gtlt Kb KW KW 7 10 gtlt10714 7 710 KT 18x10 5 75M K Step 3 Write the ionization constant expression in terms of the equilibrium concentrations Knowing the value of the equilibrium constant K3 solve for x w NHX K a 434 1581 1582 1585 1586 1587 1588 1591 CHAPTER 15 ACIDS AND BASES x2 x2 56gtlt10 10 042 7 x 042 x H15x1075M pH 710g15 X 10 5 482 Since NH4Cl is the salt of a weak base aqueous ammonia and a strong acid HCl we expect the solution to be slightly acidic which is confirmed by the calculation Hcog HHCOJ Ka 48gtlt10711 KW 7 10x10 14 7 78 Ka 42Xl077 7 24gtltlO HCO37 H20 HzCO3 OH7 Kb HCO37 has a greater tendency to hydrolyze than to ionize Kb gt K3 The solution will be basic pH gt 7 The acid and base reactions are acid HPO427aq Haq P04370111 base HPO42 aqHZOl HzPO47aqOH7aq Ka for HPO427 is 48 X 10713 Note that HPO427 is the conjugate base of H2PO4T so Kb is 16 X 1077 Comparing the two K39s we conclude that the monohydrogen phosphate ion is a much stronger proton acceptor base than a proton donor acid The solution will be basic Metal ions with high oxidation numbers are unstable Consequently these metals tend to form covalent bonds rather than ionic bonds with oxygen Covalent metal oxides are acidic while ionic metal oxides are basic The latter oxides contain the 027 ion which reacts with water as follows 02 Hzo 20H The most basic oxides occur with metal ions having the lowest positive charges or lowest oxidation um ers 21 A1203 lt BaO ltKzO b CrO3 lt CrzO3 lt CrO a 2HClaq ZnOHzs gt ZnClzaq 2HzOl b 20H aq ZnOHzs a ZnOH427aq AlOH3 is an amphoteric hydroxide The reaction is AlOH3s OH7aq gt AlOH47aq This is a Lewis acidbase reaction Can you identify the acid and base a Lewis acid see the reaction with water shown in Section 1512 of the text b Lewis base water combines with H to form H3O 1592 1593 1594 1595 1596 1597 1598 CHAPTER 15 ACIDS AND BASES 435 c Lewis base d Lewis acid 502 reacts with water to form H2503 Compare to C02 above Actually 502 can also act as a Lewis base under some circumstances e Lewis base see the reaction with H4r to form ammonium ion 1 Lewis base see the reaction with H4r to form water Lewis acid does H4r have any electron pairs to donate A a V h Lewis acid compare to the example of NH3 reacting with BF3 AlCl3 is a Lewis acid with an incomplete octet of electrons and C17 is the Lewis base donating a pair of electrons c1 c1 c1 T gt Cl Al Cl c1 c1 21 Both molecules have the same acceptor atom boron and both have exactly the same structure trigonal planar Fluorine is more electronegative than chlorine so we would predict based on electronegativity arguments that boron trifluoride would have a greater affinity for unshared electron pairs than boron trichloride b V Since it has the larger positive charge ironlH should be a stronger Lewis acid than ironH By definition Brznsted acids are proton donors therefore such compounds must contain at least one hydrogen atom In Problem 1591 Lewis acids that do not contain hydrogen and therefore are not Brznsted acids are C02 502 and BCl3 Can you name others The ionization of any acid is an endothermic process The higher the temperature the greater the Ka value Formic acid will be a stronger acid at 40 C than at 25 C We first find the number of moles of C02 produced in the reaction 1 mol NaHCO3 X 1 mol COZ 0350 g NaHCO3 gtlt 8401 g NaHCO3 1 mol NaHCO3 4l7gtlt10 3molCOz 0106 L V 7 ncoZRT 7 417gtlt10 3mol00821LatmKmol370273K CO2 P 100 atm Choice c because 070 MKOH has a higher pH than 060 MNaOH Adding an equal volume of 060 M NaOH lowers the 0H7 to 065 M hence lowering the pH If we assume that the unknown monoprotic acid is a strong acid that is 100 ionized then the H concentration will be 00642 M pH flog 00642 119 Since the actual pH of the solution is higher the acid must be a weak acid 436 CHAPTER 15 ACIDS AND BASES 1599 a For the forward reaction N39H4Jr and NH3 are the conjugate acid and base pair respectively For the reverse reaction NH3 and NH are the conjugate acid and base pair respectively b Hir corresponds to N39H439 OHT corresponds to Nsz For the neutral solution N39H4 NHf 15100 The reaction of a weak acid with a strong base is driven to completion by the formation of water Irrespective of whether the strong base is reacting with a strong monoprotic acid or a weak monoprotic acid the same number of moles of acid is required to react with a constant number of moles of base Therefore the volume of base required to react with the same concentration of acid solutions either both weak both strong or one strong and one weak will be the same 15101 Ka w HA HA m 01 M A m 01 M Therefore K Ka Hl W OH OH KW Ka 15102 High oxidation state leads to covalent compounds and low oxidation state leads to ionic compounds Therefore CrO is ionic and basic and CrO3 is covalent and acidic 15103 HCOOH HCOOT HJr Ka l7gtlt 1074 HOH HZO K VNL1410gtlt1014 Kw 10 X 10 HCOOH OH7 HCOOT H20 K K314 17 X 10 410 gtlt 1014 17 x 1010 15104 We can write two equilibria that add up to the equilibrium in the problem HCH3COO 5 CHCOOHa Ha CHCOO a K l8gtltlO 3 q q 3 q a CH3COOH v 1 1 3 Haq NOz aq HNOzaq K 22 X10 3 KaHNOZ 45 X 10 4 39 7 w a 7 7 Hl llNOz CH3COOHaq NOaq CH3COOTaq HNOzaq K W Ka X K CH3COOHNO CHAPTER 15 ACIDS AND BASES 437 The equilibrium constant for this sum is the product of the equilibrium constants of the component reactions K Kagtlt K 18 x107522gtlt103 40x 10 2 15105 21 H H20 5 OH H2 base1 acidz basez acid1 b H7 is the reducing agent and H20 is the oxidizing agent 15106 In this specific case the Ka of ammonium ion is the same as the Kb of acetate ion KaN39H4 56 X 10710 KbCH3COOi 56 X 10710 The two are of exactly to two significant figures equal strength The solution will have pH 700 What would the pH be if the concentration were 01 M in ammonium acetate 04M 15107 Kb 891 X 10 6 K Ka W 11x10 9 Kb pH 740 7740 H 10 398gtlt1078 K 7 Hconjugate base a acid Therefore conjugate base 7 Ka 7 llgtlt 1079 acid H 7 398gtlt10 8 0028 15108 The fact that uorine attracts electrons in a molecule more strongly than hydrogen should cause NF3 to be a poor electron pair donor and a poor base NH3 is the stronger base 15109 Because the PiH bond is weaker there is a greater tendency for PH4Jr to ionize Therefore PH3 is a weaker base than NH3 15110 The autoionization for deuteriumsubstituted water is D20 D ODT DOD l35gtlt10715 1 a The definition opr is pD flogD flog l35gtlt10 15 743 b To be acidic the pD must be lt 743 438 15111 15112 15113 15114 15115 CHAPTER 15 ACIDS AND BASES c Taking flog of both sides of equation 1 above flogD 7logOD7 710g135 X 10 pDpOD 1487 HF c BF3 d NH3 20 HNOz b e H2503 1 HCO and cogz The reactions for f are HCO37aq Haq gt COzg HzOl co aq 2Haq c02g H200 First we must calculate the molarity of the trifluoromethane sulfonic acid Molar mass 1501 gmol Molarity 00164 M Since trifluoromethane sulfonic acid is a strong acid and is 100 ionized the H is 00165 M pH flog00164 179 a The Lewis structure of H30 is H EIS H H Note that this structure is very similar to the Lewis structure of NH3 The geometry is trigonal pyramidal b H4O2 does not exist because the positively charged H30 has no affinity to accept the positive H4r ion If H4O2 existed it would have a tetrahedral geometry HF H F 1 The reactions are F7 HF Hin 2 Note that for equation 2 the equilibrium constant is relatively large with a value of 52 This means that the equilibrium lies to the right Applying Le Chatelier s principle as HF ionizes in the first step the F7 that is produced is partially removed in the second step More HF must ionize to compensate for the removal of the F at the same time producing more H The equations are Clzg HzOl HClaq HClOaq HClaq AgNO3aq AgCls HN03aq In the presence of OH7 ions the first equation is shifted to the right CHAPTER 15 ACIDS AND BASES 439 Ht from HC1 OH H20 Therefore the concentration of HClO increases The bleaching action is due to C107 ions 15116 21 We must consider both the complete ionization of the strong acid and the partial ionization of water HA H A H20 H OH From the above two equations the H in solution is Hi A OH 1 We can also write HOH Kw K 0H7 W Hll Substituting into Equation 1 K W H1A H 1 Ht2 mm Kw H127 A H7Kw 0 Solving a quadratic equation F 7 A39 i A392 4KW f b For the strong acid HCl with a concentration of 10 X 1077 M the C17 Will also be 10 X 1077 M HT cr hcnz 4KW 1x10 7 i 1 X 10 72 41x10 14 2 2 H 16 X 1077M or 760 X 1078A4 which is impossible pH flogl6 X 10 7 680 15117 We examine the hydrolysis of the cation and anion separately NH4CNaq NH4aq CN W Cation NH aq H200 NH3aq H3Oaq lnitial A4 200 0 0 Change A4 7x x x Equilibrium A4 200 7 x x x 440 15118 CHAPTER 15 ACIDS AND BASES K 7 NH3llH3Ol a NHX 56gtlt10 10 L m i 200 7 x 200 x 335 x10 5M 1130 Anion CN7aq HzOl HCNaq OH7aq Initial A4 200 0 0 Change A4 7x x x Equilibrium A4 200 7 x x x Kb HCNOH CN 2 2 20gtlt10 5 7 y y 2007y 200 y 632 x10 3M OH CNi is stronger as a base thanN39H4Jr is as an acid Some 0H7 produced from the hydrolysis of CN7 will be neutralized by H30 produced from the hydrolysis of N39H4 H3Oaq OH aq a 2HzOl lnitialA4 335 x 10 5 632 x 10 3 Change M 7335 x10 5 7335 x10 5 Final A4 0 629 x 10 3 OH 629 x10 3M pOH 220 pH 1180 The solution for the first step is standard H3PO4aq HYaq HzPqu Initial 0100 0000 0 000 Change A4 7x x x Equil A4 0100 7 x x x K HH2P011 al H3PO4l 2 75 gtlt1073 x 0100 7 x CHAPTER 15 ACIDS AND BASES 441 In this case we probably cannot say that 0100 7 x m 0100 due to the magnitude of K3 We obtain the quadratic equation x2 75 X 10 3x7 75 x1074 0 The positive root is x 00239M We have H HzPO4 00239M H3PO4 01007 00239M 0076M For the second ionization H2P04 aq HYaq HPof mq Initial 00239 00239 0000 Change 1V1 7y y y Equil 1V1 00239 7 y 00239 y y H HPOZ Kai 74 1 H2P04l 62X 108 00239 yy 00239y 00239 7 y 00239 y 62 X 1078M Thug Hll HzPO4 00239 M HPOZf y 62 X 10 8M We set up the problem for the third ionization in the same manner Hpo aq HYaq P043010 Initial M 62 X 10 8 00239 0 Change 1V1 72 2 2 Equil M 62 gtlt 10 s 7 2 00239 2 2 37 Ka H P gt4 1 3 HPO4 48 X 1013 00239 7 zz 0239z 62 X 10 s 7 z 62 x1078 z 12x10 18M The equilibrium concentrations are H HzPof 00239114 113104 0076M HPO42 62 x 10 8 M P0431 12 x 10 18M 442 CHAPTER 15 ACIDS AND BASES 15119 21 We carry an additional significant figure throughout this calculation to minimize rounding errors Number ofmoles NaOH Mgtlt vol L 00568Mgtlt 00138 L 7838 X 1074 mol If the acid were all dimer then mol NaOH 7 7838 X10 4mol 2 2 molofdimer 39l9gtlt10 4mol If the acetic acid were all dimer the pressure that would be exerted would be P 002896 atm nRT 7 39l9gtlt10 4mol0082lLatmKmol324 K V 0360 L However the actual pressure is 00342 atm If on mol of dimer dissociates to monomers then 2oc monomer forms CH3COOH2 2CH3COOH l 7 oc 2oc The total moles of acetic acid is moles dimer monomer l 7 on 20c 1 on Using partial pressures Pobserved P0 0 00342 atm 002896 atml on a 0181 b The equilibrium constant is 2 2oc 2 2 P observed 2 K PCH3COOH 1 0 40 Pobserved 463 X 10 3 P 1 7 on 1 7 a2 CH3COOHZ Pabserved 15120 0100MNazCO3 7 0200MNaJr 0100M C0327 First stage co3z aqH20l Hco3 aq OH aq lnitial 1V1 0100 0 0 Change ItI 7x x x Equilibrium 1V1 0100 7 x x x K 14 K1 W 10X10 2lgtlt10 4 K2 48gtlt10 11 7 HCO OH 7 30 CHAPTER 15 ACIDS AND BASES 443 x2 21x10 4 x 4 01007x 0100 x 46gtlt1073M Hcog OH Second stage HCO37aqHzOl H2C03aq OH7aq Initial M 46 X 10 3 0 46 X 10 3 Change 1V1 7y y y Equilibrium M 46 X 10 7 y y 46 gtlt 10 y K 4 W 2 7 HCO3 24 X 108 y46 x1073 y y46 x1073 46x10 37 y 46x10 3 y 24gtlt1078M At equilibrium Na 0200M HC03 46 X 10 3M7 24 X 10 8M 4 46x 10 3 M 112003 24 x 10 8 M 011 46 x1073M24 X 108 M 4 46 x 10 3 M 10gtlt10 14 3 22 gtlt 103912 M 46gtlt10 11 15121 col kP 228 x1073m01Latm320atm 730gtlt1073M COzaq 4 H200 Haq HCO aq 730x10737xM xM xM HHCO a c021 X2 x2 42x10 7 3 2 3 730gtlt10 7 x 730x10 x 55 x1075M H pH 426 444 15122 15123 15124 CHAPTER 15 ACIDS AND BASES When NaCN is treated with HCl the following reaction occurs NaCN HCl NaCl HCN HCN is a very weak acid and only partially ionizes in solution HCNaq Haq CN7aq The main species in solution is HCN which has a tendency to escape into the gas phase HCNaq HCNg Since the HCNg that is produced is a highly poisonous compound it would be dangerous to treat NaCN with acids without proper ventilation When the pH is 1000 the pOH is 400 and the concentration of hydroxide ion is 10 X 1074 M The concentration of HCN must be the same Why If the concentration of NaCN is x the table looks like CN7aq HzOl HCNaq OH7aq Initial 1V1 x 0 0 Change M 710x10 4 10gtlt1074 10gtlt1074 Equilibrium M x 710 X 10 4 10 X 10 4 10 X 104 Kb w CN 74 2 20 gtlt10 5 1390 X10 4 x 7 10 X 10 x 60gtlt10 4M CN 0 60 X 10 4 mol NaCN X 4901 g NaCN 74 x 10 3 g NaCN 1000 mL 1 mol NaCN Amount of NaCN 250 mL gtlt pH 2 53 flogH Ht 295 X 10 3M Since the concentration of H4r at equilibrium is 295 X 1073 M that means that 295 X 1073 M HCOOH ionized Let represent the initial concentration of HCOOH as I The equation representing the ionization of formic acid is HCOOHaq Haq Hcoo aq Initial 1V1 I 0 0 Change M 7295 X 10 3 295 X 10 3 295 X 10 3 Equilibrium M 17 295 X 10 295 X 10 3 295 X 10 3 7 HHCOO a HCOOH CHAPTER 15 ACIDS AND BASES 445 295 gtlt10732 17gtlt1074 3 1 7295gtlt10 I 0054M There are 0054 moles of formic acid in 1000 mL of solution The mass of formic acid in 100 mL is 0054 mol formic acid X 4603 g formic acid 100 mL gtlt 1000 mL soln 1 mol form1c acid 025 g formic acid 15125 The equilibrium is established CH3COOHaq CH3coo aq Haq Initial M 0150 0 0100 Change 111 7x x x Equilibrium 111 0150 7 x x 0100 x K CH3COO H a CH3COOH 18 X105 7 x0100 x N 0100x 01507x 0150 x 27 X 1075M 27 X 1075 M is the H contributed by CH3COOH HCl is a strong acid that completely ionizes 1t contributes a H of 0100 M to the solution Htmal 0100 27 gtlt 1075M m 0100M pH 1000 The pH is totally determined by the HCl and is independent of the CH3COOH 15126 The balanced equation is Mg 2HCl 7 MgClz H2 lmol Mg 2431 g Mg mol of Mg 187 g Mg gtlt 00769 mol From the balanced equation mol of HCl required for reaction 2 X mol Mg 200769 mol 0154 mol HCl The concentration of HCl pH 70544 thus HT 350M initial mol HCl Mgtlt Vol L 350A100800 L 0280 molHCl Moles of HCl left after reaction initial mol HCl 7 mol HCl reacted 0280 mol 7 0154 mol 0126 mol HCl 446 15127 15128 15129 15130 15131 CHAPTER 15 ACIDS AND BASES Molarity of HCl left after reaction M molL 0126 mol0080L 158M pH flogl58 020 a The pH of the solution of HA would be lower Why b The electrical conductance of the HA solution would be greater Why c The rate of hydrogen evolution from the HA solution would be greater Presumably the rate of the reaction between the metal and hydrogen ion would depend on the hydrogen ion concentration ie this would be part of the rate law The hydrogen ion concentration will be greater in the HA solution The important equation is the hydrolysis of N027 N027 HzO HNOz OH7 21 Addition of HCl will result in the reaction of the H from the HCl with the OH7 that was present in the solution The OHT will effectively be removed and the equilibrium will shift to the right to compensate more hydrolysis b Addition of NaOH is effectively addition of more OHT which places stress on the right hand side of the equilibrium The equilibrium will shift to the left less hydrolysis to compensate for the addition of OH c Addition of NaCl will have no effect d Recall thatthe percent ionization of a weak acid increases with dilution see Figure 154 of the text he same is true for weak bases Thus dilution will cause more hydrolysis shifting the equilibrium to the right Like carbon dioxide sulfur dioxide behaves as a Lewis acid by accepting a pair of electrons from the Lewis base water The Lewis acidbase adduct rearranges to form sulfurous acid in a manner exactly analogous to the rearrangement of the carbon dioxidewater adduct to form carbonic acid that is presented on page 683 of the textbook In Chapter ll we found that salts with their formal electrostatic intermolecular attractions had low vapor pressures and thus high boiling points Ammonia and its derivatives amines are molecules with dipole dipole attractions as long as the nitrogen has one direct NiH bond the molecule will have hydrogen bonding Even so these molecules will have much higher vapor pressures than ionic species Thus if we could convert the neutral ammoniatype molecules into salts their vapor pressures and thus associated odors would decrease Lemon juice contains acids which can react with neutral ammoniatype amine molecules to form ammonium salts NH3 H NH4 RNH2H RNH3 pH 1064 pOH 336 0H 44x 1074M 15132 CHAPTER 15 ACIDS AND BASES 447 OH W 4 4 X 10 4 M CH3N39Hzaq 4 H200 CH3NH3aq x744gtlt1074M 44gtlt1074M CH3NH3lOH Kb CH3NH2 44X 7 44gtlt10 444gtlt 10 4 x 7 44gtlt10 4 44 x1074x719 gtlt1077 19 gtlt1077 x 86x 104114 The molar mass of CH3NH2 is 3106 gmol The mass of CH3N39H2 in 1000 mL is 86x10 4 mol CH3NH2 3106 g CH3NH2 27 x 10393 g CH3NH2 15133 15134 1000 mL X 1000 mL 1 mol CH3NH2 HCOOH H Hcoo Initial 111 0400 0 0 Change 111 7x x x Equilibrium 111 0400 7 x x x Total concentration of particles in solution 0400 7 x x x 0400 x Assuming the molarity of the solution is equal to the molality we can write ATf Kfm 0758 1860400x x 000753 Hl HCOO HHCOO 7 000753000753 0400 7 000753 14 x 10394 HCOOH 21 NH H20 N113 OH N3 3HzO N113 3OH b N37 is the stronger base since each ion produces 3 0H7 ions SOzg HzOl Haq Hso3 aq Recall that 012 ppm 502 would mean 012 parts 502 per 1 million 106 parts of air by volume The number of particles of SO per volume will be directly related to the pressure 448 CHAPTER 15 ACIDS AND BASES PSOZ mam l2gtlt10 7atm 10 parts air We can now calculate the H from the equilibrium constant expression lH lleogl Psoz x2 l3gtlt10 2 7 12x10 x2 13 X 10 212x 10 7 x 39gtlt1075M Ht pH 710g39 X 10 5 440 Ht c10 HClO 15135 30 X 10 8 A pH of 78 corresponds to H 16 X 1078M Substitute H into the equation above to solve for the Clo ratio HClO c101 7 30x10 8 7 19 HCIO l6gtlt10 8 This indicates that to obtain a pH of 78 the C107 must be 19 times greater than the HClO We can write part C107 C10 part ClO part HClO gtlt100 igtltlOO 66 l9l0 By difference HClO 34 15136 In inhaling the smelling salt some of the powder dissolves in the basic solution The ammonium ions react with the base as follows NH aq OH aq a N39H3aq H20 It is the pungent odor of ammonia that prevents a person from fainting 15137 21 The overall equation is FezO3s 6HClaq 2FeCl3aq 3HzOl and the net ionic equation is FezO3s 6Haq 2Fe3aq 3HzOl Since HCl donates the H4r ion it is the Brznsted acid Each FezO3 unit accepts six H4r ions therefore it is the Brznsted base b CHAPTER 15 ACIDS AND BASES 449 The first stage is CaCO3s HClaq Ca2aq Hco3 aq c1 aq and the second stage is HClaq HCOgYaq COzg Cliaq HzOl The overall equation is CaC03s 2HClaq CaClzaq HzOl COzg The CaClz formed is soluble in water We need to find the concentration of the HCl solution in order to determine its pH Let39s assume a volume of 1000 L 1000 mL The mass of 1000 mL of solution is 1000 mL gtlt 1073g 1073g L The number ofmoles of HCl in a 15 percent solution is 15 HCl 44 mol HCl 100 soln X 1073 g soln 16 gtlt102 g HC1x 3646gHC1 Thus there are 44 moles of HCl in one liter of solution and the concentration is 44M The pH of the solution is pH flog44 064 This is a highly acidic solution note that the pH is negative which is needed to dissolve large quantities of rocks in the oil recovery process 15138 c does not represent a Lewis acidbase reaction In this reaction the FiF single bond is broken and single bonds are formed between P and each F atom For a Lewis acidbase reaction the Lewis acid is an electron pair acceptor and the Lewis base is an electronpair donor 15139 a b c d False A Lewis acid such as C02 is not a Brznsted acid It does not have a hydrogen ion to donate False Consider the weak acid NH4 The conjugate base of this acid is NH3 which is neutral False The percent ionization of a base decreases with increasing concentration of base in solution False A solution of barium fluoride is basic The uoride ion F7 is the conjugate base of a weak acid It will hydrolyze to produce 0H7 ions 15140 From the given pH39s we can calculate the H in each solution Solution 1 H io PH 10quotquot12 76gtlt 1075M Solution 2 H 10 53976 17gtlt1076M Solution 3 Ht 10 53934 46gtlt1076M 450 15141 15142 CHAPTER 15 ACIDS AND BASES We are adding solutions 1 and 2 to make solution 3 The volume of solution 2 is 0528 L We are going to add a given volume of solution 1 to solution 2 Let39s call this volume x The moles of H4r in solutions 1 and 2 Will equal the moles of H4r in solution 3 mol H4r soln l molHJr soln 2 mol H4r soln 3 Recall that mol Mgtlt L We have 76 X 10 5 molLx L 17 X 10 6 molL0528 L 46 X 10 6 molL0528 xL 76 x1075x 90 X 10 7 24 X 10 46 x1076x 71 X 10 5x 15 gtlt1076 x 0021 L 21 mL Given the equation HbHJr 02 HbOz Hir a From the equilibrium equation high oxygen concentration puts stress on the left side of the equilibrium and thus shifts the concentrations to the right to compensate Hb02 is favored b High acid H4r concentration places stress on the right side of the equation forcing concentrations on the left side to increase thus releasing oxygen and increasing the concentration of Hle39 c Removal of C02 decreases H4r in the form of carbonic acid thus shifting the reaction to the right More HbOz Will form Breathing into a paper bag increases the concentration of C02 rebreathing the exhaled C02 thus causing more 02 to be released as explained above The balanced equations for the two reactions are MCO3S 2HClaq gt MClzaq COzg HzOl HClaq NaOHaq NaClaq HzOl First let s find the number of moles of excess acid from the reaction with NaOH 0588 mol NaOH X lmol HCl 1 L soln 1 mol NaOH 003280 L gtlt 00193 mol HCl The original number ofmoles of acid was 0100 mol HCl 1 L soln 0500 L gtlt 00500 mol HCl The amount of hydrochloric acid that reacted with the metal carbonate is 00500 molHCl 7 00193 mol HCl 00307 mol HQ The mole ratio from the balanced equation is 1 mole MC03 2 mole HCl The moles of MC03 that reacted are 1 mol MCO3 2 mol HCl 00307 mol HCl gtlt 001535 mol MCO3 15143 15144 CHAPTER 15 ACIDS AND BASES 451 We can now determine the molar mass of MCO3 which will allow us to identify the metal 1294 g MCO3 molar mass MCO3 0 01535 mol MCO 39 3 843 gmol We subtract off the mass of C0327 to identify the metal molar massM 843 gmoli 6001 gmol 243 gmol The metal is magnesium We start with the equation for the hydrolysis of a weak acid HA H20 H3O A At equilibrium 1130 A7 H30A HA H3012 K21 HA Because this is a weak acid the concentration of HA at equilibrium is approximately equal to its initial concentration HA m HA0 Substituting into the above equation gives K H H302 a HAlo Hgoti KalHAlo By definition JK HA EOX1OO 0 i 0 ionization a l A HAlo HAlo Looking at this equation for ionization when HA0 is decreased by a factor of 10 to ionization HAlo 10 increases by m This result is in accord with LeChatelier s principle which predicts that ionization should increase with dilution Because HF is a much stronger acid than HCN we can assume that the pH is largely determined by the ionization of HF HFaqHzOl H30aqF aq Initial 1V1 100 0 0 Change ItI 7x x x Equilibrium 1V1 100 7 x x x H OJr F Ka 3 1 HF N 2 7lgtlt10 4 x m l007x 1 O 452 CHAPTER 15 ACIDS AND BASES x 0027M H30 pH 157 HCN is a very weak acid so at equilibrium HCN m 100 M H30 llCN Ka HCN 0027CN 100 49 gtlt10710 CN 18 x 10 8M In a 100 M HCN solution the concentration of CNF would be HCNaq HzOl H3oaq CN aq Initial A4 100 Change A4 7x x x Equilibrium A4 100 7 x x x K Hgot CN 3 HCN 2 49x10 10 x m x 100 7 x 100 x 22 x 10 5M CN CNF is greater in the 100 M HCN solution compared to the 100M HCNl OOM HF solution According to LeChatelier s principle the high H30 from HF shifts the HON equilibrium from right to le decreasing the ionization of HCN The result is a smaller CN in the presence of HF 15145 Both NaF and San provide F7 ions in solution NaF a Na F San a Sn2 2F Because HF is a much stronger acid than H20 it follows that F7 is a much weaker base than 0H7 The F7 ions replace OHF ions during the remineralization process 5Ca2 3PO437 F7 gt Ca5PO43F uorapatite because 0H7 has a much greater tendency to combine with H4r OH Hl H20 than F7 does F Hl HF Because F7 is a weaker base than 0H7 uorapatite is more resistant to attacks by acids compared to hydroxyapatite CHAPTER 15 ACIDS AND BASES 453 15146 The van t Hoff equation allows the calculation of an equilibrium constant at a different temperature if the value of the equilibrium constant at another temperature and AH for the reaction are known K77 R 3771 K1 AH E 1 1 First we calculate AH for the ionization of water using data in Appendix 3 of the text H200 HaqOH aq AHquot AHFH AHHOH H AHHHzO AHquot 0 7 22994 kJmol 7 72858 kJmol AHquot 559 kJmol We substitute AH and the equilibrile constant at 25 C 298 K into the van t Hoff equation to solve for the equilibrium constant at 100 C 373 K 1n10x10 14 7 559gtlt103 Jmol 1 7 1 K2 8314 JmolK 373K 298K 10 gtlt10 14 7 64537 K2 K2 93 x10 13 We substitute into the equilibrium constant expression for the ionization of water to solve for H and then pH K2 HIGH 93 x10 13 x2 x 11 96 x1o 7M pH 71og96 X104 602 Note that the water is not acidic at 100 C because If 0H7
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