Chem 116 POGIL Worksheet - Week 9 Weak Acid and Base Equilibria

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1 Chem 116 POGIL Worksheet - Week 9 Wek Acid nd Bse Equilibri Why? We hve seen tht the clcultion of [HO ] nd ph for solutions of strong cids nd bse. To crry out clcultion of ll species present in solution of pure wek cid in wter requires use of the equilibrium constnt for the cid s hydrolysis, clled K. In similr fshion, clculting the concentrtions of ll species in solution of wek bse in wter requires solving the equilibrium expression for the bses s hydrolysis, clled K b. In cid solutions we concentrte on clculting [HO ] nd ph. In bsic solutions we concentrte on finding [OH ] nd poh, insted. Some wek bses re the conjugte bses of wek cids. These re generlly the nions produced by the cid's hydrolysis. Like moleculr bses, these conjugte bses hve bse hydrolysis equilibri with ssocited K b vlues. The reltionship between K nd K b for conjugte cid-bse pir llows us to clculte the ph of solution of either species. The strengths of cids nd their conjugte bses re relted to their moleculr structures. Knowing the trends llows us to predict whether n cid is strong or wek, nd if wek how it compres in strength to other similr wek cids. Beyond solutions of pure cid or bse in wter, we need to look t the effect of dding extr mounts of the conjugte bse or cid to the solution. The shift in the position of the equilibrium, clled the common ion effect, chnges the ph nd imbues the solution with certin properties tht re the bsis for formulting buffer. Buffer solutions re importnt for regulting ph in mny chemicl nd biologicl systems. Lerning Objectives Understnd the role of solvent protolysis in cid-bse chemistry Understnd ph nd the ph scle Understnd the importnce of n cid s or bses s hydrolysis equilibrium Understnd the concept of K for wek cid nd K b for wek bse Know the reltionship between K nd K b for conjugte cid-bse pir Know the reltionships between structure types nd cid strength solution of wek cid or wek bse Understnd the principles of buffer solutions Success Criteri Be ble to use K w to clculte concentrtions of HO nd OH in cidic or bsic solutions Be ble to clculte ph nd poh Be ble to clculte concentrtions of hydronium ion nd hydroxide ion in solutions of pure strong cid or strong bse in wter Be ble to use K to clculte the concentrtions of ll species present in solution of pure wek cid in wter Be ble to use K b to clculte the concentrtions of ll species present in solution of pure wek bse in wter

2 Be ble to determine nd use K b for conjugte bse, given K of its conjugte cid, nd vice vers. Be ble to predict reltive strengths of cids or their conjugte bses from structure considertions Prerequisite Hve red ll of Chpter 16 Informtion (Wter s Autoprotolysis) Pure wter hs n equilibrium with HO nd OH clled utoprotolysis or utodissocition. H2O H2O HO OH bse cid cid bse for which we might write the following equilibrium expression: K c K w = [HO ][OH ] We omit the concentrtion of H2O(l) from the expression, becuse it is pure liquid. We cn lso ignore the concentrtion of H2O for dilute solutions, becuse [H2O] is virtully constnt. At o 14 temperture of 25 C K hs mesured vlue of 1.0 x 10 : w 14 2 o K w = [HO ][OH ] = 1.00 x 10 M t 25 C K w is clled the ion product or dissocition constnt of wter. The units on K w re generlly 2 not written, but they re implicitly M, s shown bove. For pure wter, [HO ] = [OH ]. To find the vlue of this concentrtion, let x = [HO ] = [OH ] nd substitute into the ion product expression for wter: 14 2 K w = [HO ][OH ] = 1.0 x 10 = x 7 x = 1.0 x 10 M = [HO ] = [OH ] 7 This condition of [HO ] = [OH ] = 1.0 x 10 M defines neutrl wter. The K w expression must be obeyed for ny dilute solution in wter. Suppose we dd strong cid, such s HCl, which dissocites completely into HO nd Cl ions. This will increse [HO ] in the solution, but by LeChtelier s Principle it will cuse the utoprotolysis equilibrium 14 to shift left to mintin K w = [HO ][OH ] = 1.0 x 10. This will cuse [OH ] to become smller s more moleculr H2O is formed in response to the stress of excess HO ion in the solution. A strong bse, such s NOH, which dissocites completely into N nd OH ions, would hve the reverse effect. This will increse [OH ] in the solution, but it will mke [H O ]

3 14 smller s the utoprotolysis equilibrium shifts left to mintin K w = [HO ][OH ] = 1.0 x 10. In generl, owing to these shifts in the utoprotolysis equilibrium, solution is cidic if [HO ] > M nd [OH ] < 10 M, nd solution is bsic if [H O ] < 10 M nd [OH ] > 10 M. Key Questions 1. Given the following concentrtions of HO or OH, clculte the concentrtion of OH or H O, nd indicte whether the solution is cidic or bsic. [HO ] [OH ] Acidic or bsic? 1.2 x 10 8 M 1.0 x 10 5 M 4.0 x 10 9 M Informtion (ph nd poh) It is often convenient to express [HO ] nd [OH ] by the logrithmic terms ph nd poh, respectively, defined s ph = log [HO ] poh = log [OH ] Note tht the symbol "p" in front of ny quntity mens "negtive bse-10 logrithm of"; e.g., pk = logk. The reltionship between ph nd poh cn be derived from K. w w w 14 K w = [HO ][OH ] = 1.0 x 10 Tking bse-10 logrithms of both sides of this eqution, we hve log K w = log[ho ][OH ] = log[ho ] log[oh ] = Chnging the sign cross this eqution, we hve log K w = log[ho ] log[oh ] = If we now recognize tht the negtive logrithm of ny quntity X cn be designted with the symbol px, then this eqution becomes pk = ph poh = w 7 For neutrl wter, where [HO ] = [OH ] = 1.0 x 10 M, ph = poh = For cidic solutions ph is less thn 7, nd for bsic solutions ph is greter thn 7. Note tht poh runs in the opposite sense. For cid solutions poh is greter thn 7, nd for bsic solutions poh is less thn

4 7. It is possible to hve solutions in which ph or poh re greter thn 14 or less thn 0 (i.e., negtive). Note tht the number of deciml plces in logrithm indictes the number of significnt figures 15 in the originl number. For exmple, the bse-10 logrithm of 1.45 x 10 ( sig. figs.) is 14.89, where the deciml prt indictes the three significnt figures. The 14 prt (left of the deciml) reltes to the power of ten, nd hs no relevnce to significnt figures. Why is this? Becuse we cnnot tke the logrithm of negtive deciml number, the logrithm is ctully the sum of negtive integer ( 15) nd positive deciml (0.161); i.e., = The 15, clled the chrcteristic, reltes to the 10 of the originl number, nd the 0.161, clled the mntiss, reltes to the 1.45 of the originl number, which is where the significnt digits re. Be sure when clculting ph or poh tht the number of deciml plces mtches the number of significnt figures for the concentrtion of [H O ] or [OH ]. Key Questions 2. Complete the following tble by clculting the missing entries nd indicte whether the solution is cidic or bsic. Be sure your nswers re expressed to the proper number of significnt figures. [HO ] [OH ] ph poh cidic or bsic? 5.8 x 10 5 M 2.0 M 6.2 x 10 6 M Informtion (Solutions of Strong Acids nd Bses) When strong cid,, dissolves in wter it dissocites completely into HO nd A ions, leving no moleculr in the solution: (q) H2O(l) HO (q) A (q) Thus, when we tlk bout the concentrtion of strong cid solution, sy 0.10 M HCl, we re usully referring to the number of moles per liter used to mke the solution, not wht is ctully present in the solution. Becuse of dissocition, there is virtully no moleculr HCl in 0.10 M HCl solution! We cn define the nlyticl concentrtion of the cid, symbolized C, s the number of moles per liter of cid used to mke the solution. Thus, for 0.10 M HCl, C HCl = 0.10 M. Then, using brcket symbols, we cn refer to ctul concentrtions in the solution s

5 7 [], [HO ], nd [A ]. For moderte nlyticl concentrtions of strong cid (C >> 10 M), we cn generlly ssume tht the cid s dissocition supplies virtully ll the HO in the solution; i.e., wter s contribution is insignificnt. For given nlyticl concentrtion of pure strong cid t moderte concentrtion, we would hve the following vlues initilly nd fter dissocition: (q) H2O(l) HO (q) A (q) Initil C 0 0 After dissocition 0 C C In other words, the concentrtions of hydronium ion nd conjugte bse in the solution re equivlent to the number of moles per liter of cid dded to mke the solution; i.e., [HO ] = [A ] = C. Becuse of complete dissocition, [] = 0. For 0.10 M HCl, we would hve [HCl] = 0, [H O ] = [Cl ] = 0.10 M. In similr fshion, strong bse dissocites completely on dissolving in wter. Most strong bses re ionic hydroxide compounds, with the generl formul M(OH) n, where n = 1, 2,, n depending on the chrge on the ction, M. The number of moles per liter of ionic bse used to mke the solution, the nlyticl concentrtion, cn be designted. Anlogous to the strong cid cse, for solutions of pure strong bse t moderte nlyticl concentrtion 7 ( >> 10 M), we cn generlly ssume tht the bse s dissocition supplies virtully ll the OH ion in the solution; i.e., wter s contribution is insignificnt. For given nlyticl concentrtion of pure strong bse t moderte concentrtion, we would hve the following vlues initilly nd fter dissocition: n M(OH) n(q) M (q) n OH (q) Initil 0 0 After dissocition 0 n In other words, the concentrtion of hydroxide ion nd ction in the solution re equivlent to the number of moles per liter of bse used to mke the solution, fctoring in the stoichiometry of n the dissocition; i.e., [OH ] = n, [M ] =. Becuse of complete dissocition, [M(OH) n] = 0. For ny cid solution, once [H O ] is determined, [OH ] = K /[H O ]. Likewise, for ny bse solution, once [OH ] is determined, [H O ] = K /[OH ]. w w

6 Key Questions. Determine the concentrtions of HO nd OH in the following solutions of strong cids or bses in wter. [Cution: Think bout how the cid or bse dissocites in wter.] Solution [HO ] [OH ] 2.5 x 10 M HCl M NOH x 10 M C(OH) 2 Informtion (Wek Acid Hydrolysis) All cids in wter hve n cid hydrolysis equilibrium tht produces hydronium ion nd the conjugte bse of the cid: H2O HO A With strong cids this equilibrium lies completely to the right. With wek cids, the equilibrium lies to the left, which mens tht undissocited cid,, is usully the principl species in the solution, with only reltively smll concentrtions of HO nd A produced from the hydrolysis rection. The equilibrium constnt for the hydrolysis of n cid is lbeled K, defined s K is constnt for given cid t prticulr temperture regrdless of the number of moles per liter of cid tht ws dissolved (the nlyticl concentrtion of the cid) to mke the solution. We cn use K to clculte the equilibrium concentrtions of ll species involved in the hydrolysis rection (undissocited cid, hydronium ion, nd conjugte bse) in solution of pure wek cid in wter. We will only del with solutions t moderte concentrtions, in which we cn ssume tht the dded cid is the principl source of hydronium ion. Therefore, we will 7 ssume tht prior to dding cid we hve [HO ] = 1.0 x 10 M 0 M. To set up the problem, let x be the concentrtion of hydronium ion, [HO ], produced by the cid s hydrolysis once equilibrium is estblished. Then, for n initil concentrtion of cid, symbolized here s C, t equilibrium we hve H2O HO A Initil C ~0 0 Chnge x x x Equilibrium C x x x

7 If we substitute the lgebric expressions for ll the equilibrium concentrtions into K, we hve This is gives the qudrtic eqution, 2 x Kx KC = 0 where =1, b = K, nd c = KC. Solving the qudrtic nd ignoring the negtive root, we obtin x = [HO ] = [A ]. Then, we cn solve [] = C x to obtin the concentrtion of undissocited cid in the solution. Very often we cn mke some simplifying ssumptions tht mke it unnecessry to solve qudrtic eqution. When the cid is very wek (very smll K vlue) or its nlyticl concentrtion is reltively lrge (C >> K ), the mount of dissocition of moleculr my be very smll. Put nother wy, we could sy C >> x. Thus, in such cses we cn often ignore x in the expression [] = C x, which mens [] C. H2O HO A Initil C ~0 0 Chnge x x x Equilibrium ~C x x Substituting our simplified lgebric expressions into the K expression, we hve from which it follows x = = [HO ] = [A ]. As prcticl mtter, becuse it tkes so little time to clculte, try using x = clculte the percent dissocition s, nd then % dissocition = [HO ]/C x 100% If this clcultion gives n nswer of less thn 5%, the clculted vlues of [HO ] nd [A ] re resonbly ccurte nd cn be ccepted. If this clcultion gives vlue tht is greter thn 5%, then the cid s dissocition is too gret to ignore in the lgebric expression for []. Then, go bck nd use the more exct expression [] = C x, which mens you must solve the qudrtic eqution. Usully, if C >> K by two powers of 10 or more, the cid is probbly not pprecibly dissocited nd you cn use [H O ] =. But be sure to check your method by clculting the percent dissocition. If C K (within two powers of 10), you will

8 probbly hve to solve the qudrtic eqution. By either method, once [HO ] hs been clculted, the concentrtion of hydroxide ion in the solution cn be determined by solving the 14 K expression; i.e., [OH ] = K /[H O ] = 1.00 x 10 /[H O ]. w w Key Questions 4. A M solution of wek cid hs ph of Wht is the vlue of K for the cid? [Hint: Wht is the ctul concentrtion of undissocited, [], in this solution?] 2 5. Wht is the ph of M benzoic cid (C6H5CO2H = HBz), for which K = ? Is it necessry to solve the qudrtic eqution in this cse? 6. Find the concentrtions of ll species nd ph for M HF solution. K = You will need to find vlues for [HF], [H O ], [F ], nd [OH ]. Informtion (Wek Bse Hydrolysis) When wek bse like mmoni is dded to wter it hydrolyzes to give smll mounts of hydroxide ion nd its conjugte cid in wter. If we symbolize wek bses s B, then the generl hydrolysis equilibrium is B H2O BH OH for which we cn write the equilibrium constnt expression We cn use K b to clculte the concentrtions of ll species in the solution in mnner nlogous to the wy we used K for wek cid solutions. However, with bse solutions we focus on determining [OH ] first, rther thn [HO ]. Once we hve determined [OH ], we cn 14 use K = [H O ][OH ] = to clculte [H O ] nd ph. w For solution of pure wek bse in wter with moderte nlyticl concentrtion, C B, we would hve the following concentrtions initilly nd t equilibrium: B H2O BH OH Initil C B 0 ~0 Chnge x x x Equilibrium C x x x Substituting into the K b expression, we obtin B

9 2 This is qudrtic eqution of the form x Kbx KbC B = 0. Notice tht this is virtully the sme eqution we obtined in the wek cid cse, except here x = [BH ] = [OH ] nd of course we hve K b insted of K. As with the cid cse, if the nlyticl concentrtion of the bse is reltively lrge (C B >> K b), hydrolysis my be miniml nd we cn ignore x in the expression for [B]; i.e., [B] = C x C. Then, the eqution bove simplifies to B B from which it follows Anlogous to the cid cse, if C B >> K b by two or more powers of 10, then this simplified eqution probbly cn be used. If C B K b, within two powers of 10, then the qudrtic eqution will probbly need to be solved. If the simplified eqution is used, check its vlidity by clculting the percent hydrolysis of the bse: If this clcultion gives 5% or greter, use the more exct expression [B] = C B x nd solve the qudrtic eqution. By either method, once [OH ] hs been clculted, use K w = [HO ][OH ] = to clculte [H O ], nd ph poh = to clculte ph. Key Questions 5 7. Wht is the ph of M solution of NH (q)? For mmoni, K b = Is it necessry to solve the qudrtic eqution? Justify the pproch you took, either wy. Informtion (K of Conjugte Bses nd K of Conjugte Acids) b The conjugte bse of strong cid hs little tendency to cquire H nd reform the cid. Thus, Cl, the conjugte bse of HCl, hs virtully no rel bse strength. But the conjugte bses of wek cids do hve tendency to cquire H, nd so they re rel wek bses. Like ny other wek bse, they hve bse hydrolysis equilibrium nd relted K b. For exmple, consider wek cid nd its conjugte bse A. The two re relted to ech other by the cid's hydrolysis equilibrium: H2O HO A But they re lso relted to ech other by the conjugte bse's hydrolysis equilibrium: A H2O OH

10 How is the vlue of K of relted to the vlue of K b of A? Recll tht when two equilibri re dded together the equilibrium constnt for the sum is the product of the individul equilibrium constnts. If we dd the cid hydrolysis equilibrium of nd the bse hydrolysis equilibrium of A, the sum is 2 H2O HO OH, which is wter's utoprotolysis equilibrium, governed by K : w H2O HO A A H2O OH 2 H2O HO OH Therefore, in generl we cn write Similrly, if we hve moleculr bse B with known K b, its conjugte cid BH would hve K given by As these reltionships show, the lrger the vlue of K for n cid, the smller the vlue of K will be for its conjugte bse, nd vice vers. Conjugte bses of wek cids re true bses. For exmple, the cette ion (often bbrevited ) OAc, which is the conjugte bse of cetic cid (often bbrevited HOAc), cn mke bsic solution s result of its bse hydrolysis. The K b vlue for the cette ion cn be clculted 5 from cetic cid's K = , using : b OAc H2O HOAc OH But we cnnot simply dd cette ions to solution. They must be supplied in the form of n ionic slt, such s sodium cette, which dissocites completely in wter to give n equivlent mount of cette ion nd sodium ion in solution. Sodium ion hs no cid-bse chrcter, so the ph of the solution is the result of the cette ion's bse hydrolysis: NOAc N OAc OAc H O HOAc OH ph > 7 In similr mnner, we cn crete solution of the conjugte cid of moleculr wek bse by dding one of its soluble slts, such s chloride. For exmple, mmonium chloride, NH4Cl, when dissolved in wter supplies n equivlent mount of mmonium ion, NH, the conjugte 2 4

11 cid of mmoni, nd chloride ion. The chloride ion, being the conjugte bse of strong cid (HCl) hs no rel cid-bse chrcter, so the ph of the solution is the result of the mmonium ion's cid hydrolysis: NH Cl Cl NH NH H O NH H O ph < The K for the mmonium ion cn be clculted from K = for mmoni. b NH 4 H2O NH OH Key Questions 8. Sodium hypochlorite, NOCl, is the ctive ingredient in chlorine blech (e.g., Chlorox ). 8 For hypochlorous cid, HOCl, K = Write the eqution for the bse hydrolysis equilibrium of the hypochlorite ion. b. Wht is the vlue of K b for the hypochlorite ion? c. Wht is the ph of 0.10 M solution of sodium hypochlorite? 5 9. Trimethylmine, (CH ) N, is wek bse (K b = ) tht hydrolyzes by the following equilibrium: (CH ) N H2O (CH ) NH OH. Write the eqution for the cid hydrolysis equilibrium of trimethylmmonium ion, (CH ) NH. b. Wht is the vlue of K for the trimethylmmonium ion, (CH ) NH? c. Wht is the ph of 0.12 M solution of trimethylmmonium chloride, ((CH ) NH)Cl? Informtion (Acid Strength nd Structure) The tendency of n cid to lose proton to solvent wter, s indicted by its K vlue, depends upon two principl fctors: The strength of the H A bond. The weker the bond, the greter is the tendency to lose protons, nd the lrger K will be. The stbility of the conjugte bse in solution. The more stble the conjugte bse A, the greter is the tendency of to lose protons, nd the lrger K will be.

12 For simple inorgnic cids, the first fctor is more importnt. The second fctor is often importnt in comprisons mong orgnic cids. Acid strengths of binry cids, with the generl formul HnX, increse s the H X bond wekens. This results in the following trends: The lrger the X tom, the weker the H X bond is nd the stronger the cid is. HF < HCl < HBr < HI H2O < H2S < H2Se < H2Te Among cids with similr H X bond strengths, the more electronegtive the X tom, the more polr the H X bond will be. This mkes the H tom more positive, fcilitting its removl by solvent wter. Thus, cross period, NH < H2O < HF One of the most studied groups of cids is the fmily of oxycids, with the generl formul HOXO n, where X is centrl tom. In these cids, the cidic hydrogen is lwys bonded to n oxygen tom, with linkge of the type X O H. Any hydrogen toms bonded directly to the centrl X tom (i.e., X H) re not cidic. These cids my lso hve one or more terminl oxygen toms, bonded directly to X but not bering hydrogen tom (i.e. X O). In most oxycids (with some exceptions) there re no X H bonds, so the number of terminl oxygen toms is just the totl number of oxygen toms minus the number of hydrogen toms in the formul. Thus, HClO hs two terminl oxygen toms, with the third oxygen tom bonded to the cidic hydrogen tom. The number of these terminl oxygen toms ffects the cid s strength. In generl, two fctors re importnt in judging the strengths of oxycids: Among cids of the sme structure type, cid strength increses with the electronegtivity of the centrl X tom. For exmple, HOI < HOBr < HOCl H2SeO 4 < H2SO4 As electronegtivity of X increses, the polrity of the O H bond increses, mking removl of H by solvent wter more fvorble. Acidity increses s the number of terminl oxygen toms (those without H ttched to them) increses. For exmple, HOCl < HOClO < HOClO 2 < HOClO

13 The electron withdrwing bility of the terminl O toms increses the forml chrge on the centrl X tom, decresing electron density in the O H bond, thereby wekening it. This is n exmple of n inductive effect. The number of terminl oxygen toms in the oxycids bers strong correltion to K. O toms K rnge Strength very wek wek 1 2 >10 strong >>1 very strong An importnt group of orgnic cids contins the crboxyl group, -CO2H. Accordingly, such cids re clled crboxylic cids. HCO2H CHCO2H CHCH2CO2H CHCH2CH2CO2H HO2CCO2H formic cetic propionic butyric oxolic (diprotic) The cidity of the crboxyl group is prtly due to the electronegtivity of the double-bonded oxygen, which withdrws electron density from the O H bond, thereby fcilitting the loss of H to solvent wter. Furthermore, when crboxyl group loses H, it forms resonnce stbilized nion, clled crboxylte ion: (The symbol R represents the rest of the orgnic molecule.) The cidity of crboxylic cid cn be enhnced by dding electronegtive toms to the rest of the molecule. The electronwithdrwing bility of such toms wekens the O H bond (inductive effect), thereby enhncing the bility of solvent wter to remove H. For exmple, ClCCO2H is much stronger cid thn CH CO H. 2

14 Key Questions 10. Explin the following observtions:. H2Se is stronger cid thn H2S b. HNO 2 is wek cid, but HNO is strong cid c. HPO 4 is stronger cid thn so4 d. FCH2CO2H is stronger cid thn CHCO2H