[ HOCl] Chapter 16. Problem. Equilibria in Solutions of Weak Acids. Equilibria in Solutions of Weak Acids

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1 Equilibri in Solutions of Wek Acids Chpter 16 Acid-Bse Equilibri Dr. Peter Wrburton The dissocition of wek cid is n equilibrium sitution with n equilibrium constnt, clled the cid dissocition constnt, bsed on the eqution HA (q) H O (l) H O (q) A - (q) [ H O [ A [ HA Equilibri in Solutions of Wek Acids The cid dissocition constnt, is lwys bsed on the rection of one mole of the wek cid with wter. If you see the symbol, it lwys refers to blnced eqution of the form HA (q) H O (l) H O (q) A - (q) The ph of 0.10 mol/l HOCl is 4.. Clculte for hypochlorous cid. HOCl (q) H O (l) H O (q) ClO - (q) [ H O [ ClO [ HOCl 4 1

2 Clculting Equilibrium Concentrtions in Solutions of Wek Acids We cn clculte equilibrium concentrtions of rectnts nd products in wek cid dissocition rections with known vlues for. To do this, we will often use the ICE tble technique we sw in the lst chpter on equilibrium. Clculting Equilibrium Concentrtions in Solutions of Wek Acids We need to figure out wht is n cid nd wht is bse in our system. For exmple, if we strt with 0.10 mol/l HCN, then HCN is n cid, nd wter is bse. HCN (q) H O (l) H O (q) CN - (q) 4.9 x Like our previous equilibrium problems, we then crete tble of the initil concentrtions of ll chemicls, the chnge in their concentrtion, nd their equilibrium concentrtions in terms of known nd unknown vlues. (ll in mol/l) HCN (q) H O (l) H O (q) CN - (q) Initil conc N/A Conc. chnge -x N/A x x Equil. conc x N/A x x 4.9x10 10 [HO [CN [HCN so 4.9x10 10 (0.10 x) (ll in mol/l) HCN (q) H O (l) H O (q) CN - (q) Initil conc N/A Conc. chnge -x N/A x x Equil. conc x N/A x x 4.9 x [HO [CN [HCN so 4.9 x 10 (0.10 x) We cn ALWAYS solve this eqution using the qudrtic formul nd get the right nswer, but it might be possible to do it more simply

3 4.9 x [HO [CN [HCN so 4.9 x (0.10 x) 4.9 x [HO [CN [HCN so 4.9 x (0.10 x) Every time we do n wek cid equilibrium problem, divide the initil concentrtion of the cid by. For this exmple 0.10 / 4.9 x x / 4.9 x x 10 8 Since this vlue is greter thn 100, we cn ssume tht the initil concentrtion of the cid nd the equilibrium concentrtion of the cid re the sme. This ssumption will led to nswers with less thn 5% error since this pre-check is greter thn x [HO [CN [HCN so 4.9 x (0.10 x) 4.9 x [HO [CN [HCN so 4.9 x (0.10 x) The ssumption we will mke is tht x << [HCN i so [HCN eqm [HCN I 4.9 x x / 0.10 x (4.9 x )(0.10) x 4.9 x x ±7.0 x 10-6 mol/l Bsed on the ssumption we ve mde, t equilibrium x [H O eqm [CN - eqm 7.0 x 10-6 mol/l (-ve vlue isn t physiclly possible) [HCN eqm 0.10 mol/l. 11 1

4 Any time we mke n ssumption, we MUST check it. We ssumed x << [HCN i To check the ssumption, we divide x by [HCN i nd express it s percentge -6 x 7.0 x 10 M x % HCN 0.10 M [ i 1 As long s the ssumption check is less thn 5%, then the ssumption is vlid! If the ssumption ws not vlid, we would hve to go bck nd use the qudrtic formul! 14 Remember! H O (l) H O (l) H O (q) OH - (q) is lwys tking plce in wter whether or not we hve dded n cid or bse. This rection lso contributes H O (q) nd OH - (q) to our system t equilibrium 15 Remember! Since t 5 C w [H O [OH x it turns out tht if our cid-bse equilibrium we re interested in gives ph vlue between bout 6.8 nd 7. then the uto-dissocition of wter contributes significnt mount of [H O nd [OH - to our system nd the rel ph would not be wht we clculted in the problem. 16 4

5 Acetic cid CH COOH (or HAc) is the solute tht gives vinegr its chrcteristic odour nd sour tste. Clculte the ph nd the concentrtion of ll species present in: ) 1.00 mol/l CH COOH b) mol/l CH COOH 17 ) (ll in mol/l) CH COOH (q) H O (l) H O (q) CH COO - (q) Initil conc N/A Conc. chnge -x N/A x x Equil. conc x N/A x x 1.8 x 10 5 [HO [CHCOO [CH COOH so 1.8 x 10 5 (1.00 x) Let s check the initil cid concentrtion / rtio / 1.8 x is lrger thn ) 1.8 x 10 5 [HO [CHCOO [CH COOH so 1.8 x 10 5 We cn probbly ssume tht x << [HAc i so [HAc eqm [HAc i (1.00 x) 1.8 x 10-5 x / 1.00 x (1.8 x 10-5 )(1.00) x 1.8 x 10-5 x ±4. x 10 - mol/l (but must be vlue since x [H O ) 19 ) So t equilibrium, [H O [CH COO - 4. x 10 - mol/l [CH COOH 1.00 mol/l. x [ CH COOH i - 4. x 10 M - 4. x % 1.00 M The ssumption ws vlid nd so ph - log [H O ph - log 4. x 10 - ph.8 0 5

6 b) (ll in mol/l) CH COOH (q) H O (l) H O (q) CH COO - (q) Initil conc N/A Conc. chnge -x N/A x x Equil. conc x N/A x x 1.8 x 10 5 [HO [CHCOO [CH COOH so 1.8 x 10 5 ( x) Let s check the initil cid concentrtion / rtio / 1.8 x is smller thn 100. b) (ll in mol/l) CH COOH (q) H O (l) H O (q) CH COO - (q) Initil conc N/A Conc. chnge -x N/A x x Equil. conc x N/A x x 5 [HO [CHCOO x 10 so 1.8 x 10 [CHCOOH ( x) We cn probbly CAN NOT ssume tht x << [HAc i so [HAc eqm [HAc i 5 [1.8 x 10 ( x) x 5 0 x 1.8 x 10 x -1.8 x b) b) b ± b 4c x (1.8 x 10 ) ± (1.8 x 10 ) 4(1)(-1.8 x 10 ) sox (1) x 10.4 x x x 10 x or x x x x x 10 x or x x x 10 x or x 4 so x 1. x 10 mol/l or x x x x 10 4 mol/l 8 Since [H O x we must use the positive vlue, so [H O [CH COO - 1. x 10-4 mol/l [CH COOH mol/l 1. x 10-4 mol/l mol/l. 4 6

7 b) Let s confirm tht x << [HAc i IS NOT TRUE x [ CH COOH i x 10 M % M ph - log [H O ph - log 1. x 10-4 ph.89 A vitmin C tblet contining 50 mg of scorbic cid (C 6 H 8 O 6 ; 8.0 x 10-5 is dissolved in 50 ml glss of wter to give solution where [C 6 H 8 O x 10 - mol/l. Wht is the ph of the solution? 5 6 (ll in mol/l) C 6H 8O 6 (q) H O (l) H O (q) C 6H 7O - 6 (q) Initil conc x 10 - N/A Conc. chnge -x N/A x x Equil. conc x 10 - x N/A x x 5 [HO [C6H7O x 10 so 8.0 x 10 - [C H O (5.68 x 10 x) Check the initil cid concentrtion / rtio x 10 - / 8.0 x which is not lrger thn 100 so x 10 (5.68 x 10 x) x 5 0 x 8.0 x 10 x x b ± b 4c x 5 (8.0 x 10 ) ± sox 5 7 (8.0 x 10 ) 4(1)(-4.5 x 10 ) (1) x x 10 (1.8 x 10 ) -8.0 x x10 x or x x x x x 10 x or x 1.4x x 10 x or x so x 7.1x 10 mol/l or x 6.x (1.8 x10 6 ) mol/l 7 8 7

8 Since [H O x the nswer must be the positive vlue [H O [C 6 H 7 O 6-6. x 10-4 mol/l [C 6 H 8 O 6 (5.68 x x 10-4 )mol/l 5.05 x 10 - mol/l ph - log [H O ph - log 6. x 10-4 ph.0 Degree of ioniztion The ph of solution of wek cid like cetic cid will depend on the initil concentrtion of the wek cid nd. Therefore, we cn define second mesure of the strength of wek cid by looking of the degree (or percent) ioniztion of the cid. % ioniztion [HA ionized / [HA initil x 100% 9 0 Percent ioniztion In prt ) of n erlier problem n cetic cid solution with initil concentrtion of 1.00 mol/l t equilibrium hd [H O eqm [HA ionized 4. x 10 - mol/l % ioniztion [HA ionized / [HA initil x 100% % ioniztion 4. x 10 - mol/l / 1.00 mol/l x 100% % ionized 0.4% Percent ioniztion In prt b) of n erlier problem n cetic cid solution with initil concentrtion of mol/l t equilibrium hd [H O [HA ionized 1. x 10-4 mol/l % ioniztion [HA ionized / [HA initil x 100% % ioniztion 1. x 10-4 mol/l / mol/l x 100% % ioniztion 1% 1 8

9 Figure Equilibri in Solutions of Wek Bses The dissocition of wek bse is n equilibrium sitution with n equilibrium constnt, clled the bse dissocition constnt, b bsed on the eqution B (q) H O (l) BH (q) OH - (q) b [BH [OH [B 4 Equilibri in Solutions of Wek Bses The bse dissocition constnt, b is lwys bsed on the rection of one mole of the wek bse with wter. If you see the symbol b, it lwys refers to blnced eqution of the form B (q) H O (l) BH (q) OH - (q) Equilibri in Solutions of Wek Bses Our pproch to solving equilibri problems involving bses is exctly the sme s for cids. 1. Set up the ICE tble. Estblish the equilibrium constnt expression. Mke simplifying ssumption when possible 4. Solve for x, nd then for eq m mounts 5 6 9

10 Strychnine (C 1 H N O ), dedly poison used for killing rodents, is wek bse hving b 1.8 x Clculte the ph if [C 1 H N O initil 4.8 x 10-4 mol/l (ll in mol/l) C 1H N O (q) H O (l) C 1H N O (q) OH - (q) Initil conc. 4.8 x 10-4 N/A Conc. chnge -x N/A x x Equil. conc. 4.8 x 10-4 x N/A x x 6 [C1H N O [OH 6 b 1.8 x 10 so 1.8 x 10-4 [C H N O (4.8 x 10 x) 1 Check the initil bse concentrtion / b rtio 4.8 x 10-4 / 1.8 x which is greter thn 100 We re probbly good to mke simplifying ssumption tht x << [C 1 H N O i [C1HNO [OH 6 b 1.8 x 10 so 1.8 x 10-4 [C H N O (4.8 x 10 x) 1 The ssumption we will mke is tht x << [C 1 H N O i so [C 1 H N O eqm [C 1 H N O I 1.8 x 10-6 x / 4.8 x 10-4 x (1.8 x 10-6 )(4.8 x 10-4 ) x x x ±.9 4 x 10-5 mol/l Since x [OH -, the nswer must be the positive vlue, x [C 1 H N O [OH -.9 x 10-5 mol/l [C 1 H N O 4.8 x 10-4 mol/l.9 x 10-5 mol/l 4.5 x 10-4 mol/l. We should check the ssumption! x [ C H N O 1 i.9 x x M % M

11 x [ C H N O 1 i.9 x x 10 M % M In this cse, the error is more thn 5%. I will leve it to you to go bck nd use the qudrtic formul. Compre the two nswers -5-4 To continue towrds the nswer of the problem AS IF the ssumption WERE VALID poh - log [OH - poh - log.9 x 10-5 poh 4.54 ph poh ph poh ph (4.54) ph Reltion Between nd b Reltion Between nd b The strength of n cid in wter is expressed through, while the strength of bse cn be expressed through b Since Brønsted-Lowry cid-bse rections involve conjugte cid-bse pirs there should be connection between the vlue nd the b vlue of conjugte cid-bse pir. HA (q) H O (l) H O (q) A - (q) [ H O [ A [ HA A - (q) H O (l) OH - (q) HA (q) b [ OH [ HA [ A

12 Since these rections tke plce in the sme beker t the sme time let s dd them together HA (q) H O (l) A- (q) H O (l) H O (q) A- (q) OH - (q) HA (q) H O (l) H O (q) OH - (q) The sum of the rections is the dissocition of wter rection, which hs the ion-product constnt for wter HA (q) H O (l) A- (q) H O (l) H O (q) A- (q) OH - (q) HA (q) H O (l) H O (q) OH - (q) x w [H O [OH x t 5 C Closer inspection shows us tht [ HO [ A [ OH [ HA 14 HO OH w 1.0x10 [ HA A b [ o [ [ t 5 C As the strength of n cid increses (lrger ) the strength of the conjugte bse must decrese (smller b ) becuse their product must lwys be the dissocition constnt for wter w. Strong cids lwys hve very wek conjugte bses. Strong bses lwys hve very wek conjugte cids. Since x b w then w / b nd b w /

13 ) Piperidine (C 5 H 11 N) is n mine found in blck pepper. Find b for piperidine in Appendix C, nd then clculte for the C 5 H 11 NH ction. b 1. x 10 - b) Find for HOCl in Appendix C, nd then clculte b for OCl -..5 x 10-8 Acid-Bse Properties of Slts When cids nd bses rect with ech other, they form ionic compounds clled slts. Slts, when dissolved in wter, cn led to cidic, bsic, or neutrl solutions, depending on the reltive strengths of the cid nd bse we derive them from. Strong cid Strong bse Neutrl slt solution Strong cid Wek bse Acidic slt solution Wek cid Strong bse Bsic slt solution Slts tht Yield Neutrl Solutions Strong cids nd strong bses rect to form neutrl slt solutions. When the slt dissocites in wter, the ction nd nion do not pprecibly rect with wter to form H O or OH -. Slts tht Yield Neutrl Solutions Strong bse ctions like the lkli metl ctions (Li, N, ) or lkline erth ctions (C, Sr, B, but NOT Be ) nd strong cid nions such s Cl -, Br -, I -, NO -, nd ClO 4- will combine together to give neutrl slt solutions with ph

14 Slts tht Yield Neutrl Solutions Sodium chloride (NCl) will dissocite into N nd Cl - in wter. Cl - hs no cidic or bsic tendencies. Cl - (q) H O (l) no rection Chloride ions DO NOT HAVE hydrolysis rections with wter since it is the conjugte of strong cid, which mkes it very, very wek. Slts tht Yield Neutrl Solutions N hs no cidic or bsic tendencies. N (q) H O (l) no rection Sodium ions DO NOT HAVE hydrolysis rections with wter since it is the conjugte of strong bse, which mkes it very, very wek Slts tht Yield Acidic Solutions The rection of strong cid with nions like Cl -, Br -, I -, NO -, nd ClO - 4 with wek bse will led to n cidic slt solution. The solution is cidic becuse the nion shows no cidic or bsic tendencies, but the ction does, s it is the conjugte cid of wek bse. Slts tht Yield Acidic Solutions Ammonium chloride (NH 4 Cl) will dissocite into NH 4 nd Cl - in wter. Cl - hs no cidic or bsic tendencies. Cl - (q) H O (l) no rection Chloride ions DO NOT HAVE hydrolysis rections with wter since it is the conjugte of strong cid, which mkes it very, very wek

15 Slts tht Yield Acidic Solutions NH 4 hs cidic tendencies. Tht is: NH 4 (q) H O (l) NH (q) H O (q) Ammonium ions hydrolyze in wter becuse it is the conjugte cid of the wek bse NH, which mens mmonium is wek cid. Slts tht Yield Bsic Solutions The rection of strong bse with ctions like Li, N,, C, Sr, nd B with wek cid will led to n bsic slt solution. The solution is cidic becuse the ction shows no cidic or bsic tendencies, but the nion does, s it is the conjugte bse of wek cid Slts tht Yield Bsic Solutions Sodium fluoride (NF) will dissocite into N nd F - in wter. N (q) H O (l) no rection Sodium ions DO NOT HAVE hydrolysis rections with wter since it is the conjugte of strong bse, which mkes it very, very wek. Slts tht Yield Bsic Solutions F - hs bsic tendencies. Tht is: F - (q) H O (l) HF (q) OH - (q) Fluoride ions hydrolyze in wter becuse it is the conjugte bse of the wek cid HF, which mens fluoride is wek bse

16 Predict whether the following slt solution is neutrl, cidic, or bsic nd clculte the ph. 0.5 mol/l NH 4 Br NH hs b vlue of 1.8 x 10-5 (ll in mol/l) NH 4 (q) H O (l) H O (q) NH (q) Initil conc. 0.5 N/A Conc. chnge -x N/A x x Equil. conc. 0.5 x N/A x x 5.5 x [HO [NH [NH 4 so 5.5 x (0.5 x) Initil cid [HA / rtio is 0.5 / x x 10 8 we cn probbly ssume 0.5 >> x x x / 0.5 x (5.5 6 x )(0.5) x 1. 9 x x 1. 9 x x x 10-5 mol/l 61 6 Negtive nswer not physiclly possible so therefore, [H O x 10-5 mol/l [ NH x 10 M x x M x - 4 i Since we ve shown the ssumption is vlid ph -log [H O - log x % Slts tht Contin Acidic Ctions nd Bsic Anions If slt is composed of n cidic ction nd bsic nion, the cidity or bsicity of the slt solution depends on the reltive strengths of the cid nd bse

17 Slts tht Contin Acidic Ctions nd Bsic Anions If the cid ction is stronger thn the bse nion, it wins nd the slt solution is cidic. If the bse nion is stronger thn the cid ction, it wins nd the slt solution is bsic. 65 Slts tht Contin Acidic Ctions nd Bsic Anions > b the cid ction is stronger nd the slt solution is cidic. < b the bse nion is stronger nd the slt solution is bsic. b the slt solution is close to neutrl. 66 Clssify ech of the following slts s cidic, bsic, or neutrl: ) Br b) NNO c) NH 4 Br d) NH 4 F for HF 6.6 x 10-4 b for NH 1.8 x 10-5 The Common-Ion Effect Solutions consisting of both n cid nd its conjugte bse re very importnt becuse they re very resistnt to chnges in ph. Such buffer solutions regulte ph in vriety of biologicl systems

18 The Common-Ion Effect Let s consider solution mde of 0.10 moles of cetic cid nd 0.10 moles of sodium cette with totl volume of 1.00 L, mking the initil [CH COOH [CH COO mol/l. First we must identify ll potentil cids nd bses in the system. CH COOH CH COO - N H O cid bse neutrl cid or bse Point of view of the cid Our rection will be CH COOH (q) H O (l) H O (q) CH COO - (q) 1.8 x 10-5 (ll in mol/l) CH COOH (q) H O (l) H O (q) CH COO - (q) Initil conc N/A Conc. chnge -x N/A x x Equil. conc x N/A x 0.10 x Note tht the initil concentrtion of our product CH COO - is NOT ZERO! [HO [CHCOO [CH COOH 1.8 x 10 5 (x)(0.10 x) (0.10 x) Let s check the initil cid concentrtion / rtio / 1.8 x It s probbly sfe to ssume tht x << [HAc i so [HAc eqm [HAc I nd x << [Ac - i so [Ac - eqm [Ac - i 1.8 x 10-5 x (0.10 x) / (0.10 x) 1.8 x 10-5 x (0.10) / (0.10) x 1.8 x 10-5 mol/l 71 [HO [CHCOO 1.8 x 10 [CH COOH At equilibrium, [H O 1.8 x 10-5 mol/l [CH COO x mol/l [CH COOH x mol/l Assumption ws vlid! Check for yourself! ph - log [H O ph - log 1.8 x ph 4.74 (x)(0.10 x) (0.10 x) 7 18

19 If we hd strted out with only 0.10 mol/l cetic cid, the ph would be found from (ll in mol/l) CH COOH (q) H O (l) H O (q) CH COO - (q) Initil conc N/A Conc. chnge -x N/A x x Equil. conc x N/A x x [HO [CHCOO [CH COOH 1.8 x 10 5 (0.10 x) The initil cid concentrtion / will still be the sme, so we cn ssume x << [HAc i so [HAc eqm [HAc i 1.8 x 10-5 x / (0.10 x) 1.8 x 10-5 x / (0.10) x 1.8 x 10-6 mol/l x ±1. x 10 - mol/l (cn t be ve) ph - log [H O ph - log 1. x 10 - ph CH COOH (q) H O (l) H O (q) CH COO - (q) Without the cette ion the ph of 0.10 M cetic cid is.89. With n equl concentrtion of cette ion present, the ph of 0.10 M cetic cid 0.10 M cette is 4.74 The cette ion mkes lrge difference on the equilibrium ph! Adding the conjugte bse ( stress!) to the equilibrium system of n cid dissocition shows the commonion effect, where the ddition of common ion cuses the equilibrium to shift. This is n exmple of Le Chtlier s Principle. Addition of the wek bse to the cid dissocition

20 Clculte the concentrtions of ll species present, nd the ph in solution tht is 0.05 mol/l HCN nd mol/l NCN. ( of HCN 4.9 x ) (ll in mol/l) HCN (q) H O (l) H O (q) CN - (q) Initil conc N/A Conc. chnge -x N/A x x Equil. conc x N/A x x [HO [CN [HCN 4.9 x (x)(0.010 x) (0.05 x) The initil bse concentrtion / rtio is / 4.9 x x 10 7 It s probbly sfe to ssume tht x << [HCN i so [HCN eqm [HCN I nd x << [CN - i so [CN - eqm [CN - i x x (0.010 x) / (0.05 x) 4.9 x x (0.010) / (0.05) x 1. x 10-9 mol/l So t equilibrium, [H O 1. x 10-9 mol/l [CN x mol/l [HCN x mol/l. Assumption ws vlid! Check this for yourself! 79 ph - log [H O ph - log 1. x 10-9 ph

21 Buffer Solutions Solutions tht contin both wek cid nd its conjugte bse re buffer solutions. These solutions re resistnt to chnges in ph. Buffer Solutions If more cid (H O ) or bse (OH - ) is dded to the system, the system hs enough of the originl cid nd conjugte bse molecules in the solution to rect with the dded cid or bse, nd so the new equilibrium mixture will be very close in composition to the originl equilibrium mixture Buffer solutions A 0.10 mol L -1 cetic cid 0.10 mol L -1 cette mixture hs ph of 4.74 nd is buffer solution! CH COOH (q) H O (l) H O (q) CH COO - (q) (ll in mol/l) CH COOH (q) H O (l) H O (q) CH COO - (q) Initil 0.10 N/A Conc. chnges -x N/A x x Equil. conc x N/A x 0.10 x 1.8 x 10 5 [HO [CHCOO [CH COOH Buffer solutions If we rerrnge the expression to solve for [H O [H O [CHCOOH 1.8 x 10 [CH COO [CHCOOH [CH COO

22 Buffer solutions [H O [CHCOOH 1.8 x 10 [CH COO [CHCOOH [CH COO 5 Assume x << [HAc i so [HAc eqm [HAc I nd x << [Ac - i so [Ac - eqm [Ac - i, nd we should see If [CH COOH i [CH COO - i, then [H O 1.8 x 10-5 M nd ph p Buffer solutions Wht hppens if we dd 0.01 mol of NOH (strong bse) to 1.00 L of the cetic cid cette buffer solution? CH COOH (q) OH - (q) H O (l) CH COO - (q) This rection goes to completion nd keeps occurring until we run out of the limiting regent OH - (ll in moles) CH COOH (q) OH - (q) H O (l) CH COO - (q) Initil N/A 0.10 Chnge -x -x N/A x Finl (where x 0.01 due to limiting OH - ) 0.10 x x 0.00 N/A 0.10 x 0.11 New [CH COOH 0.09 M nd new [CH COO M 86 Buffer solutions (ll in mol/l) CH COOH (q) H O (l) H O (q) CH COO - (q) Initil conc N/A Conc. chnge -x N/A x x Equil. conc x N/A x 0.11 x [HO [CHCOO 5 (x)(0.11 x) 1.8 x 10 With the ssumption [CH COOH tht x is much (0.09 smller x) thn 0.09 mol (n ssumption we lwys need to [CHCOOH [HO 1.8 x x 10 M check fter [CH clcultions COO re done!), 0.11 we find Note we ve mde the ssumption tht x << 0.09 M! ph - log [H O ph - log 1.5 x 10-5 ph 4.8 Buffer solutions Adding 0.01 mol of OH - to 1.00 L of wter would hve given us ph of 1.0 becuse there is no significnt mount of cid in wter for the bse to rect with. Our buffer solution resisted this chnge in ph becuse there is significnt mount of cid (cetic cid) for the dded bse to rect with

23 Buffer solutions Wht hppens if we dd 0.01 mol of HCl (strong cid) to 1.00 L of the cetic cid cette buffer solution? CH COO - (q) H O (q) H O (l) CH COOH (q) This rection goes to completion nd keeps occurring until we run out of the limiting regent H O (ll in moles) CH COO - (q) H O (q) H O (l) CH COOH (q) Initil N/A 0.10 Chnge -x -x N/A x Finl (where x 0.01 due limiting H O ) 0.10 x x 0.00 N/A 0.10 x 0.11 New [CH COOH 0.11 M nd new [CH COO M Buffer solutions (ll in mol/l) CH COOH (q) H O (l) H O (q) CH COO - (q) Initil conc N/A Conc. chnge -x N/A x x Equil. conc x N/A x 0.09 x [HO [CHCOO 5 (x)(0.09 x) 1.8 x 10 With the ssumption [CH COOH tht x is much smller thn ( x) mol (n ssumption we lwys need to check fter [CHCOOH [HO clcultions re 1.8done!), x 10 we find. x 10 M [CH COO 0.09 Note we ve mde the ssumption tht x << 0.09 M! ph - log [H O ph - log. x 10-5 ph Buffer solutions Adding 0.01 mol of H O to 1.00 L of wter would hve given us ph of.0 becuse there is no significnt mount of cid in wter for the bse to rect with. Our buffer solution resisted this chnge in ph becuse there is significnt mount of bse (cette) for the dded cid to rect with. 91 9

24 Buffer cpcity Buffer cpcity is the mesure of the bility of buffer to bsorb cid or bse without significnt chnge in ph. Lrger volumes of buffer solutions hve lrger buffer cpcity thn smller volumes with the sme concentrtion. Buffer solutions of higher concentrtions hve lrger buffer cpcity thn buffer solution of the sme volume with smller concentrtions. 9 Clculte the ph of L buffer solution tht is 0.5 mol/l in HF nd 0.50 mol/l in NF. (ll in mol/l) HF (q) H O (l) H O (q) F - (q) Initil conc. 0.5 N/A Conc. chnge -x N/A x x Equil. conc. 0.5 x N/A x 0.50 x [HO [F 4 (x)(0.50 x).5 x 10 [HF (0.5 x) Assume With the x ssumption << [HF tht x is much smller i so [HF eqm [HF i thn 0.5 mol (n ssumption we lwys need to check nd fter x << clcultions [F - i so [F - re eqm done!), [F - i we find 94 [H O [HF [F.5 x x ph - log [H O ph - log x 10-4 ph.76 [HF [F 5 x 10 4 M ) Wht is the chnge in ph on ddition of 0.00 mol of HNO? (ll in moles) F - (q) H O (q) H O (l) HF (q) Initil N/A 0.05 Chnge -x -x N/A x Finl (where x x 0.00 x N/A 0.05 x due to limiting H O ) New [HF 0.7 M nd new [F M (ll in mol/l) HF (q) H O (l) H O (q) F - (q) Initil conc. 0.7 N/A Conc. chnge -x N/A x x Equil. conc. 0.7 x N/A x 0.48 x [HO [F [HF.5 x 10 4 (x)(0.48 x) (0.7 x)

25 [H O [HF.5 x 10 [F x 10 4 Notice we ve mde the ssumption tht x << 0.7 M. We should check this! ph - log [H O ph - log x 10-4 ph.71 M b) Wht is the chnge in ph on ddition of mol of OH? (ll in moles) HF (q) OH - (q) H O (l) F - (q) Initil N/A Chnge -x -x N/A x Finl (where x x x N/A x due to limiting OH - ) New [HF 0.1 M nd new [F M (ll in mol/l) HF (q) H O (l) H O (q) F - (q) Initil conc. 0.1 N/A Conc. chnge -x N/A x x Equil. conc. 0.1 x N/A x 0.54 x.5x10 4 [HO [F [HF so.5x10 4 (x)(0.54 x) (0.1 x) [H O [HF.5 x 10 [F x 10 4 Notice we ve mde the ssumption tht x << 0.1 M. We should check this! ph - log [H O ph - log 1. 6 x 10-4 ph.87 M The Henderson-Hsselblch Eqution We ve seen tht, for buffer solutions contining members of conjugte cidbse pir, tht [cid [H O [bse [cid log [HO log log log [bse ph p log [bse / [cid This is clled the Henderson- Hsselblch Eqution. [cid [bse

26 The Henderson-Hsselblch Eqution If we hve buffer solution of conjugte cid-bse pir, then the ph of the solution will be close to the p of the cid. This p vlue is modified by the logrithm of rtio of the concentrtions of the bse nd cid in the solution to give the ctul ph. Use the Henderson-Hsselblch Eqution to clculte the ph of buffer solution prepred by mixing equl volumes of 0.0 mol/l NHCO nd 0.10 mol/l N CO. We need the nd the concentrtions of the cid (HCO - ) nd the bse (CO - ). 5.6 x (we use the for the second proton of H CO!) NOTE: The concentrtions we re given for the cid nd the bse re the concentrtions before the mixing of equl volumes! 10 If we mix equl volumes, the totl volume is TWICE the volume for the originl cid or bse solutions. Since the number of moles of cid or bse DON T CHANGE on mixing, the initil concentrtions we use will be hlf the given vlues. ph p log [bse / [cid ph (-log 5.6 x ) log (0.05) / (0.10) ph ph