Chapter 16 Acid Base Equilibria
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1 Chpter 16 Acid Bse Equilibri 16.1 Acids & Bses: A Brief Review Arrhenius cids nd bses: cid: n H + donor HA(q) H(q) A(q) bse: n OH donor OH(q) (q) OH(q) Brønsted Lowry cids nd bses: cid: n H + donor HA(q) H(q) A(q) bse: n H + cceptor H(q) B(q) BH(q) 16. Brønsted Lowry Acids & Bses The H + Ion in Wter H3O is clled the hydronium ion nd is wht truly hppens when H + is in wter since ll queous solution re in wter we will use H 3 O + & H + interchngebly Conjugte Acid Bse Pirs conjugte cid: the cid tht is creted fter the Brønsted Lowry bse hs ccepted the proton, BH + conjugte bse: the bse tht is creted fter the Brønsted Lowry cid hs donted the proton, A exmples: bsic: NH3( g) H O( l) NH4( q) OH( q) Kb bse cid conjcid conjbse bse conjbse conjcid cidic: HCl H O Cl H O cid Reltive Strengths of Acids & Bses strong cid will completely dissocite/ionize in solution: HNO H O H O NO 3( q) ( l ) 3 ( q) 3( q) s soon s strong cid is plced in wter it will ionize completely ll the rectnt goes to product product is very wek conjugte cid nd bse other strong cids: HCl, HBr, H SO 4, HI, HClO 4, HClO 3, HNO 3 the sme is the cse for strong bse
2 exmples of strong bses: OH (where is lkli metl), NH, H wek cid will only prtilly dissocite 4 HNO ( q) H ( q) NO ( q) the eq constnt is clled K for cid in this cse some of the rectnt is present t eq the lrger the K the more n cid dissocites, the stronger the cid e.g. for nitric cid K >> 1 we tlk bout K 's t length little lter 16.3 The Autoioniztion of Wter s we will soon lern wter my ct s either n cid or bse mphoteric wter s bse: HA(q) H O(l) H3 O(q) A(q) wter s n cid: B(q) H O(l) BH(q) OH(q) Autoioniztion occurs when species cn ionize itself this occurs for wter s shown below: HO ( l) OH ( q) HO 3 ( q) Kc [ OH ( q) [ HO 3 ( q) since wter is in excess (it s the solvent) the concentrtion is constnt Kw [ OH( q) [ H3 O( q) reltionship between [ OH ( q) nd [ HO 3 ( q) [ H3 O( q) [ OH( q) [ H3 O( q) [ OH( q) cidic neutrl [ H O [ OH bsic 3 ( q) ( q) the vlue for both [ OH ( q) nd [ HO 3 ( q) in pure wter re 1.0 x 10 7 t 98K K w (1.010 )(1.010 ) Exmple: Determine the hydroxide concentrtion in solution with 4 [ H O ( q) K K H O OH OH 14 w 11 w [ 3 ( q) [ ( q) [ ( q) [ HO 3 ( q) The ph Scle ph is log scle which describes the "power of hydrogen" the ph is relted to the [ HO 3 ( q) : ph log[ H3 O ( q) 7 for pure wter: ph log[ H3 O ( q) log(1.010 ) 7 this is neutrl ph when solution hs ph < 7.0 it is cidic when solution hs ph > 7.0 it is bsic poh & Other p Scles ll p scles men log of poh is exctly like ph except it is dependent upon [ OH ( q) : poh = log[ OH ( q) In Chpter 17 we will hve pk nd pk b mening the log(k ) etc. the ph nd poh re relted to K w
3 Kw [ OH( q) [ H3 O( q) log Kw log{[ OH( q) [ H3 O( q) } log Kw log[ OH( q) log[ H3 O( q) 14 pk poh ph log(1.010 ) 14 w esuring ph you my hve used litmus pper in GenChem I lb to see if something ws more or less bsic phenolphthlein is pink in bsic solution nd ppers cler in cidic s you cn see bove This will become more relevnt in Chpter 17 when we tlk bout titrtions when we tlk bout color indictors 16.5 Strong Acids & Bses s we previously discussed strong cids nd bses completely dissocite in wter therefore, whtever the concentrtion of our strong cid or bse will be the concentrtion of the H + nd OH, respectively Exmples: Write the blnced eqution for ech of the following nd determine the ph..) HClO 4(q) HClO4(q) H O(l) H3 O(q) ClO4(q) [H3 O (q) [HClO 4(q) ph log( ) 0.30 b.) LiOH (q) LiOH(q) Li(q) OH(q) [OH(q) [LiOH (q) ph 14 poh 14 ( log(0.056 )) 1.41 Exmple: Determine the hydronium ion concentrtion for C(OH). C(OH) (q) C(q) OH(q) OH (q) C(OH) (q) K w [H3 O (q) OH (q) 16.6 Wek Acids wek cids nd bses do not completely dissocite in wter therefore t eq they re still present: ( q) ( l) 3 ( q) ( q) ( q) ( q) ( q) HA H O H O A or HA H A K HO 3 ( q) A ( q) H ( q) A ( q) HA HA the K is clled the cid dissocition constnt nd gives us n ide of cid strength
4 the lrger the K the more strongly the equilibrium will lie towrd product the more likely the cid is to dissocite nd rise the cidity of the solution the stronger the cid Clculting K from ph Similr to wht we did lst chpter with the %dissocition, we cn get the K from the ph Exmple: Wht is the K of solution of HC 7 H 5 O if the ph of this solution is.75? We begin with n ICE Tble HC H O C H O H 7 5 ( q) 7 5 ( q) ( q) HC7H5O ( q) CHO 7 5 ( q) H ( q) Initil Chnge x +x +x Eq x +x +x Next, we write down symbolic representtion of the equilibrium/cid dissocition expression: [ CHO 7 5 ( q) [ H( q) [ HC H O 7 5 ( q) Now we remember we don t know K but need to find it which mens we must know x: ph.75 3 [ H ( q) Finlly we plug in x for our equilibrium expression nd solve [ CHO q [ H q x [ HC7H5O( q) x Percent Ioniztion degree of ioniztion/dissocition: percentge tht n cid ionizes 7 5 ( ) ( ) 5 [ H HA( q) H( q) A( q) 100% [ HA Exmple: Determine the percent dissocition of of benzoic cid. HC7H5O( q) C7H5O( q) H( q) We lredy will find [ H therefore 5 3 [ H % 3.54% HA It should be smll since our K is so smll Using K to Clculte ph/poh we use ICE tbles
5 Ex: Clculte [H + nd the poh of of benzoic cid. HC7H5O( q) C7H 5O( q) H( q) HC7H5O ( q) 5 CHO 7 5 ( q) H ( q) Initil Chnge x +x +x Eq x +x +x [ CHO 7 5 ( q) [ H( q) x [ HC7H5O( q) x x x x x x x x [ H ph poh poh 14 ph 14 log[ H 11.5 Polyprotoic Acids for cids contining 1 proton we cll them monoprotic for two they re diprotic (e.g. H SO 4 ) for three they re clled triprotic (H 3 PO 4 ) Exmple: Clculte the [H + of of sulfuric cid. HSO 4( q) H( q) HSO4( q) 1 HSO H SO K ( q) ( q) 4( q) Initilly ll of the H SO 4 dissocites completely into H( q) HSO4( q) HSO H 4( q) ( q) SO 4( q) Initil Chnge x +x +x Eq x x +x [ H( q) [ SO4( q) (0.050 xx ) 1.10 [ HSO4( q) (0.050 x) xx x 4 x 0.06x x [H + = = x x Why is K 1 > K? HSO H HSO K 1 4( q) ( q) 4( q) HSO H SO K ( q) ( q) 4( q)
6 electrostticlly, it is more difficult to remove H+ from SO 4( q) thn HSO 4( q) therefore, the second K is lwys smller thn the first 16.7 Wek Bses we hndle them just like we do wek cids this time our eq const is the bse dissocition constnt or K b Exmple: Clculte the ph of NH 3. NH3( q) H O OH ( q) NH 4( q) Kb NH 3( q) 5 OH ( q) NH 4( q) Initil Chnge x +x +x Eq x +x +x [ OH( q) [ NH 4( q) x [ NH3( q) x x x x x x 4 x [ OH ph 14 log[ OH Reltionship Between & Kb K w = K *K b flls out of the reltionship tht we hve lredy discussed K totl = K 1 *K * Exmple: Determine the K b of HCN if K = 4.9 x Kw Kw Kb Kb K Acid Bse Properties of Slt Solutions the stronger prtner lwys domintes: strong cid + strong bse = neutrl soln strong cid + wek bse = cidic soln wek cid + strong bse = bsic soln Exmple: Clssify the following solutions s bsic, cidic, or neutrl..) KBr b.) NNO c.) NH 4 Cl Answer:.) Neutrl b.) bsic c.) cidic Exmple: Clculte the K for the ction & the K b for the nion in n queous solution contining NH 4 CN. Is the solution cidic, bsic, or neutrl?
7 b Kw b Kb CN we will hve to use the of HCN to get its Kb b? Kw b b 4 the soln is bsic NH CN NH CN for NH : NH H O NH H O? we will not find this K in tble BUT we cn find the K of NH to it: NH H O NH OH K K for CN H O HCN OH K HCN H O CN H O K K K CN K NH Exmple: Clculte the ph of 0.5 NC H 3 O, K = 1.76x Kw Kb K ( q) ( l) 3 ( q) ( q) CHO HO HCHO OH CHO HCH3 O ( q) 3 ( q) OH q Initil Chnge x +x +x Eq 0.50 x +x +x [ HCH3 O( q) [ OH x 10 Kb [ CHO 0.50 x 3 ( q) becuse we hve lrge concentrtion of cette & smll K b we will try nd ssume 0.50 >> x x x x x ck : 100% 0.005% 5% 0.50 therefore our ssumption is vlid nd [OH = 4.77x10 5 ph = 14 poh = 14 + log[oh = Acid Bse Behvior & Chemicl Structure electronegtivity drw trend cid strength down column in the periodic tble s we go down column we decrese EN more EN tom will hve stronger bond to H + if the H X bond is strong then it will be hrd to dissocite HX this will led to weker cid in solution cid strength: HF < HCl < HBr < HI cid strength going cross row in the periodic tble
8 s we go cross row from left to right we increse EN we lso increse polrity of the H X bond this mens tht if the H X bond is broken it does not necessrily led to H + (e.g. CH 4 ) if we don't hve H + then we hve weker cid cid strength: CH 4 < NH 3 < H O < HF oxocids: H n XO m electronegtivity effects for oxocids of the hlogens the trend is: HOI < HOBr < HOCl this trend occurs becuse s we increse the EN of the hlogen we re pulling electron density wy from the O tom if we tke electron density wy from the O tom we weken the O H bond this will in turn llow H + the freedom to brek wy nd go into soln number of oxygens in the cid (oxocids) if we increse the number of O toms in n cid then we increse its strength this becuse we increse the oxidtion number of the centrl tom Exmple: Oxocids of Chlorine Acid Oxidition Stte of Cl K HClO +1.9 x 10 8 HClO x 10 HClO HClO x Lewis Acids & Bses Definition: bse is substnce tht dontes pirs of electrons n cid is substnce tht ccepts pirs of electrons e.g. NH3BF3 H3NBF3 where NH 3 dontes e 's (bse) nd B is the cceptor (cid)
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