The nature acids: taste sour bases: taste bitter, feel slippery

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1 7 Acids and bases The nature acids: taste sur bases: taste bitter, feel slippery Definitin Arrhenius ( ) acid and base (1887) Acid Prduces hydrgen ins (H + ) in aqueus slutin Base Prduces hydride ins (OH ) in aqueus slutin Brønsted acid and base (19 by Brønsted and Lry) Acid: A prtn (H + ) dnr Base: A prtn acceptr Hydrnium in HA(aq) + H O(l) H O (aq) + A (aq) acid base cnjugate acid cnjugate base cnjugate ( 共軛 ) acid-base pair Can be represented as HA(aq) H (aq) + A (aq) Dissciatin cnstant a [ HO [A [H [A [HA [HA 1

2 E. NH (g) + HCl(g) NH Cl(s) Brønsted base Prtn acceptr Brønsted acid Prtn dnr [NH + [Cl Leis acid and base Acid: An electrn-pair acceptr Base: An electrn-pair dnr E. BF + :NH F BNH Leis acid Leis base Acid strength Recall: HA(aq) + H O(l) H O (aq) + A (aq) r HA(aq) H (aq) + A (aq) Dissciatin cnstant a [ HO [A [H [A [HA [HA

3 HA is a strng acid hen a is large r equilibrium lies far t the right a is a measurement f acid strength Strng acid gives eak cnjugate base Weak acid gives strng cnjugate base E. p a = lg a H O 15.7 NH H O H SO HCl 6 ~ 7 HI 9.5 ~ 10 Autinizatin f ater H O(l) H H O H (aq) + O H(aq) Amphteric: behaves as an acid r a base O H H H O acid H base (curved arrs represent fl f e s) = [H O + [OH In-prduct cnstant r dissciatin cnstant At 5 C = [H + = [OH = M neutral cnditin

4 The ph ph = lg[h + At neutral cnditin [H + = ph = 7.00 ph < 7.00 ph > 7.00 acidic basic Nte abut significant figures lg (1/5) = lg5 = lg (.5 10) = [( lg.5) + 1 eact number (frm the rder) 0.65 t significant figures = 1.65 E. [H + = ph = 7.00 t significant figures t significant figures Overall: three significant figures

5 Calculating ph Strng acid: cmplete dissciatin The ph f a miture f eak acids E M HCN a = M HNO a =.0 10 ph? H O(l) H (aq) + OH (aq) = HNO is the strngest acid, the ther t can be neglected HNO (aq) H + (aq) + NO (aq) ini (M) final 5.00 a Small (assumptin) =.5 10 M = [H + ph = 1.5 Check: [H + is indeed small cmpared t 5.00 M (tlerance: ±5%) 5

6 HCN (aq) H + (aq) + CN (aq) final (M) Cntrlled by HNO (assumptin) a 10 (.5 10 ) (.5 10 ) = M = [CN Check: [CN is small indeed als [H + is mainly frm HNO Percent dissciatin amt dissciated 100% initial cnc Fr eak acid: mre dilute higher percent dissciatin HA(aq) H (aq) + A (aq) a [HA percent dissciatin [HA a [HA [HA 100% a [HA [HA percent dissciatin 100% 6

7 Bases NaOH(aq), OH(aq) dissciate cmpletely in H O Strng bases Ca(OH) : similar but slubility is small des nt prduce high [OH Slaked lime ( 熟石灰 ) In acidic sln: Ca(OH) (aq) + H + (aq) Ca + (aq) + H O(l) Remval f Ca + in hard ater CaO + Na CO (the lime-sda prcess) CaO(s) + H O(l) Ca(OH) (aq) CO (aq) + H O(l) HCO (aq) + OH (aq) Ca(OH) (aq) + Ca + (aq) + HCO (aq) CaCO (s) + H O(l) in hard ater Other bases NH (aq) + H O(l) NH (aq) + OH (aq) ammnia Amine bases p a = 9.5 H C N H H C N N C H C 5 H 5 H HCH C H 5 methylamine dimethylamine triethylamine N pyridine 7

8 In general B(aq) + H O(l) BH (aq) + OH (aq) b [BH [OH [B b base strength = a b In practice the base strength is ften measured by the p a f its cnjugate acid p a base strength E. NH (aq) + H O(l) NH + (aq) + OH (aq) [ NH [OH 5 b [NH NH (aq) + H O(l) NH + (aq) + OH (aq) Ini 0.15 M 0 0 Final ~ M Abut 1% cnversin In a buffer f ph = 7. [OH ~ 10 7 M [NH [NH [NH [NH + ~ 0.15 M 7 [ NH [10 5 8

9 Plyprtic acids H SO, H PO Dissciate stepise H PO (aq) H (aq) + H PO (aq) H PO (aq) H (aq) + HPO (aq) HPO (aq) H (aq) + PO (aq) a1 = a = a = In general a1 >> a >> a E. Fr a 5.0 M H PO slutin H PO (aq) H (aq) + H PO (aq) a ini (M) final 5.0 a = M Within 5% errr H 8 [ [ HPO ( 0. 19)[ HPO [ HPO [ H PO ( 0. 19) M a [H [PO [HPO (0.19)[PO ( ) [PO = M 9

10 E. Fractins f H CO, HCO, and CO at ph 9.00 CO : [CO [H CO [HCO [CO 1 [HCO [HCO 1 [CO [CO a1 [H [HCO [H CO a1 [HCO H [H CO [ a [H [CO [HCO a [CO H [HCO [ a1 a [HCO [CO [CO H [H [H CO [HCO [H CO [ F CO 1 [HCO [HCO [H 1 [CO [CO a1 a 1 [H 1 a HCO : [H CO [HCO [HCO [CO 1 [HCO [CO 1 [HCO [HCO F HCO 1 1 [HCO [CO [H a 1 1 [HCO [HCO [H a1 At ph 9.0 F HCO = 0.95 ( a1 = ; a = ) 10

11 E. ph f a M H SO Cmplete dissciatin fr the 1st step HSO (aq) H (aq) + SO (aq) ini (M) final a 1.10 ( ) ( ) Assume >> = (. 10 ) (1. 10 ) = 0 incrrect =.5 10 M [H + = = ph = 1.80 Cf. 1.0 M H SO HSO (aq) H (aq) + SO (aq) ini (M) final a 1.10 (1.0 ) (1.0 ) Very small cntributin frm the secnd inizatin 11

12 Acid-base prperties f salts Cnjugate base f strng acid Cnjugate acid f strng base n effect n ph Cnjugate base f a eak acid NaC H O, CN, F E. 0.0 M NaF(aq) a fr HF: = a b b = F (aq) + H O(l) HF(aq) + OH (aq) ini (M) final 0.0 b = M = [OH Cnjugate acid f a eak base E. Ammnium salt NH Cl NH (aq) H (aq) + NH (aq) a = A 0.10 M slutin ph = 5.1 With highly charged metal in + E. Al(H O) 6 AlCl (s) + 6H O(l) Al(H O) + 6 (aq) + Cl (aq) Al(H O) + 6 (aq) Al(OH)(H O) + 5 (aq) + H (aq) a = A M slutin ph =. 1

13 Bth ins affect the ph E. NH + C H O ammnium acetate Basic; b = Acidic; a = a > b a < b a = b acidic basic neutral E. NH + CN b = basic E. ph f a M NH CN NH (aq) H (aq) + NH (aq) a = CN (aq) + H O(l) HCN(aq) + OH (aq) b = H O(l) H + (aq) + OH (aq) = NH (aq) + CN (aq) NH (aq) + HCN(aq) [NH [HCN [NH [H [HCN [NH [CN [NH [H [CN a(nh ) a(hcn) 0.90 The dminate prcess 1

14 NH (aq) + CN (aq) NH (aq) + HCN(aq) Final =.9 10 M = [NH = [HCN (0.100 ) [NH + = [CN = = M a(nh ) a(hcn) [H [NH [NH [H + = M [H (0.09) [H [CN [H (0.051) [HCN 0.09 [H + = M ph = 9. Anther ay [H [NH [H [CN a(nh ) a(hcn) [H [NH [HCN [H 10 a(nh ) a(hcn)

15 Dilute slutins The cntributin f H O is imprtant E. A M slutin f HA, a = HA(aq) H (aq) + A (aq) Final a [H + = M Can nt be true! Cnsider H O alne [H + = M H abut: = M Frm HA Frm ater incrrect Shuld effect each ther Epect: < [H + < M HA and H O shuld be cnsidered at the same time HA(aq) H (aq) + A (aq) H O(l) H (aq) + OH (aq) 15

16 Relatinships must be flled: 1. = [H + [OH.. Charge balance [H + = [OH + [A a. Material balance (mass balance) [HA = [HA + [A Initial cncentratin f HA nn:, a, [HA Need t kn: [H +, [OH, [HA, [A [H [A [HA Epress in terms f [H + Frm OH [H [ Frm charge balance [ A [H [OH [H [H Frm mass balance [ HA [HA [A [HA [H [H Frm a [H [H [H [A [H [H a [HA [HA [H [H [HA [H [H 16

17 a [H [H [HA [H Can be slved by successive apprimatins Validity check If assume [H + >> a [H [HA [H That is t mit H O Ges back t cnsider HA alne Assume this cnditin is met hen [H + > 100 [H + shuld be > 10 6 M E. 1.0 M HCN, a = Sln Nrmal calculatin [H + = M > 10 6, O E M HCN Sln Nrmal calculatin [H + = M N gd Slve a [H [H [H Epect: < [H + < M Try M first 17

18 a [ H [H + = M (cf ) Ne guess: Find: [H + = M Final anser = M A better apprimatin a [H [H [HA [H This term is small Assume [HA [H [H a [HA [ H [ H a [HA True unless [HA is very small E M HCN [H ( )( ) ( M 1 ) 18

19 The incrrect methd HCN(aq) H (aq) + CN (aq) Ini (M) final a 7 10 ( ) 6.10 ( ) Assume ~ M = M [H + = M ~15% larger than the crrect anser! E. Miing equal vlumes f M NH and M HCl Q: ph =? Sln: Start ith M NH + NH + (aq) NH (aq) + H (aq) final (M) 10 a = M 5 ( ) Can nt mit H O Since [NH + = M > 10 6 M [H 7 a [NH M 19

20 Dilute strng acid slutin E M HNO Frm HNO alne [H + = M The cntributin f H O can nt be ignred Charge balance [H + = [NO + [OH = [H + [OH nn: [NO = M Unknn: [H + and [OH Sln: [OH = /[H + Frm charge balance: [H + = /[H + [H + = M Mre eamples A slutin f NaHCO The principal equilibrium is HCO (aq) + HCO (aq) H CO (aq) + CO (aq) Q: =? Sln: [HCO[CO [HCO [HCO Q: ph at equilibrium? Sln: a1 a [HCO [H [CO [HCO [H [HCO [H [HCO [H [CO [H [CO [H CO [HCO [H CO At equilibrium: [H CO = [CO [ H a1 a a a1 0

21 At 1 atm, 5 C, HA has a vapr density f 5.11 g/l. A ml slutin f 1.50 g HA has a ph f Q: a =? Sln: The mlar vlume f a gas is.6 L at 1 atm, 5 C Mlar mass f HA shuld be 5.11 g/l.6 L/ml = 16 g/ml The mle number f 1.50 g f HA is 1.50 g 16 g/ml = ml = 11.9 mml The cncentratin is 11.9 mml M ml lg[h + = 1.80 lg[h + = 1.80 = ning lg(1.6) = 0.0 [H + = M HA(aq) H (aq) + A (aq) Equil. ( ) a =.5 10 A M salt (BHX) slutin has a ph f 8.00 This salt is a cmbinatin f BH + and X B(aq) + H O(l) BH + (aq) + OH (aq) Q: HX =? Sln: [ BH [OH b [B HX(aq) H (aq) + X (aq) The principal equilibrium is BH (aq) + X (aq) B(aq) + HX(aq) At equilibrium: [B = [HX [BH = [X [BH [X [B [HX 1

22 ph 8.00 [H + = M [OH = M 6 [ BH [OH [BH ( ) [B [B [ BH [B [X [HX [ H [X 8 5 HX ( )( ) [HX M HCOOH, a(hcooh) = M AcOH, a(acoh) = Q: ph? Sln: [ H [HCOO a(hcooh) [HCOOH [ H [AcO 5 a(acoh) [AcOH Charge balance [H + = [HCOO + [AcO + [OH Assuming [OH is very small: [H + = [HCOO + [AcO Mass balance [HCOOH = [HCOOH + [HCOO = M [AcOH = [AcOH + [AcO = M

23 [H [HCOO [ HCOOH [H [HCOO [HCOO M [H [ HCOO ( 1) M [H [AcO [ AcOH 5 [H [AcO [AcO M [H [ AcO ( 1) M Frm charge balance: [H + = [HCOO + [AcO [H [H [H H [H [H [ 5 Make a guess first [H + frm HCOOH alne:.9 10 [H + frm AcOH alne: > [H + >.9 10 Start frm.6 10

24 [H M Ne guess: ( ) [H + =. 10 M Anther quick slutin [H + = [HCOO + [AcO [H + = [H + [HCOO + [H + [AcO = ( )[HCOOH + ( )[AcOH Assume [HCOOH ~ [HCOOH = M [AcOH ~ [AcOH = M [H + =.0 10 M Check the assumptin:.0 10 is 5.% f Calculate [OH in a M slutin f Ca(OH) Sln: In slutin, Ca(OH) is cmpletely inized [Ca + = M The cncentratin f [OH is small and shuld cnsider the inizatin f H O Charge balance [Ca + + [H + = [OH Frm: = [H + [OH [Ca [OH [OH [OH = M