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S  6z `P  z P*    6\ `  z R*  H  0޽h ? ̙3380___PPT10.D  0X(   $+ X X 0P P    X*  X 0h     Z*  X 6 `P   X*  X 6  `   Z* H X 0޽h ? ̙3380___PPT10.  0L0 P!"(    0 ,ACIDS AND BASES A guide for A level studentsD-(26f6ff  0 OKNOCKHARDY PUBLISHING(2fJ   C "A KTREj x S _ ! 00"`>B 2008 SPECIFICATIONS<G$G   B  s *޽h ? ̙33 0L0 y`(    0Se \ Acid & Bases4 8_2 (f<f  0`p 'INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... www.knockhardy.org.uk/sci.htm Navigation is achieved by... either clicking on the grey arrows at the foot of each page or using the left and right arrow keys on the keyboard (2 2(2 2  75,  0,dB  <D?  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0B  s *޽h ? ̙33t  0L0    (  j  0 H+(aq) + Cl(aq) BASE proton acceptor NH3 (aq) + H+(aq)   > NH4+(aq) FHt@  f f     )             p 0q~ ]ACIDS AND BASES2(2ffB p s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 t)(  tdB t <D)P0Q t 0Pp@ 0 0dB t@ <D)PXP t 0X0@ 0 0L t 0nN  BRNSTED-LOWRY THEORY ACID proton donor HCl   > H+(aq) + Cl(aq) BASE proton acceptor NH3 (aq) + H+(aq)   > NH4+(aq) Conjugate systems Acids are related to bases ACID PROTON + CONJUGATE BASE Bases are related to acids BASE + PROTON CONJUGATE ACID H<@P(  f f      )             t 0 nq~ ]ACIDS AND BASES2(2ffF ? t a 8N   t fB  t 6DolB  t <BDo{>T   t# L)?fB  t 6DolB t <BDo{F ? t 0 8N  t fB t 6DolB t <BDo{>T  t# L)?fB t 6DolB t <BDo{B t s *޽h ? ̙33y___PPT10Y+D=' = @B +~ 0L0 ##x#(  xdB x <D)P0Q x 0Pp@ 0 0dB x@ <D)PXP x 0X0@ 0 0 x 00n . BRNSTED-LOWRY THEORY ACID proton donor HCl   > H+(aq) + Cl(aq) BASE proton acceptor NH3 (aq) + H+(aq)   > NH4+(aq) Conjugate systems Acids are related to bases ACID PROTON + CONJUGATE BASE Bases are related to acids BASE + PROTON CONJUGATE ACID For an acid to behave as an acid, it must have a base present to accept a proton... HA + B BH+ + A acid base conjugate conjugate acid base example CH3COO + H2O CH3COOH + OH base acid acid base H<@P(@  f f      ,fffb33 33333M)           }  x 0?nq~ ]ACIDS AND BASES2(2ffF ? x   F  8N   x fB  x 6D3olB  x <BD3o{>T   x# L)?fB  x 6D3olB x <BD3o{F ? x a 8N  x fB x 6DolB x <BDo{>T  x# L)?fB x 6DolB x <BDo{F ? x 0 8N  x fB x 6DolB x <BDo{>T  x# L)?fB x 6DolB x <BDo{F ? x Ww 8N  x fB x 6D3olB  x <BD3o{>T  !x# L)?fB "x 6D3olB #x <BD3o{B x s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0    l ( l4 dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0}  04Un WLEWIS THEORY ACID lone pair acceptor BF3 H+ AlCl3 BASE lone pair donor NH3 H2O  HX<)!@  ffffffffffff X   0Pnq~ ]ACIDS AND BASES2(2ffL  C $A NH4+g(x P   C (Adatcovg l   6dln̙ ?pz  YLONE PAIR DONOR.(2     6qn̙ ?`(  ZLONE PAIR ACCEPTOR,     6\vn̙ ?0O YLONE PAIR DONOR.(2     6zn̙ ?X ` ZLONE PAIR ACCEPTOR,  B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 |D(  | | 0n0 STRONG ACIDS completely dissociate (split up) into ions in aqueous solution e.g. HCl   > H+(aq) + Cl(aq) MONOPROTIC 1 replaceable H HNO3   > H+(aq) + NO3(aq) H2SO4   > 2H+(aq) + SO42-(aq) DIPROTIC 2 replaceable H s  @&PG@ f: fff"ff ff fffffff ff ffff"t`       /  Y dB | <D)P0Q | 0Pp@ 0 0dB |@ <D)PXP | 0X0@ 0 0 | 0̠nq~ dSTRONG ACIDS AND BASES2(2ffB | s *޽h ? ̙33y___PPT10Y+D=' = @B +{  0L0    (    0$n0  STRONG ACIDS completely dissociate (split up) into ions in aqueous solution e.g. HCl   > H+(aq) + Cl(aq) MONOPROTIC 1 replaceable H HNO3   > H+(aq) + NO3(aq) H2SO4   > 2H+(aq) + SO42-(aq) DIPROTIC 2 replaceable H s STRONG BASES completely dissociate into ions in aqueous solution e.g. NaOH   > Na+(aq) + OH(aq)@&P@ f: fff"ff ff fffffff ff ffff%f-33f3f`       /          dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0 nq~ dSTRONG ACIDS AND BASES2(2ffB  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 X(    0r0  DWeak acids partially dissociate into ions in aqueous solution e.g. ethanoic acid CH3COOH(aq) CH3COO(aq) + H+(aq) When a weak acid dissolves in water an equilibrium is set up HA(aq) + H2O(l) A(aq) + H3O+(aq) The water stabilises the ions To make calculations easier the dissociation can be written... HA(aq) A(aq) + H+(aq) ?@PGFPF f<333333 33?333333333b33 33I  0  @    r        dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0F ?  :h8N   fB   6D3olB   <BD3o{>T   # L)?fB   6D3olB   <BD3o{F ?   N8N   fB  6D3olB  <BD3o{>T  # L)?fB  6D3olB  <BD3o{F ?  @ n  8N   fB  6D3olB  <BD3o{>T  # L)?fB  6D3olB  <BD3o{  0rq~ X WEAK ACIDS2 (2 ffB  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 z(  `  0dr0 Weak acids partially dissociate into ions in aqueous solution e.g. ethanoic acid CH3COOH(aq) CH3COO(aq) + H+(aq) When a weak acid dissolves in water an equilibrium is set up HA(aq) + H2O(l) A(aq) + H3O+(aq) The water stabilises the ions To make calculations easier the dissociation can be written... HA(aq) A(aq) + H+(aq) The weaker the acid the less it dissociates the more the equilibrium lies to the left. The relative strengths of acids can be expressed as Ka or pKa values The dissociation constant for the weak acid HA is Ka = [H+(aq)] [A(aq)] mol dm-3 [HA(aq)]?@PGFPF@BP< f<333333 33?333333333b33 33   fff fff  : f*I  0  @    r         I    I   dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0F ?  :h8N   fB   6D3olB   <BD3o{>T   # L)?fB   6D3olB   <BD3o{F ?   N8N   fB  6D3olB  <BD3o{>T  # L)?fB  6D3olB  <BD3o{F ?  @ n  8N   fB  6D3olB  <BD3o{>T  # L)?fB  6D3olB  <BD3o{  0jrq~ X WEAK ACIDS2 (2 ff^B  6DfoB  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0   \ (    0xr0  $:Partially react with water to give ions in aqueous solution e.g. ammonia When a weak base dissolves in water an equilibrium is set up NH3 (aq) + H2O (l) NH4+ (aq) + OH (aq) as in the case of acids it is more simply written NH3 (aq) + H+ (aq) NH4+ (aq) ;@t333 333333 33433333333  #     @        dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0drq~ X WEAK BASES2 (2 ffF ?  ' e8N    fB   6D3olB   <BD3o{>T   # L)?fB   6D3olB  <BD3o{F ?  - [ 8N   fB  6D3olB  <BD3o{>T  # L)?fB  6D3olB  <BD3o{B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 q(    0r0  9Partially react with water to give ions in aqueous solution e.g. ammonia When a weak base dissolves in water an equilibrium is set up NH3 (aq) + H2O (l) NH4+ (aq) + OH (aq) as in the case of acids it is more simply written NH3 (aq) + H+ (aq) NH4+ (aq) The weaker the base the less it dissociates the more the equilibrium lies to the left The relative strengths of bases can be expressed as Kb or pKb values.v@t333 333333 33433333333     #     @           dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0rq~ X WEAK BASES2 (2 ffF ?  ' e8N    fB   6D3olB   <BD3o{>T   # L)?fB   6D3olB  <BD3o{F ?  - [ 8N   fB  6D3olB  <BD3o{>T  # L)?fB  6D3olB  <BD3o{B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 %BL( w L L 0r #Hydrogen ion concentration [H+(aq)]R$(2f fff@ L 6tr̙ ?y Introduction hydrogen ion concentration determines the acidity of a solution hydroxide ion concentration determines the alkalinity for strong acids and bases the concentration of ions is very much larger than their weaker counterparts which only partially dissociate.@      dB ?L <D)@0A @L 0Pp@ 0 0dB AL@ <D)@X@ BL 0X0@ 0 0B L s *޽h ? ̙333 0L0 ~3v3 =3(    0r #Hydrogen ion concentration [H+(aq)]R$(2f fff" ~ 6d t̙ ?y hpH hydrogen ion concentration can be converted to pH pH = - log10 [H+(aq)] to convert pH into hydrogen ion concentration [H+(aq)] = antilog (-pH) pOH An equivalent calculation for bases converts the hydroxide ion concentration to pOH pOH = - log10 [OH(aq)] in both the above, [ ] represents the concentration in mol dm-3 n%@DH7 5 a  B  K  C      Q    =   o+F `       6$t̙ ?w   GSTRONGLY ACIDIC(2  Bt̙ ?`} Y100:(2   Bt̙ ?} Z10-1:(2   BHt̙ ? Z10-2:(2   BP't̙ ? Z10-3:(2   B,t̙ ? Z10-4:(2   BXt̙ ?  Z10-5:(2   B6t̙ ? -  Z10-6:(2   B;t̙ ?- K  f10-7F(2  B@t̙ ?K h  Z10-8:(2   BFt̙ ?h  Z10-9:(2   B2t̙ ? [10-10:(2   B$ t̙ ? [10-11:(2   B      +   FF  j  fvtT @ #  *B  TDfV?@@B  TDfV?@tT @ # F jB   TDfV?@@B   TDfV?@FF  j    nitT @  #  *B   TDfV?@@B  TDfV?@tT @ # F jB  TDfV?@@B  TDfV?@|B  TDo?dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙33 0L0 A9( w   0@ju Ionic product of water - Kw@(2f ffV  6ru̙ ?y Despite being covalent, water conducts electricity to a very small extent. This is due to the slight ionisation ... H2O(l) + H2O(l) H3O+(aq) + OH(aq) or, more simply H2O(l) H+(aq) + OH(aq) Applying the equilibrium law to the second equation gives Kc = [H+(aq)] [OH(aq)] [ ] is the equilibrium concentration in mol dm-3 [H2O(l)] As the dissociation is small, the water concentration is very large compared with the dissociated ions and any changes to its value are insignificant; its concentration can be regarded as constant. This  constant is combined with (Kc) to get a new constant (Kw). Kw = [H+(aq)] [OH(aq)] mol2 dm-6 = 1 x 10-14 mol2 dm-6 (at 25C) Because the constant is based on an equilibrium, Kw VARIES WITH TEMPERATUREL@c@F@F@?@@!I@4@PFL@uff ffffffffffffB/  D    f4fff     6     >      +              c   FF  j  fvtT @ #  *B  TDfV?@@B  TDfV?@tT @ # F jB   TDfV?@@B   TDfV?@FF  j    nitT @  #  *B   TDfV?@@B  TDfV?@tT @ # F jB  TDfV?@@B  TDfV?@|B  TDo?dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙33  0L0 Y Q  (    0u Ionic product of water - Kw@(2f fff  6u̙ ?yc The value of Kw varies with temperature because it is based on an equilibrium. Temperature / C 0 20 25 30 60 Kw / 1 x 10-14 mol2 dm-6 0.11 0.68 1.0 1.47 5.6 H+ / x 10-7 mol dm-3 0.33 0.82 1.0 1.27 2.37 pH 7.48 7.08 7 6.92 6.63|P@.@2@W@ A/         3f Z   n  C  A   6u̙ ?1  l^What does this tell you about the equation H2O(l) H+(aq) + OH(aq) ? _@* fffffNI       FF  j  i tT @ #  *B  TDfV?@@B  TDfV?@tT @  # F jB   TDfV?@@B   TDfV?@dB   <D)@0A   0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙33 0L0 @%(    0u Ionic product of water - Kw@(2f fff  6<x̙ ?yc The value of Kw varies with temperature because it is based on an equilibrium. Temperature / C 0 20 25 30 60 Kw / 1 x 10-14 mol2 dm-6 0.11 0.68 1.0 1.47 5.6 H+ / x 10-7 mol dm-3 0.33 0.82 1.0 1.27 2.37 pH 7.48 7.08 7 6.92 6.63|P@.@2@W@ A/         3f Z   n  C  A   6x̙ ?13 What does this tell you about the equation H2O(l) H+(aq) + OH(aq) ? " Kw gets larger as the temperature increases " this means the concentration of H+ and OH ions gets greater " this means the equilibrium has moved to the right " if the concentration of H+ increases then the pH decreases " pH decreases as the temperature increases ._@@@* fffff O o NhI          FF  j  i tT @ #  *B   TDfV?@@B   TDfV?@tT @  # F jB   TDfV?@@B   TDfV?@dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙33 0L0 me(    088x Ionic product of water - Kw@(2f fff  64@x̙ ?yc The value of Kw varies with temperature because it is based on an equilibrium. Temperature / C 0 20 25 30 60 Kw / 1 x 10-14 mol2 dm-6 0.11 0.68 1.0 1.47 5.6 H+ / x 10-7 mol dm-3 0.33 0.82 1.0 1.27 2.37 pH 7.48 7.08 7 6.92 6.63|P@.@2@W@ A/         3f Z   n  C  A   6T[x̙ ?1w What does this tell you about the equation H2O(l) H+(aq) + OH(aq) ? " Kw gets larger as the temperature increases " this means the concentration of H+ and OH ions gets greater " this means the equilibrium has moved to the right " if the concentration of H+ increases then the pH decreases " pH decreases as the temperature increases Because the equation moves to the right as the temperature goes up, it must be an ENDOTHERMIC processJ_@@@08@4OoNgZI        x FF  j  i tT @ #  *B  TDV?@@B  TDV?@tT @  # F jB   TDV?@@B   TDV?@dB   <D)@0A   0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙33 0L0 F>P T(  T T 0xx {Relationship between pH and pOH (2 f T 6~x̙ ?y  3mBecause H+ and OH ions are produced in equal amounts when water dissociates [H+] = [OH] = 1 x 10-7 mol dm-3 their concentrations will be the same. Kw = [H+] [OH] = 1 x 10-14 mol2 dm-6 take logs of both sides log[H+] + log[OH] = -14 multiply by minus - log[H+] - log[OH] = 14 change to pH and pOH pH + pOH = 14 (at 25C) .n  C1 tg  1       dB T <D)@0A T 0Pp@ 0 0dB  T@ <D)@X@  T 0X0@ 0 0B T s *޽h ? ̙33  0L0 ; 3  (    0x {Relationship between pH and pOH (2 f  6xx̙ ?ye (Because H+ and OH ions are produced in equal amounts when water dissociates [H+] = [OH] = 1 x 10-7 mol dm-3 their concentrations will be the same. Kw = [H+] [OH] = 1 x 10-14 mol2 dm-6 take logs of both sides log[H+] + log[OH] = -14 multiply by minus - log[H+] - log[OH] = 14 change to pH and pOH pH + pOH = 14 (at 25C) N.B. As they are based on the position of equilibrium and that varies with temperature, the above values are only true if the temperature is 25C (298K) Neutral solutions may be regarded as those where [H+] = [OH]. Therefore a neutral solution is pH 7 only at a temperature of 25C (298K) Kw is constant for any aqueous solution at the stated temperaturen@n  C1 \ @g  1        O  @ dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙33c  0L0 ` X( w X X 08x %Buffer solutions - Brief introduction^&(2ff f* X 6$x̙ ?y2   Definition  Solutions which resist changes in pH when small quantities of acid or alkali are added. Acidic Buffer (pH < 7) made from a weak acid + its sodium or potassium salt ethanoic acid sodium ethanoate Alkaline Buffer (pH > 7) made from a weak base + its chloride ammonia ammonium chloride Uses Standardising pH meters Buffering biological systems (eg in blood) Maintaining the pH of shampoos@ \b%d"lZ       + dB X <D)@0A X 0Pp@ 0 0dB  X@ <D)@X@  X 0X0@ 0 0B X s *޽h ? ̙33 0L0 jbp(    0Dx JREVISION CHECK (2  0x ZWhat should you be able to do? (2   0$z+  Recall the definition of acids and bases in the Brnsted-Lowry system Recall the definition of acids and bases in the Lewis system Recall and explain the difference between strong and weak acids Recall and explain the difference between strong and weak bases Recall the definition of pH Recall the definition of the ionic product of water Explain how and why pH varies with temperature Recall the relationship between pH, [H+], [OH], pOH and Kw  2fAff7ff.ff.ffff.ff)ff fffff3fR/ u   04z   ~,CAN YOU DO ALL OF THESE? YES NO6- 2!    0 H @P 0 0   0f h @U 0 0H  0޽h ? ̙33 0L0  ;(  9  0!z  YYou need to go over the relevant topic(s) again Click on the button to return to the menudG(282/$f f) f  s * x  P @S 0 0dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@   0X0@ 0 0H  0޽h ? ̙33 0L0  D(    084zxK (WELL DONE! Try some past paper questions@)(2 <fffdB  <D)@0A  0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0H  0޽h ? ̙33 0L0 1(    0D wACIDS AND BASES THE ENDD(26f6ff  0 h. 2008 JONATHAN HOPTON & KNOCKHARDY PUBLISHING/(2/fJ  C "A KTREj x S _B  s *޽h ? ̙33r@(+d!f'O/k E`8WF ~ b"KSQcMМh',p|,I=UrU2Oh+'0T hp   No Slide TitleHOPTONJONATHAN HOPTON536Microsoft PowerPoint@p@@hNx@GSg  )'    """)))UUUMMMBBB999|PP3f333f3333f3ffffff3f̙3ff333f333333333f33333333f33f3ff3f3f3f3333f33̙33333f333333f3333f3ffffff3f33ff3f3f3f3fff3ffffffffff3ffff̙fff3fffff3fff333f3f3ff3ff33f̙̙3̙ff̙̙̙3f̙3f333f3333f3ffffff3f̙3f3f3f333f3333f3ffffff3f̙3f3ffffffffff!___www4'A x(xKʦ 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   On-screen Show nx #Times New Roman Arial BlackArial Courier NewTahomaDefault DesignSlide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9 Slide 10 Slide 11 Slide 12 Slide 13 Slide 14 Slide 15 Slide 16 Slide 17 Slide 18 Slide 19 Slide 20 Slide 21 Slide 22 Slide 23 Slide 24 Slide 25 Slide 26 Slide 27 Slide 28 Slide 29  Fonts UsedDesign Template Slide Titles 8@ _PID_HLINKSA8N314,30,Slide 30267,2,Slide 2315,30,Slide 30455,5,Slide 5465,8,Slide 8458,9,Slide 9460,11,Slide 11462,13,Slide 13403,15,Slide 15402,17,Slide 17405,19,Slide 19406,20,Slide 20313,21,Slide 21'_OJONATHAN HOPTONJONATHAN HOPTON  "#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      !"#$%&')*+,-./1234567<Root EntrydO)PicturesACurrent User0SummaryInformation(TPowerPoint Document(!sDocumentSummaryInformation8(