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Q   $ $ $        !`$$$$b$^ZP @x|$b$bpe# zt O|R$FE-Qڛb>" 0e0e    ̙ A@  A5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E|| fff3f3f@d g4KdKd; 0pppX <4BdBd@w 0t /zg4:d:d$n 0p@ pp<4!d!d@w 0t /ʚ;ʚ;<4dddd@ x 0@r0___PPT10 pp2___PPT9/ 0? %O =H O <^3/0124,-. !9;:5P4   0` ̙33` ` ff3333f` 333MMM` f` f` 3>?" dd@,|?" dd@   " @ ` n?" dd@   @@``PR    @ ` ` p>>L0 VN (    6ࡍ P  T Click to edit Master title style! !  0   RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  0t ``  X*  0 `   Z*  0ȵ `   Z*z  bA޽h @ ?Parchment ̙33 Default Design 0 zr@ ( )   0H]v P   v P*    0v    v R*  d  c $ ?  v  0v  @ v RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  6v `P  v P*    6v `  v R*  H  0޽h ? ̙3380___PPT10.0L 0X(   $+ X X 0 P    X*  X 0     Z*  X 6  `P   X*  X 6$ `   Z* H X 0޽h ? ̙3380___PPT10. 4L 0L0 PG( IF6   0 -BUFFER SOLUTIONS A guide for A level studentsD.(26f6ffJ  C "A KTREz ^  0$ h. 2004 JONATHAN HOPTON & KNOCKHARDY PUBLISHING/(2/fB  s *޽h ? ̙33E  0L0 `s(    08Ae `Buffer solutions48_2(f<fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  0Fp 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.argonet.co.uk/users/hoptonj/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 7\dB  <D?  B  s *޽h ? ̙33j  0L0   p (    0_8  pCONTENTS What is a buffer solution? Uses of buffer solutions Acidic buffer solutions Alkaline buffer solutions Buffer solutions - ideal concentration Calculating the pH of a buffer solution Salt hydrolysis Check listl (202 2   0ke `Buffer solutions48_2(f<fdB   <D)@0A   0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0   <?I @@ 0 0  <? 0@ 0 0  <?  @ 0 0  <?i   @ 0 0  <?` h @ 0 0  <?P H@ 0 0  <?F  @ 0 0  <?  @  0 0B  s *޽h ? ̙33 0L0 h` l(  l l 0}[ vBefore you start it would be helpful to& know that weak acids and bases are only partly ionised in solution be able to calculate pH from hydrogen ion concentration be able to construct an equation for the dissociation constant of a weak acidH*(2 2)[dB l <D)@0A l 0Pp@ 0 0dB l@ <D)@X@ l 0X0@ 0 0  l 0̇e `Buffer solutions48_2(f<fB l s *޽h ? ̙33 0L0 H@P X( w X X 0 %Buffer solutions - Brief introduction^&(2ff fdB X <D)@0A X 0Pp@ 0 0dB  X@ <D)@X@  X 0X0@ 0 0  X 6@Ê̙ ?Z  muAcidic 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 shampooszv@_%d"eZY       +   X 6ʊ̙ ?? 6Definition  Solutions which resist changes in pH when small quantities of acid or alkali are added. 8h@ \  h B X s *޽h ? ̙33  0L0 i a  | (  | | 0м QBuffer solutions - uses(2f | 6׊̙ ?? 6Definition  Solutions which resist changes in pH when small quantities of acid or alkali are added. 8h@ \  h dB | <D)@0A | 0Pp@ 0 0dB  |@ <D)@X@  | 0X0@ 0 0B  | 6ފ̙ ? Biological Uses In biological systems (saliva, stomach, and blood) it is essential that the pH stays  constant in order for any processes to work properly. e.g. If the pH of blood varies by 0.5 it can lead to unconsciousness and coma Most enzymes work best at particular pH values. Other Uses Many household and cosmetic products need to control their pH values. Shampoo Buffer solutions counteract the alkalinity of the soap and prevent irritation Baby lotion Buffer solutions maintain a pH of about 6 to prevent bacteria multiplying Others Washing powder, eye drops, fizzy lemonadeh KU N2@$  3    B | s *޽h ? ̙33 0L0 ` (    0 z Acidic buffer solutions - action>!(2ff  6 7̙ ?V  * It is essential to have a weak acid for an equilibrium to be present so that ions can be removed and produced. The dissociation is small and there are few ions. CH3COOH(aq) CH3COO(aq) + H+(aq) relative concs. HIGH LOW LOW NB A strong acid can t be used as it is fully dissociated and cannot remove H+(aq) HCl(aq)   > Cl(aq) + H+(aq) Adding acid Any H+ is removed by reacting with CH3COO ions to form CH3COOH via the equilibrium. Unfortunately, the concentration of CH3COO is small and only a few H+ can be  mopped up . A much larger concentration of CH3COO is required. To build up the concentration of CH3COO ions, sodium ethanoate is added."@fffffff$R +fff     B  6 6 &*          }              .     FF  j  X | FtT @ #  *B   TDfV?@@B   TDfV?@tT @  # F jB   TDfV?@@B   TDfV?@dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙33 0L0 jb x(  x x 6'7̙ ? n It is essential to have a weak acid for an equilibrium to be present so that ions can be removed and produced. The dissociation is small and there are few ions. CH3COOH(aq) CH3COO(aq) + H+(aq) relative concs. HIGH LOW LOW NB A strong acid can t be used as it is fully dissociated and cannot remove H+(aq) HCl(aq)   > Cl(aq) + H+(aq) Adding alkali This adds OH ions which react with H+ ions H+(aq) + OH(aq) H2O(l) Removal of H+ from the weak acid equilibrium means that, according to Le Chatelier s Principle, more CH3COOH will dissociate to form ions to replace those being removed. CH3COOH(aq) CH3COO(aq) + H+(aq) As the added OH ions remove the H+ from the weak acid system, the equilibrium moves to the right to produce more H+ ions. Obviously, there must be a large concentration of undissociated acid molecules to be available.@@2@fffffff$R +fff% ff"fff  Z Bfffffff" P g          }             D     ]  `         ! FF  j x X | FtT @ x#  *B  x TDfV?@@B  x TDfV?@tT @  x# F jB  x TDfV?@@B  x TDfV?@FF  j x  X | f tT @ x#  *B x TDfV?@@B x TDfV?@tT @ x# F jB x TDfV?@@B x TDfV?@FF  j x 2 < tT @ x#  *B x TDfV?@@B x TDfV?@tT @ x# F jB x TDfV?@@B x TDfV?@ x 0f7 z Acidic buffer solutions - action>!(2ffdB x <D)@0A x 0Pp@ 0 0dB x@ <D)@X@  x 0X0@ 0 0B x s *޽h ? ̙33  0L0  x p (    6p7̙ ?yC  ,Alkaline buffer Very similar but is based on the equilibrium surrounding a weak base; AMMONIA NH3(aq) + H2O(l) OH(aq) + NH4+(aq) relative concs. HIGH LOW LOW but one needs ; a large conc. of OH(aq) to react with any H+(aq) added a large conc of NH4+(aq) to react with any OH(aq) added There is enough NH3 to act as a source of OH but one needs to increase the concentration of ammonium ions by adding an ammonium salt. Use AMMONIA (a weak base) + AMMONIUM CHLORIDE (one of its salts)-@Tff ff%ffffB    : {ff*h  "       <                 FF  j   tT @ #  *B   TDfV?@@B   TDfV?@tT @  # F jB   TDfV?@@B   TDfV?@  07 |"Alkaline buffer solutions - action>#(2ffdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙33  0L0 2 *   ( w   07 &Buffer solutions - ideal concentration>'(2ff  67̙ ?yQ  The concentration of a buffer solution is also important If the concentration is too low, there won t be enough CH3COOH and CH3COO to cope with the ions added. Summary For an acidic buffer solution one needs ... large [CH3COOH(aq)] - for dissociating into H+(aq) when alkali is added large [CH3COO(aq)] - for removing H+(aq) as it is added This situation can t exist if only acid is present; a mixture of the acid and salt is used. The weak acid provides the equilibrium and the large CH3COOH(aq) concentration. The sodium salt provides the large CH3COO(aq) concentration. One uses a WEAK ACID + its SODIUM OR POTASSIUM SALT*@s   %7 & $   = P  E  !  '      <  I dB  <D)@0A  0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0B  s *޽h ? ̙33 0L0 _W ( w   07 i/Calculating the pH of an acidic buffer solution0(20f  67̙ ?y ^XCalculate the pH of a buffer whose [HA] is 0.1 mol dm-3 and [A] of 0.1 mol dm-3. RH@5\.       dB  <D)@0A  0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0B  s *޽h ? ̙33@ 0L0 n(    0h7 i/Calculating the pH of an acidic buffer solution0(20f  67̙ ?y? SCalculate the pH of a buffer whose [HA] is 0.1 mol dm-3 and [A] of 0.1 mol dm-3. Ka = [H+(aq)] [A(aq)] [HA(aq)] RH)@#@@5   4.                |B  TDo?z dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙33 0L0 NF (    0o i/Calculating the pH of an acidic buffer solution0(20f  6 o̙ ?y  ECalculate the pH of a buffer whose [HA] is 0.1 mol dm-3 and [A] of 0.1 mol dm-3. Ka = [H+(aq)] [A(aq)] [HA(aq)] re-arrange [H+(aq)] = [HA(aq)] x Ka [A(aq)]  RH)@#@,@5   B  .                        |B  TDo?z |B  TDo? > dB  <D)@0A  0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0B  s *޽h ? ̙33  0L0 '   ( w   0!o i/Calculating the pH of an acidic buffer solution0(20f  6̙ ?y)  *Calculate the pH of a buffer whose [HA] is 0.1 mol dm-3 and [A] of 0.1 mol dm-3. Ka = [H+(aq)] [A(aq)] [HA(aq)] re-arrange [H+(aq)] = [HA(aq)] x Ka [A(aq)] from information given [A] = 0.1 mol dm-3 [HA] = 0.1 mol dm-3 ~RH)@#@,@E@5   B  A  R.                      '       |B  TDo?z |B  TDo? > dB  <D)@0A  0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0B  s *޽h ? ̙33  0L0    3 ( w   0XRo i/Calculating the pH of an acidic buffer solution0(20fv  6dWo̙ ?y! Calculate the pH of a buffer whose [HA] is 0.1 mol dm-3 and [A] of 0.1 mol dm-3. Ka = [H+(aq)] [A(aq)] [HA(aq)] re-arrange [H+(aq)] = [HA(aq)] x Ka [A(aq)] from information given [A] = 0.1 mol dm-3 [HA] = 0.1 mol dm-3 If the Ka of the weak acid HA is 2 x 10-4 mol dm-3. [H+(aq)] = 0.1 x 2 x 10-4 = 2 x 10-4 mol dm-3 0.1 hRH)@#@,@E6@J@@5   B  A           .                      '           "      |B  TDo?z |B  TDo? > |B  TDo?p F p dB  <D)@0A  0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0B  s *޽h ? ̙33  0L0 d \   (    0o i/Calculating the pH of an acidic buffer solution0(20f=  6 |B  TDo?p F p dB  <D)@0A  0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0B  s *޽h ? ̙33  0L0  ;( w   0o i/Calculating the pH of an acidic buffer solution0(20f   6o̙ ?y Calculate the pH of the solution formed when 500cm3 of 0.1 mol dm-3 of weak acid HX is mixed with 500cm3 of a 0.2 mol dm-3 solution of its salt NaX. Ka = 4 x 10-5 mol dm-3. B@@2$ :  3          dB   <D)@0A   0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0B  s *޽h ? ̙33 0L0 \T(    0o i/Calculating the pH of an acidic buffer solution0(20f=  6\s̙ ?y  Calculate the pH of the solution formed when 500cm3 of 0.1 mol dm-3 of weak acid HX is mixed with 500cm3 of a 0.2 mol dm-3 solution of its salt NaX. Ka = 4 x 10-5 mol dm-3. Ka = [H+(aq)] [X(aq)] [HX(aq)] @,@.@@2$     ?:  3                 (    |B  TDo?xx xdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙33U  0L0     ( =^=^   0,s i/Calculating the pH of an acidic buffer solution0(20fJ  6$s̙ ?y-  \Calculate the pH of the solution formed when 500cm3 of 0.1 mol dm-3 of weak acid HX is mixed with 500cm3 of a 0.2 mol dm-3 solution of its salt NaX. Ka = 4 x 10-5 mol dm-3. Ka = [H+(aq)] [X(aq)] [HX(aq)] re-arrange [H+(aq)] = [HX(aq)] Ka [X(aq)] @,@.@-@&@2$     K  (:  3                 (              |B  TDo?xx x|B  TDo?0 dB  <D)@0A  0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0B  s *޽h ? ̙33  0L0 < 4   ( w   0X7s i/Calculating the pH of an acidic buffer solution0(20f  6Us̙ ?y  3Calculate the pH of the solution formed when 500cm3 of 0.1 mol dm-3 of weak acid HX is mixed with 500cm3 of a 0.2 mol dm-3 solution of its salt NaX. Ka = 4 x 10-5 mol dm-3. Ka = [H+(aq)] [X(aq)] [HX(aq)] re-arrange [H+(aq)] = [HX(aq)] Ka [X(aq)] The solutions have been mixed; the volume is now 1 dm3 therefore [HX] = 0.05 mol dm-3 and [X] = 0.10 mol dm-3 ^@,@.@-@@"P@@2$     K  ^ * $ :  3                 (             _  "    |B  TDo?xx x|B  TDo?0 dB  <D)@0A  0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0B  s *޽h ? ̙33  0L0 W O   ( w   04s i/Calculating the pH of an acidic buffer solution0(20f0  6s̙ ?y Calculate the pH of the solution formed when 500cm3 of 0.1 mol dm-3 of weak acid HX is mixed with 500cm3 of a 0.2 mol dm-3 solution of its salt NaX. Ka = 4 x 10-5 mol dm-3. Ka = [H+(aq)] [X(aq)] [HX(aq)] re-arrange [H+(aq)] = [HX(aq)] Ka [X(aq)] The solutions have been mixed; the volume is now 1 dm3 therefore [HX] = 0.05 mol dm-3 and [X] = 0.10 mol dm-3 Substituting [H+(aq)] = 0.05 x 4 x 10-5 = 2 x 10-5 mol dm-3 0.1 pH = - log10 [H+(aq)] = 4.699@,@.@-@@"P@L@M@2$     K  ^ * $     1  :  3                 (             _  "    -  =   |B  TDo?xx x|B  TDo?0 |B  TDo?P ~ P dB  <D)@0A  0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0B  s *޽h ? ̙33/ 0L0 ]( w   0,s ISALT HYDROLYSIS(2fL  6s̙ ??0E bMany salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate.:c@&   dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙33y 0L0   ( w     0s ISALT HYDROLYSIS(2fH   6t̙ ??0c ^Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of strong acids and strong bases - SODIUM CHLORIDE NaCl dissociates completely in water Na+ Cl   > Na+ + Cl Water only ionises to a very small extent H2O OH + H+ Na+ and OH are ions of a strong base so remain apart H+and Cl are ions of a strong acid so remain apart all the OH and H+ ions remain in solution therefore [H+] = [OH] and the solution will be NEUTRAL6c@PF@c8*  5    -f3*    9  "        Z    @  -  t FF  j   E  tT @  #  *B   TDV?@@B   TDV?@tT @  # F jB   TDV?@@B   TDV?@dB   <D)@0A   0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0B   s *޽h ? ̙33q 0L0  ( w   0d"t ISALT HYDROLYSIS(2f  6Xet̙ ??0 Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of strong acids and weak bases - AMMONIUM CHLORIDE NH4Cl dissociates completely in water NH4+ Cl   > NH4+ + Cl Water only ionises to a very small extent H2O OH + H+ NH3 + H2O Na+ and OH are ions of a strong base so tend to be associated H+and Cl are ions of a strong acid so remain apart all the H+ ions remain in solution therefore [H+] > [OH] and the solution will be ACIDICd@P2@c9 %   5       -*    e       P           H   FF  j   Y tT @ #  *B  TDV?@@B  TDV?@tT @ # F jB   TDV?@@B   TDV?@FF  Dg     ' tT @  # M  g B   TDV?@@B  TDV?@tT @ # M% Df B  TDV?@@B  TDV?@dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙339 0L0 g( w   0@t ISALT HYDROLYSIS(2f  6qt̙ ??0 T Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of weak acids and strong bases - SODIUM ETHANOATE CH3COONa dissociates completely in water CH3COO Na +   > Na+ + CH3COO Water only ionises to a very small extent H2O OH + H+ CH3COOH Na+ and OH are ions of a strong base so remain apart H+and CH3OO are ions of a weak acid so tend to be associated all the OH ions remain in solution therefore [OH] > [H+] and the solution will be ALKALINE^d@P2@c8 +   ;    % *    b           ]      "   FF  j   Q tT @ #  *B  TDV?@@B  TDV?@tT @ # F jB   TDV?@@B   TDV?@FF  Dg     l tT @  # M  g B   TDV?@@B  TDV?@tT @ # M% Df B  TDV?@@B  TDV?@dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙33  0L0  9( --   0Dt ISALT HYDROLYSIS(2f>  6t̙ ??0D Many salts dissolve in water to produce solutions which are not neutral. This is because the ions formed react with the hydroxide and hydrogen ions formed when water dissociates. There are four distinct systems. All dissociated ions are aqueous ions. When mixed, the ions of strong acids and bases remain apart. Ions of weak acids and bases associate. Salts of weak acids and weak bases - AMMONIUM ETHANOATE CH3COONH4 dissociates completely in water CH3COO NH4 +   > NH4+ + CH3COO Water only ionises to a very small extent H2O OH + H+ NH3 + H2O CH3COOH Na+ and OH are ions of a weak base so tend to be associated H+and CH3OO are ions of a weak acid so tend to be associated the solution might be alkaline or acidic i.e. APPROXIMATELY NEUTRALd@P2@c8  %   ;      0f3    c       FF  j   Q tT @ #  *B  TDV?@@B  TDV?@tT @ # F jB   TDV?@@B   TDV?@FF  Dg     / tT @  # M  g B   TDV?@@B  TDV?@tT @ # M% Df B  TDV?@@B  TDV?@FF  Dg    4 tT @ # M  g B  TDV?@@B  TDV?@tT @ # M% Df B  TDV?@@B  TDV?@dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0B  s *޽h ? ̙33O 0L0  w(    0Dv JREVISION CHECK (2  0v ZWhat should you be able to do? (2   0<vx 51Recall the definition of a buffer solution Recall the difference between an acidic and an alkaline buffer solution Recall the uses of buffer solutions Understand the action of buffer solutions Calculate the pH of an acidic buffer solution Recall and understand the reactions due to salt hydrolysis 2 2f%ffBff"f f f f'ff$ff   0v   ~,CAN YOU DO ALL OF THESE? YES NO6- 2!    0 H @P 0 0   0f h @U 0 0H  0޽h ? ̙335 0L0 0](  9  0#v  YYou need to go over the relevant topic(s) again Click on the button to return to the menudG(282/$f f) f"  s *fX@ 0 0  0P@Q 0 0  s * x  P @S 0 0H  0޽h ? ̙33 0L0 @(    0!vxK (WELL DONE! 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