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S  0( ``  X*  0̳ `   Z*  0| `   Z*z  bA޽h @ ?Parchment ̙33 Default Design 0 zr@ ( )   0 P    P*    0     R*  d  c $ ?    0  @  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  6P `P   P*    60 `   R*  H  0޽h ? ̙3380___PPT10. 0X(   $+ X X 0  P    X*  X 0     Z*  X 6 `P   X*  X 6 `   Z* H X 0޽h ? ̙3380___PPT10.1 0L0 P/(    0 &AN INTRODUCTION TO SOLUBILITY PRODUCTST(2 82 f0f<f " <̙? x @ 0 0 ( <̙? x @L 0 0 - 0(  OKNOCKHARDY PUBLISHING(2fJ . C "A KTREj x S _ / 0"`>B 2008 SPECIFICATIONS<G$G   B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   P (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0Y   0`a 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, dB   <D? xP    0v{0P( OKNOCKHARDY PUBLISHING(2    0{X aSOLUBILITY PRODUCTS2(2$f<fB  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 ` (  T  0Pu   CONTENTS Solubility of ionic compounds Saturated solutions Solubility products Calculations Common Ion EffectN 2o 2ndB  <D)P0Q  0Pp@ 0 0dB  @ <D)PXP   0X0@ 0 0  6C"?YH @x 0 0  6C"?4( @ 0 0  0(,X aSOLUBILITY PRODUCTS2(2$f<f  6C"?I( @ 0 0  6C"? @ 0 0  6C"? @ 0 0B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 p U( w dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0   0q~ kSOLUBILITY OF IONIC COMPOUNDS2(2ffn   0X  l" ionic compounds tend to be insoluble in non-polar solvents " ionic compounds tend to be soluble in water " water is a polar solvent and stabilises the separated ions r} !!!,  R  C *Akspsolm+c R  C *Akspsolx-  B  s *޽h ? ̙33y___PPT10Y+D=' = @B +"  0L0 1) L(  LdB L <D)P0Q L 0Pp@ 0 0dB L@ <D)PXP L 0X0@ 0 0 L 0Ԗq~ kSOLUBILITY OF IONIC COMPOUNDS2(2ff L 0x~ v" ionic compounds tend to be insoluble in non-polar solvents " ionic compounds tend to be soluble in water " water is a polar solvent and stabilises the separated ions " some ionic compounds are very insoluble (AgCl, PbSO4, PbS) " even soluble ionic compounds have a limit as to how much dissolves<} !!![ N@     I R L C *Akspsolm+c R  L C *Akspsolx-  B L s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 P ,)(  ,dB , <D)P0Q , 0Pp@ 0 0dB ,@ <D)PXP , 0X0@ 0 0 , 0q~ aSATURATED SOLUTIONS2(2ff , 0ː0 X" solutions become saturated when solute no longer dissolves in a solvent " solubility varies with temperature " most solutes are more soluble at higher temperatures}  B , s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 # X(  XdB X <D)P0Q X 0Pp@ 0 0dB X@ <D)PXP X 0X0@ 0 0 X 0q~ aSATURATED SOLUTIONS2(2ff X 00 X" solutions become saturated when solute no longer dissolves in a solvent " solubility varies with temperature " most solutes are more soluble at higher temperatures}  P X C (Akspsol10* 0  X 0| j Ionic crystal lattices can dissociate (break up) when placed in water. The ions separate as they are stabilised by polar water molecules.   B X s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0  TI(  TdB T <D)P0Q T 0Pp@ 0 0dB T@ <D)PXP T 0X0@ 0 0 T 0q~ aSATURATED SOLUTIONS2(2ff T 0D?0 X" solutions become saturated when solute no longer dissolves in a solvent " solubility varies with temperature " most solutes are more soluble at higher temperatures}  P T C (Akspsol10* R  T C *Akspsol12h  4  T 00?| j Ionic crystal lattices can dissociate (break up) when placed in water. The ions separate as they are stabilised by polar water molecules.   *  T 08 ?| 0  Eventually, no more solute dissolves and the solution becomes saturated. There is a limit to the concentration of ions in solution.  B T s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0  PE(  PdB P <D)P0Q P 0Pp@ 0 0dB P@ <D)PXP P 0X0@ 0 0 P 0?q~ aSATURATED SOLUTIONS2(2ff P 0\?0 X" solutions become saturated when solute no longer dissolves in a solvent " solubility varies with temperature " most solutes are more soluble at higher temperatures}  P P C (Akspsol10* R  P C *Akspsol12h  0  P 0?| j Ionic crystal lattices can dissociate (break up) when placed in water. The ions separate as they are stabilised by polar water molecules.   *  P 0 ?| 0  Eventually, no more solute dissolves and the solution becomes saturated. There is a limit to the concentration of ions in solution.  B P s *޽h ? ̙33y___PPT10Y+D=' = @B +C 0L0 RJ(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0)?q~ `SOLUBILITY PRODUCT2(2ff,  05? :Even the most insoluble ionic compounds dissolve to a small extent. An equilibrium exists between the undissolved solid and its aqueous ions; (i) MX(s) M+(aq) + X(aq) (ii) BaSO4(s) Ba2+(aq) + SO42-(aq) (iii) PbCl2(s) Pb2+(aq) + 2Cl(aq) }ffffffffffffffffffftf   '       y F ?   8N    fB   6DfolB   <BDfo{>T   # L)?fB   6DfolB  <BDfo{F ?   8N   fB  6DfolB  <BDfo{>T  # L)?fB  6DfolB  <BDfo{F ?   '8N   fB  6DfolB  <BDfo{>T  # L)?fB  6DfolB  <BDfo{B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 \(  \dB \ <D)P0Q \ 0Pp@ 0 0dB \@ <D)PXP \ 0X0@ 0 0 \ 0H\?q~ `SOLUBILITY PRODUCT2(2ff \ 0n? )Even the most insoluble ionic compounds dissolve to a small extent. An equilibrium exists between the undissolved solid and its aqueous ions; (i) MX(s) M+(aq) + X(aq) (ii) BaSO4(s) Ba2+(aq) + SO42-(aq) (iii) PbCl2(s) Pb2+(aq) + 2Cl(aq) Applying the equilibrium law to (i) and assuming the concentration of MX(s) is constant in a saturated solution. [M+(aq)] [X(aq)] = a constant, Ksp [ ] is the concentration in mol dm-3 Ksp is known as the SOLUBILITY PRODUCTF}'fffffffffffffffffff$S!!!!!!" ffffff   '         ,      -  $ F ? \  8N   \ fB  \ 6DfolB  \ <BDfo{>T   \# L)?fB  \ 6DfolB \ <BDfo{F ? \  8N  \ fB \ 6DfolB \ <BDfo{>T  \# L)?fB \ 6DfolB \ <BDfo{F ? \  '8N  \ fB \ 6DfolB \ <BDfo{>T  \# L)?fB \ 6DfolB \ <BDfo{B \ s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 ?^(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0|?q~ `SOLUBILITY PRODUCT2(2ff  0?j  Bj AgCl(s) Ag+(aq) + Cl(aq) Ksp = [Ag+(aq)] [Cl(aq)] BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = [Ba2+(aq)] [SO42-(aq)] PbCl2(s) Pb2+(aq) + 2Cl(aq) Ksp = [Pb2+(aq)] [Cl(aq)]2 Notice that the concentration of Cl(aq) is raised to the power of 2 because there are two Cl(aq) ions in the equation x                          "!!9!!!              .  I  >  4  ! F ?  o8N    fB   6D8clB   <BD8c{>T   # L)?fB   6D8clB  <BD8c{F ?  8N   fB  6D8clB   <BD8c{>T  !# L)?fB " 6D8clB # <BD8c{F ? $ e8N  % fB & 6D8clB ' <BD8c{>T  (# L)?fB ) 6D8clB * <BD8c{B  s *޽h ? ̙33y___PPT10Y+D=' = @B +5" 0L0 D!  _     +  G  H   F ? 4 o8N   4 fB  4 6D8clB  4 <BD8c{>T   4# L)?fB  4 6D8clB 4 <BD8c{F ? 4 8N  4 fB 4 6D8clB 4 <BD8c{>T  4# L)?fB 4 6D8clB 4 <BD8c{F ? 4 e8N  4 fB 4 6D8clB 4 <BD8c{>T  4# L)?fB 4 6D8clB 4 <BD8c{F ? 4  ;5 8N  4 fB 4 6D8clB  4 <BD8c{>T  !4# L)?fB "4 6D8clB #4 <BD8c{F ? $4 ? ? 8N  %4 fB &4 6D8clB '4 <BD8c{>T  (4# L)?fB )4 6D8clB *4 <BD8c{F ? +4  C 8N  ,4 fB -4 6D8clB .4 <BD8c{>T  /4# L)?fB 04 6D8clB 14 <BD8c{ 24 VAParchmentS" ?  B 4 s *޽h ? ̙33y___PPT10Y+D=' = @B +C" 0L0 R!J!22< (  <dB < <D)P0Q < 0Pp@ 0 0dB <@ <D)PXP < 0X0@ 0 0 < 0\yq~ `SOLUBILITY PRODUCT2(2ff < 0Dly   AgCl(s) Ag+(aq) + Cl(aq) Ksp = [Ag+(aq)] [Cl(aq)] BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = [Ba2+(aq)] [SO42-(aq)] PbCl2(s) Pb2+(aq) + 2Cl(aq) Ksp = [Pb2+(aq)] [Cl(aq)]2 Notice that the concentration of Cl(aq) is raised to the power of 2 Complete the equilibrium equation and write an expression for Ksp for& PbS(s) Pb2+(aq) + S2-(aq) Ksp = [Pb2+(aq)] [S2-(aq)] Fe(OH)2(s) Fe2+(aq) + 2OH(aq) Ksp = [Fe2+(aq)] [OH(aq)]2 Fe(OH)3(s) Fe3+(aq) + 3OH(aq) Ksp = [Fe3+(aq)] [OH(aq)]3 *FG        #$?!!!ff ff ff ffff fff           x             .  I  >  _     +  G  H   F ? < o8N   < fB  < 6D8clB  < <BD8c{>T   <# L)?fB  < 6D8clB < <BD8c{F ? < 8N  < fB < 6D8clB < <BD8c{>T  <# L)?fB < 6D8clB < <BD8c{F ? < e8N  < fB < 6D8clB < <BD8c{>T  <# L)?fB < 6D8clB < <BD8c{F ? <  ;5 8N  < fB < 6Df8clB  < <BDf8c{>T  !<# L)?fB "< 6Df8clB #< <BDf8c{F ? $< ? ? 8N  %< fB &< 6D8clB '< <BD8c{>T  (<# L)?fB )< 6D8clB *< <BD8c{F ? +<  C 8N  ,< fB -< 6D8clB .< <BD8c{>T  /<# L)?fB 0< 6D8clB 1< <BD8c{ 2< VAParchmentS" ?  B < s *޽h ? ̙33y___PPT10Y+D=' = @B +'" 0L0 6!.!22@ (  @dB @ <D)P0Q @ 0Pp@ 0 0dB @@ <D)PXP @ 0X0@ 0 0 @ 0yq~ `SOLUBILITY PRODUCT2(2ff @ 0$z  p  AgCl(s) Ag+(aq) + Cl(aq) Ksp = [Ag+(aq)] [Cl(aq)] BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = [Ba2+(aq)] [SO42-(aq)] PbCl2(s) Pb2+(aq) + 2Cl(aq) Ksp = [Pb2+(aq)] [Cl(aq)]2 Notice that the concentration of Cl(aq) is raised to the power of 2 Complete the equilibrium equation and write an expression for Ksp for& PbS(s) Pb2+(aq) + S2-(aq) Ksp = [Pb2+(aq)] [S2-(aq)] Fe(OH)2(s) Fe2+(aq) + 2OH(aq) Ksp = [Fe2+(aq)] [OH(aq)]2 Fe(OH)3(s) Fe3+(aq) + 3OH(aq) Ksp = [Fe3+(aq)] [OH(aq)]3 FG       #$?!!!ff ff ff ffff ffffff fffffff     x             .  I  >  _     +  G  H   F ? @ o8N   @ fB  @ 6D8clB  @ <BD8c{>T   @# L)?fB  @ 6D8clB @ <BD8c{F ? @ 8N  @ fB @ 6D8clB @ <BD8c{>T  @# L)?fB @ 6D8clB @ <BD8c{F ? @ e8N  @ fB @ 6D8clB @ <BD8c{>T  @# L)?fB @ 6D8clB @ <BD8c{F ? @  ;5 8N  @ fB @ 6Df8clB  @ <BDf8c{>T  !@# L)?fB "@ 6Df8clB #@ <BDf8c{F ? $@ ? ? 8N  %@ fB &@ 6Df8clB '@ <BDf8c{>T  (@# L)?fB )@ 6Df8clB *@ <BDf8c{F ? +@  C 8N  ,@ fB -@ 6D8clB .@ <BD8c{>T  /@# L)?fB 0@ 6D8clB 1@ <BD8c{ 2@ VAParchmentS" ?  B @ s *޽h ? ̙33y___PPT10Y+D=' = @B +! 0L0   12DV (  DdB D <D)P0Q D 0Pp@ 0 0dB D@ <D)PXP D 0X0@ 0 0 D 0l&zq~ `SOLUBILITY PRODUCT2(2ff D 0P8z   AgCl(s) Ag+(aq) + Cl(aq) Ksp = [Ag+(aq)] [Cl(aq)] BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = [Ba2+(aq)] [SO42-(aq)] PbCl2(s) Pb2+(aq) + 2Cl(aq) Ksp = [Pb2+(aq)] [Cl(aq)]2 Notice that the concentration of Cl(aq) is raised to the power of 2 Complete the equilibrium equation and write an expression for Ksp for& PbS(s) Pb2+(aq) + S2-(aq) Ksp = [Pb2+(aq)] [S2-(aq)] Fe(OH)2(s) Fe2+(aq) + 2OH(aq) Ksp = [Fe2+(aq)] [OH(aq)]2 Fe(OH)3(s) Fe3+(aq) + 3OH(aq) Ksp = [Fe3+(aq)] [OH(aq)]3 DFG        #$?!!!ff ff ff ffff ffffff ffffffffff fffffx             .  I  >  _     +  G  H   F ? D o8N   D fB  D 6D8clB  D <BD8c{>T   D# L)?fB  D 6D8clB D <BD8c{F ? D 8N  D fB D 6D8clB D <BD8c{>T  D# L)?fB D 6D8clB D <BD8c{F ? D e8N  D fB D 6D8clB D <BD8c{>T  D# L)?fB D 6D8clB D <BD8c{F ? D  ;5 8N  D fB D 6Df8clB  D <BDf8c{>T  !D# L)?fB "D 6Df8clB #D <BDf8c{F ? $D ? ? 8N  %D fB &D 6Df8clB 'D <BDf8c{>T  (D# L)?fB )D 6Df8clB *D <BDf8c{F ? +D  C 8N  ,D fB -D 6Df8clB .D <BDf8c{>T  /D# L)?fB 0D 6Df8clB 1D <BDf8c{B D s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 , $ 28(  8dB 8 <D)P0Q 8 0Pp@ 0 0dB 8@ <D)PXP 8 0X0@ 0 0 8 0zq~ `SOLUBILITY PRODUCT2(2ff 8 0|z : Units The value of Ksp has units and it varies with temperature AgCl Ksp = [Ag+(aq)] [Cl(aq)] units of& mol2 dm-6 BaSO4 Ksp = [Ba2+(aq)] [SO42-(aq)] mol2 dm-6 PbCl2 Ksp = [Pb2+(aq)] [Cl(aq)]2 mol3 dm-9 !! +                         +        $  2  * B 8 s *޽h ? ̙33y___PPT10Y+D=' = @B +W 0L0 f ^ 0  (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0zq~ `SOLUBILITY PRODUCT2(2ff  0|D  t : Units The value of Ksp has units and it varies with temperature AgCl Ksp = [Ag+(aq)] [Cl(aq)] units of& mol2 dm-6 BaSO4 Ksp = [Ba2+(aq)] [SO42-(aq)] mol2 dm-6 PbCl2 Ksp = [Pb2+(aq)] [Cl(aq)]2 mol3 dm-9 Work out the units of Ksp for the following& PbS Ksp = [Pb2+(aq)] [S2-(aq)] Fe(OH)2 Ksp = [Fe2+(aq)] [OH(aq)]2 Fe(OH)3 Ksp = [Fe3+(aq)] [OH(aq)]3 -!! +     !!!      f     f     fD  +        $  2  @      +  ,    B  s *޽h ? ̙33y___PPT10Y+D=' = @B +] 0L0 ldH(  HdB H <D)P0Q H 0Pp@ 0 0dB H@ <D)PXP H 0X0@ 0 0 H 0 |q~ `SOLUBILITY PRODUCT2(2ff H 0.|  z t Units The value of Ksp has units and it varies with temperature AgCl Ksp = [Ag+(aq)] [Cl(aq)] units of& mol2 dm-6 BaSO4 Ksp = [Ba2+(aq)] [SO42-(aq)] mol2 dm-6 PbCl2 Ksp = [Pb2+(aq)] [Cl(aq)]2 mol3 dm-9 Work out the units of Ksp for the following& PbS Ksp = [Pb2+(aq)] [S2-(aq)] mol2 dm-6 Fe(OH)2 Ksp = [Fe2+(aq)] [OH(aq)]2 mol3 dm-9 Fe(OH)3 Ksp = [Fe3+(aq)] [OH(aq)]3 mol4 dm-12 -!! +     !!!     fffff     fffff     fffffD  +        $  2  @      4  5  + B H s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 pt^(  tdB t <D)P0Q t 0Pp@ 0 0dB t@ <D)PXP t 0X0@ 0 0 t 0}|q~ Z CALCULATIONS2 (2 ff< t 04|0 Solubility products can be used to calculate the solubility of compounds. At 25C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. K!Bf3f3f3f3f3f33f3Zm     6    B t s *޽h ? ̙33y___PPT10Y+D=' = @B +A  0L0 P H `p(  pdB p <D)P0Q p 0Pp@ 0 0dB p@ <D)PXP p 0X0@ 0 0 p 0Г|q~ Z CALCULATIONS2 (2 ff p 0|0 Solubility products can be used to calculate the solubility of compounds. At 25C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq) K!Bf3f3f3f3f3f33f3<   tm     6  3  ' F ? p %8N   p fB  p 6D8clB  p <BD8c{>T   p# L)?fB  p 6D8clB p <BD8c{B p s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 '  Pl (  ldB l <D)P0Q l 0Pp@ 0 0dB l@ <D)PXP l 0X0@ 0 0 l 0|q~ Z CALCULATIONS2 (2 ff l 0|0u _mSolubility products can be used to calculate the solubility of compounds. At 25C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq) The expression for the solubility product is Ksp = [Pb2+(aq)] [S2-(aq)] @nK!Bf3f3f3f3f3f33f3< 2    m     6  3  U   F ? l %8N   l fB  l 6D8clB  l <BD8c{>T   l# L)?fB  l 6D8clB l <BD8c{B l s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   @h (  hdB h <D)P0Q h 0Pp@ 0 0dB h@ <D)PXP h 0X0@ 0 0 h 0|q~ Z CALCULATIONS2 (2 ff h 0d|0  N@Solubility products can be used to calculate the solubility of compounds. At 25C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq) The expression for the solubility product is Ksp = [Pb2+(aq)] [S2-(aq)] According to the equation, you get one Pb2+(aq) for every one S2-(aq); the concentrations will be equal [Pb2+(aq)] = [S2-(aq)] Substituting and rewriting the expression for Ksp Ksp = [Pb2+(aq)]2 (AK!Bf3f3f3f3f3f33f3< 6 )  ,   7     m     6  3  U       F ? h %8N   h fB  h 6D8clB  h <BD8c{>T   h# L)?fB  h 6D8clB h <BD8c{B h s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 xt(  xdB x <D)P0Q x 0Pp@ 0 0dB x@ <D)PXP x 0X0@ 0 0 x 0H~q~ Z CALCULATIONS2 (2 ff x 0~00  Solubility products can be used to calculate the solubility of compounds. At 25C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq) The expression for the solubility product is Ksp = [Pb2+(aq)] [S2-(aq)] According to the equation, you get one Pb2+(aq) for every one S2-(aq); the concentrations will be equal [Pb2+(aq)] = [S2-(aq)] Substituting and rewriting the expression for Ksp Ksp = [Pb2+(aq)]2 Re-arranging; [Pb2+(aq)] = Ksp = 4 x 10-28 = 2 x 10-14 mol dm-3 K!Bf3f3f3f3f3f33f3< 6 0, 7       m     6  3  U      6  4 F ? x %8N   x fB  x 6D8clB  x <BD8c{>T   x# L)?fB  x 6D8clB x <BD8c{F  b  x   rB x BDo?< b rB xB BDo? @ xB x HDo? M z F  b  x  } rB x BDo?< b rB xB BDo? @ xB x HDo? M z B x s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 0d(  ddB d <D)P0Q d 0Pp@ 0 0dB d@ <D)PXP d 0X0@ 0 0 d 0I~q~ Z CALCULATIONS2 (2 ff d 0PP~0 Solubility products can be used to calculate the solubility of compounds. At 25C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq) The expression for the solubility product is Ksp = [Pb2+(aq)] [S2-(aq)] According to the equation, you get one Pb2+(aq) for every one S2-(aq); the concentrations will be equal [Pb2+(aq)] = [S2-(aq)] Substituting and rewriting the expression for Ksp Ksp = [Pb2+(aq)]2 Re-arranging; [Pb2+(aq)] = Ksp = 4 x 10-28 = 2 x 10-14 mol dm-3 As you get one Pb2+ from one PbS, the solubility of PbS = 2 x 10-14 mol dm-3 |K!Bf3f3f3f3f3f33f3< 6 0, 7   -ff ffm     6  3  U      6  Q     F ? d %8N   d fB  d 6D8clB  d <BD8c{>T   d# L)?fB  d 6D8clB d <BD8c{F  b  d   rB d BDo?< b rB dB BDo? @ xB d HDo? M z F  b  d  } rB d BDo?< b rB dB BDo? @ xB d HDo? M z B d s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 & `(  `dB ` <D)P0Q ` 0Pp@ 0 0dB `@ <D)PXP ` 0X0@ 0 0 ` 0Ȑ~q~ Z CALCULATIONS2 (2 ff* ` 0~0 fSolubility products can be used to calculate the solubility of compounds. At 25C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq) The expression for the solubility product is Ksp = [Pb2+(aq)] [S2-(aq)] According to the equation, you get one Pb2+(aq) for every one S2-(aq); the concentrations will be equal [Pb2+(aq)] = [S2-(aq)] Substituting and rewriting the expression for Ksp Ksp = [Pb2+(aq)]2 Re-arranging; [Pb2+(aq)] = Ksp = 4 x 10-28 = 2 x 10-14 mol dm-3 As you get one Pb2+ from one PbS, the solubility of PbS = 2 x 10-14 mol dm-3 Mr for PbS is 239; the solubility is 239 x 2 x 10-14 g dm-3 = 5.78 x 10-12 g dm-3 [mass = moles x molar mass]gK!Bf3f3f3f3f3f33f3< 6 0, 7  3  0    ffff*m     6  3  U      6  Q    &  n F ? ` %8N   ` fB  ` 6D8clB  ` <BD8c{>T   `# L)?fB  ` 6D8clB ` <BD8c{F  b  `   rB ` BDo?< b rB `B BDo? @ xB ` HDo? M z F  b  `  } rB ` BDo?< b rB `B BDo? @ xB ` HDo? M z B ` s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 `(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0~q~ Z CALCULATIONS2 (2 ff  0~0 n fSolubility products can be used to calculate the solubility of compounds. At 25C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq) The expression for the solubility product is Ksp = [Pb2+(aq)] [S2-(aq)] According to the equation, you get one Pb2+(aq) for every one S2-(aq); the concentrations will be equal [Pb2+(aq)] = [S2-(aq)] Substituting and rewriting the expression for Ksp Ksp = [Pb2+(aq)]2 Re-arranging; [Pb2+(aq)] = Ksp = 4 x 10-28 = 2 x 10-14 mol dm-3 As you get one Pb2+ from one PbS, the solubility of PbS = 2 x 10-14 mol dm-3 Mr for PbS is 239; the solubility is 239 x 2 x 10-14 g dm-3 = 5.78 x 10-12 g dm-3 [mass = moles x molar mass]gK!Bf3f3f3f3f3f33f3&ff fff1ff fffff)  *ff fff2 ff fffffffffffff fff -ff fff $ fffff !!!!*m     6  3  U      6  Q    &  n F ?  %8N    fB   6D3f8clB   <BD3f8c{>T   # L)?fB   6D3f8clB  <BD3f8c{F  b     rB  BD3fo?< b rB B BD3fo? @ xB  HD3fo? M z F  b    } rB  BD3fo?< b rB B BD3fo? @ xB  HD3fo? M z B  s *޽h ? ̙33y___PPT10Y+D=' = @B +" 0L0 1)0$(  $dB $ <D)P0Q $ 0Pp@ 0 0dB $@ <D)PXP $ 0X0@ 0 0 $ 0@4q~ Z CALCULATIONS2 (2 ff $ 09 EThe solubility of ionic compound MY at 25C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25C. xx6f3f3f3f3]f3   B $ s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 p (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0Fq~ Z CALCULATIONS2 (2 ff  0(LO RFThe solubility of ionic compound MY at 25C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200 x6f3f3f3f3]f3 /  S  G RB  s *D1 RB   s *D1B  s *޽h ? ̙33y___PPT10Y+D=' = @B +5 0L0 D< (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0bq~ Z CALCULATIONS2 (2 ff  0h rThe solubility of ionic compound MY at 25C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200 = 2.5 x 10-12 mol dm-3 xD6f3f3f3f3]f3 /  a   s RB  s *D1 RB   s *D1B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   ` (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ Z CALCULATIONS2 (2 ff  0\V  The solubility of ionic compound MY at 25C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200 = 2.5 x 10-12 mol dm-3 The equation for its solubility is MY(s) M+(aq) + Y-(aq) ^x6f3f3f3f3]f3/aM   Z        RB  s *D1 RB   s *D1F ?   hh8N    fB   6D8clB   <BD8c{>T  # L)?fB  6D8clB  <BD8c{B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 "  @ (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ Z CALCULATIONS2 (2 ff  0  The solubility of ionic compound MY at 25C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200 = 2.5 x 10-12 mol dm-3 The equation for its solubility is MY(s) M+(aq) + Y-(aq) The expression for the solubility product is Ksp = [M+(aq)] [Y-(aq)] x6f3f3f3f3]f3/aM 2           2        RB  s *D1 RB   s *D1F ?   hh8N    fB   6D8clB   <BD8c{>T  # L)?fB  6D8clB  <BD8c{B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   P (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0ُq~ Z CALCULATIONS2 (2 ff  0  kThe solubility of ionic compound MY at 25C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200 = 2.5 x 10-12 mol dm-3 The equation for its solubility is MY(s) M+(aq) + Y-(aq) The expression for the solubility product is Ksp = [M+(aq)] [Y-(aq)] According to the equation; moles of M+(aq) = moles of Y-(aq) = moles of dissolved MY(s) x=6f3f3f3f3]f3/aM 6 &  #       2       (       RB  s *D1 RB   s *D1F ?   hh8N    fB   6D8clB   <BD8c{>T  # L)?fB  6D8clB  <BD8c{B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 0+(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ Z CALCULATIONS2 (2 ffy  0  The solubility of ionic compound MY at 25C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200 = 2.5 x 10-12 mol dm-3 The equation for its solubility is MY(s) M+(aq) + Y-(aq) The expression for the solubility product is Ksp = [M+(aq)] [Y-(aq)] According to the equation; moles of M+(aq) = moles of Y-(aq) = moles of dissolved MY(s) Substituting values Ksp = [2.5 x 10-12 ] [2.5 x 10-12 ] The value of the solubility product Ksp = 6.25 x 10-24 mol2 dm-6x6f3f3f3f3]f3/aM 6 ,#    0    *       2       (        P   RB  s *D1 RB   s *D1F ?   hh8N    fB   6D8clB   <BD8c{>T  # L)?fB  6D8clB  <BD8c{B  s *޽h ? ̙33y___PPT10Y+D=' = @B +* 0L0 91 (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0HJq~ Z CALCULATIONS2 (2 ff  0P The solubility of ionic compound MY at 25C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200 = 2.5 x 10-12 mol dm-3 The equation for its solubility is MY(s) M+(aq) + Y-(aq) The expression for the solubility product is Ksp = [M+(aq)] [Y-(aq)] According to the equation; moles of M+(aq) = moles of Y-(aq) = moles of dissolved MY(s) Substituting values Ksp = [2.5 x 10-12 ] [2.5 x 10-12 ] The value of the solubility product Ksp = 6.25 x 10-24 mol2 dm-6vx6f3f3f3f3]f3 )ffff8fffff:ff fff1ff fffff%  =ffff fff+!!!!!!!!*       2       (        P   RB  s *D1 RB   s *D3f1F ?   hh8N    fB   6D3f8clB   <BD3f8c{>T  # L)?fB  6D3f8clB  <BD3f8c{B  s *޽h ? ̙33y___PPT10Y+D=' = @B +T 0L0 c[ (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ cTHE COMMON ION EFFECT2(2ff  0 nAdding a common ion, (one which is present in the solution), will result in the precipitation of a sparingly soluble ionic compound. eg Adding a solution of sodium chloride to a saturated solution of silver chloride will result in the precipitation of silver chloride.&   B  s *޽h ? ̙33y___PPT10Y+D=' = @B +)  0L0 80 (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  00q~ cTHE COMMON ION EFFECT2(2ff  0 nAdding a common ion, (one which is present in the solution), will result in the precipitation of a sparingly soluble ionic compound. eg Adding a solution of sodium chloride to a saturated solution of silver chloride will result in the precipitation of silver chloride.&     0ܬ\  eAdding the ionic compound MA to a solution of MX increases the concentration of M+(aq). M+(aq) is a common ion as it is already in solution.tQ    !@P    0 T   C ,A cieffect2?a"    VAParchmentS" ?X  B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0    + (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ cTHE COMMON ION EFFECT2(2ff  0ƀ nAdding a common ion, (one which is present in the solution), will result in the precipitation of a sparingly soluble ionic compound. eg Adding a solution of sodium chloride to a saturated solution of silver chloride will result in the precipitation of silver chloride.&     0̀\  eAdding the ionic compound MA to a solution of MX increases the concentration of M+(aq). M+(aq) is a common ion as it is already in solution.tQ    !@P    0    0׀\  The extra M+ ions means that the solubility product is exceeded. To reduce the value of [M+(aq)][X-(aq)] below the Ksp, some ions are removed from solution by precipitating.  N   8@Y      8 T   C ,A cieffect2?a" B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0     (   dB   <D)P0Q   0Pp@ 0 0dB  @ <D)PXP   0X0@ 0 0   0xq~ cTHE COMMON ION EFFECT2(2ff   0  C7Adding a common ion, (one which is present in the solution), will result in the precipitation of a sparingly soluble ionic compound. eg Adding a solution of sodium chloride to a saturated solution of silver chloride will result in the precipitation of silver chloride. Silver chloride dissociates in water as follows AgCl(s) Ag+(aq) + Cl(aq) The solubility product at 25C is Ksp = [Ag+(aq)] [Cl(aq)] = 1.2 x 10-10 mol2 dm-6 If the value of the solubility product is exceeded, precipitation will occur. &NG  3   +   O!        .       F ?   z'U8N    fB   6D8clB   <BD8c{>T   # L)?fB   6D8clB   <BD8c{B   s *޽h ? ̙33y___PPT10Y+D=' = @B +( 0L0 7 /  (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ cTHE COMMON ION EFFECT2(2ff  0 ftAdding a common ion, (one which is present in the solution), will result in the precipitation of a sparingly soluble ionic compound. eg Adding a solution of sodium chloride to a saturated solution of silver chloride will result in the precipitation of silver chloride. Silver chloride dissociates in water as follows AgCl(s) Ag+(aq) + Cl(aq) The solubility product at 25C is Ksp = [Ag+(aq)] [Cl(aq)] = 1.2 x 10-10 mol2 dm-6 If the value of the solubility product is exceeded, precipitation will occur. The value can be exceeded by adding EITHER of the two soluble ions. If sodium chloride solution is added, the concentration of Cl(aq) will increase and precipitation will occur. Likewise, addition of silver nitrate solution AgNO3(aq) would produce the same effect as it would increase the concentration of Ag+(aq).N> 3 +O!e O         .           F ?  z'U8N    fB   6D8clB   <BD8c{>T   # L)?fB   6D8clB  <BD8c{B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 T(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0(@q~ cTHE COMMON ION EFFECT2(2ff)  0@I If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3) solutions are mixed, will AgCl be precipitated? Ksp = 1.2 x10-10 mol2 dm-6 @f3f3 f3f3f3f3 f3f3f3f3f3f3f3f3f3f3 f3f3f3f3f3f3f3Z6  0     B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   - (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0aq~ cTHE COMMON ION EFFECT2(2ff  0 n 2If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3) solutions are mixed, will AgCl be precipitated? Ksp = 1.2 x10-10 mol2 dm-6 " in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3 in NaCl the concentration of Cl is 2 x 10-5 mol dm-3 @f3f3 f3f3f3f3 f3f3f3f3f3f3f3f3f3f3 f3f3f3f3f3f3f3   ffff ffff6  0    \     B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   c (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ cTHE COMMON ION EFFECT2(2ff8  0  ,If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3) solutions are mixed, will AgCl be precipitated? Ksp = 1.2 x10-10 mol2 dm-6 " in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3 in NaCl the concentration of Cl is 2 x 10-5 mol dm-3 " when equal volumes are mixed, the concentrations are halved [Ag+] = 1 x 10-5 mol dm-3 [Cl] = 1 x 10-5 mol dm-3 A@<@f3f3 f3f3f3f3 f3f3f3f3f3f3f3f3f3f3 f3f3f3f3f3f3f3    F ffffffffff6  0    \    y   B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0    i (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0ԃq~ cTHE COMMON ION EFFECT2(2ff>  0T  If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3) solutions are mixed, will AgCl be precipitated? Ksp = 1.2 x10-10 mol2 dm-6 " in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3 in NaCl the concentration of Cl is 2 x 10-5 mol dm-3 " when equal volumes are mixed, the concentrations are halved [Ag+] = 1 x 10-5 mol dm-3 [Cl] = 1 x 10-5 mol dm-3 " [Ag+] [Cl] = [1 x 10-5 mol dm-3] x [1 x 10-5 mol dm-3] = 1 x 10-10 mol2 dm-6 A@@f3f3 f3f3f3f3 f3f3f3f3f3f3f3f3f3f3 f3f3f3f3f3f3f3    G   ffff ffffffffffff6  0    \    y  &  Q B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0  (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  04q~ cTHE COMMON ION EFFECT2(2ffZ  0\: ~If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3) solutions are mixed, will AgCl be precipitated? Ksp = 1.2 x10-10 mol2 dm-6 " in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3 in NaCl the concentration of Cl is 2 x 10-5 mol dm-3 " when equal volumes are mixed, the concentrations are halved [Ag+] = 1 x 10-5 mol dm-3 [Cl] = 1 x 10-5 mol dm-3 " [Ag+] [Cl] = [1 x 10-5 mol dm-3] x [1 x 10-5 mol dm-3] = 1 x 10-10 mol2 dm-6 " because this is lower than the Ksp for AgCl... NO PRECIPITATION OCCURS bA@@f3f3 f3f3f3f3 f3f3f3f3f3f3f3f3f3f3 f3f3f3f3f3f3f3    G   ffff ffffffffffff% !6  0    \    y  &  r     B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 @((  (dB ( <D)P0Q ( 0Pp@ 0 0dB (@ <D)PXP ( 0X0@ 0 0 ( 0{q~ cTHE COMMON ION EFFECT2(2ffj ( 0ȍ ~If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3) solutions are mixed, will AgCl be precipitated? Ksp = 1.2 x10-10 mol2 dm-6 " in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3 in NaCl the concentration of Cl is 2 x 10-5 mol dm-3 " when equal volumes are mixed, the concentrations are halved [Ag+] = 1 x 10-5 mol dm-3 [Cl] = 1 x 10-5 mol dm-3 " [Ag+] [Cl] = [1 x 10-5 mol dm-3] x [1 x 10-5 mol dm-3] = 1 x 10-10 mol2 dm-6 " because this is lower than the Ksp for AgCl... 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   On-screen Show / 5Times New Roman Arial BlackTahomaArial Courier NewDefault 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 Slide 30 Slide 31 Slide 32 Slide 33 Slide 34 Slide 35 Slide 36 Slide 37 Slide 38 Slide 39 Slide 40 Slide 41 Slide 42 Slide 43 Slide 44 Slide 45 Slide 46 Slide 47  Fonts UsedDesign Template Slide Titles/ 8@ _PID_HLINKSA0316,3,Slide 3316,3,Slide 3316,3,Slide 3426,4,Slide 4539,6,Slide 6535,10,Slide 10568,35,Slide 35556,19,Slide 19'_JONATHAN HOPTONJONATHAN HOPTON  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|~      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~     Root EntrydO)Pictures<Current UserSummaryInformation(TPowerPoint Document(}DocumentSummaryInformation8