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Fractions of Dissociating Species in Polyligand Complexes When polyligand complexes are dissociated in solution, metal ions, ligand, and intermediates are obtained in equilibrium with the complex. For example, look at the following equilibria Ag + + NH 3 Ag(NH 3 ) + k f1 = [Ag(NH 3 ) + ]/[Ag + ][NH 3 ] Ag(NH 3 ) + + NH 3 Ag(NH 3 ) 2 + k f2 = [Ag(NH 3 ) 2 + ]/[Ag(NH 3 ) + ][NH 3 ] We have Ag +, NH 3, Ag(NH 3 ) +, and Ag(NH 3 ) 2 + all present in solution at equilibrium where C Ag = [Ag + ] + [Ag(NH 3 ) + ] + [Ag(NH 3 ) 2 + ]

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The fraction of each Ag + species can be defined as: 0 = [Ag + ]/ C Ag 1 = [Ag(NH 3 ) + ]/ C Ag 2 = [Ag(NH 3 ) 2 + ]/ C Ag As seen for fractions of a polyprotic acid dissociating species, one can look at the values as 0 for the fraction with zero ligand (free metal ion, Ag + ), 1 as the fraction of the species having one ligand (Ag(NH 3 ) + ) while 2 as the fraction containing two ligands (Ag(NH 3 ) 2 + ). The sum of all fractions will necessarily add up to unity ( 0 + 1 2 = 1)

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For the case of 0, we make all terms as a function of Ag + since 0 is a function of Ag +. We use the equilibrium constants of each step: C Ag = [Ag + ] + [Ag(NH 3 ) + ] + [Ag(NH 3 ) 2 + ] k f1 = [Ag(NH 3 ) + ]/[Ag + ][NH 3 ] [Ag(NH 3 ) + ] = k f1 [Ag + ][NH 3 ] K f1 x k f2 = [Ag(NH 3 ) 2 + ]/[Ag + ][NH 3 ] 2 [Ag(NH 3 ) 2 + ] = K f1 x k f2 [Ag + ][NH 3 ] 2 Substitution in the C Ag relation gives: C Ag = [Ag + ] + k f1 [Ag + ][NH 3 ] + K f1 x k f2 [Ag + ][NH 3 ] 2 C Ag = [Ag + ]( 1 + k f1 [NH 3 ] + K f1 k f2 [NH 3 ] 2 ) C Ag /[Ag + ] = ( 1 + k f1 [NH 3 ] + K f1 k f2 [NH 3 ] 2 ) The inverse of this equation gives: 0 = 1/ ( 1 + k f1 [NH 3 ] + K f1 k f2 [NH 3 ] 2 )

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If we use the same procedure for the derivation of relations for other fractions we will get the same denominator but the nominator will change according to the species of interest: 0 = 1/ ( 1 + k f1 [NH 3 ] + K f1 k f2 [NH 3 ] 2 ) 1 = k f1 [NH 3 ] / ( 1 + k f1 [NH 3 ] + K f1 k f2 [NH 3 ] 2 ) 2 = K f1 k f2 [NH 3 ] 2 / ( 1 + k f1 [NH 3 ] + K f1 k f2 [NH 3 ] 2 )

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Example Calculate the concentration of the different ion species of silver for M Ag + in a 0.10 M NH 3 solution. Assume that ammonia concentration will not change. K f1 = 2.5x10 3, k f2 = 1.0x10 4 Solution The concentration of silver ion species can be obtained as a function of ammonia concentration and formation constants from the relations above: 0 = 1/ ( 1 + k f1 [NH 3 ] + K f1 k f2 [NH 3 ] 2 ) Substitution in the above equation yields:

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0 = 1/ ( x10 3 * x10 3 * 1.0x10 4 *( 0.10) 2 ) 0 = 4.0x10 -6 0 = [Ag + ]/ C Ag 4.0x10 -6 = [Ag + ]/0.010 [Ag + ] = 4.0x10 -8 M In the same manner calculations give: 1 = 1.0x10 -3 1 = [Ag(NH 3 ) + ]/ C Ag 1.0x10 -3 = [Ag(NH 3 ) + ]/ [Ag(NH 3 ) + ] = 1.0x10 -5 M

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2 = 1.0 2 = [Ag(NH 3 ) 2 + ]/ C Ag 1.0 = [Ag(NH 3 ) 2 + ]/ [Ag(NH 3 ) 2 + ] = M Therefore, it is clear that most Ag + will be in the complex form Ag(NH 3 ) 2 + since the formation constant is large for the overall reaction: K f = k f1 *k f2 K f = 2.5x10 3 * 1.0x10 4 = 2.5x10 7 The actual concentration of ammonia is 0.08 M not 0.10 M. The value of 0.08 M NH 3 should have been used for the calculation of the fractions.

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Precipitation Reactions and Titrations

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We had previously looked at precipitation equilibria in solution and we are familiar with calculations of solubility of sparingly soluble salts in pure water, in presence of a common ion, and in presence of diverse ions. However, we should remember that a salt is formed from a metal ion (a very weak conjugate Lewis acid) that will not react with water, and a conjugate base that may be weak and will not react with water (like Cl -, Br -, I - ) or a strong conjugate base that will react with water (all conjugate bases of weak acids). Calculations presented so far deal with the first situation where the conjugate base is a very weak base. Let us now look at sparingly soluble salts of weak acids:

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Effect of Acidity on the Solubility of Precipitates The conjugate base of a weak acid is strong enough to react with water upon dissociation of the precipitate. At low pH values, excessive amounts of the conjugate base will be converted to the weak acid and thus forces the precipitate to further dissociate to produce more anions to satisfy the equilibrium constant (k sp ). Look at the general example assuming A - is the conjugate base of a weak acid and in presence of acid solution (low pH):

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MA (s) M + + A - k sp = [M + ][A - ] A - + H + HA k a = [HA]/[A - ][H + ] Since some of the formed A - is converted to HA, [M + ] no longer equals [A - ]. However, we can write: C T = [A - ] + [HA], where C T = [M + ] [A - ] = 1 C T Recall EDTA equilibria in solution when its chelate dissociates in water. Therefore, for the equilibrium reaction: MA (s) M + + A -

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We can write: K sp = s * 1 s S = (k sp / 1 ) 1/2 00SolidBefore equil A-A- M+M+ MA(s)Equation 1s1s sSolidAfter Equil

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Example Calculate the solubility of CaC 2 O 4 (k a1 = 6.5x10 -2, k a2 = 6.1x10 -5, k sp = 2.6x10 -9 ) in a M HCl solution. Solution CaC 2 O 4 Ca 2+ + C 2 O 4 2- The released oxalate will form H 2 C 2 O 4, HC 2 O 4 -, and some will remain as C 2 O 4 2- where: C C2O4 2- = s = [Ca 2+ ] = [H 2 C 2 O 4 ] + [HC 2 O 4 - ] + [C 2 O 4 2- ] and: [C 2 O 4 2- ] = 2 s

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K sp = s * 2 s S = (k sp / 2 ) 1/2 Therefore, we must calculate 2 2 = k a1 k a2 / ([H + ] 2 + k a1 [H + ] + k a1 k a2 ) 2 = (6.5x10 -2 * 6.1x10 -5 )/( (0.001) 2 + (6.5x10 -2 * 0.001) + (6.5x10 -2 * 6.1x ) ) 2 = 5.7x10 -2 s = (2.6x10 -9 /5.7x10 -2 ) 1/2 = 2.1x10 -4 M Compare with solubility in absence of acidity!! 00SolidBefore equil C 2 O 4 2- Ca 2+ CaC 2 O 4 (s)Equation 2s2s sSolidAfter Equil

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Example Find the solubility of Ag 3 PO 4 at pH 3. 3 at pH 3=3.3x10 -14, k sp = 1.3x Solution Ag 3 PO 4 3 Ag + + PO 4 3- Solid 3s 3 s K sp = (3 s) 3 * 3 s S = (k sp /27 3 ) 1/4 S = (1.3x /27x3.3x ) 1/4 S = 1.1x10 -2 M

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Effect of Complexation on Solubility We have seen earlier that acidity affects the solubility of precipitates that contains the conjugate base of a weak acid. In the same manner, if a sparingly soluble precipitate was placed in a solution that contains a ligand which can form a complex with the metal ion of the precipitate, solubility of the precipitate will increase due to complex formation. For example, if solid AgCl is placed in an ammonia solution, the following equilibria will take place:

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AgCl (s) Ag + + Cl - Ag + + NH 3 Ag(NH 3 ) + Ag(NH 3 ) + + NH 3 Ag(NH 3 ) 2 + The dissociated Ag + ions will form complexes with ammonia, forcing AgCl to further dissolve. Now, the solubility of AgCl will increase and will equal [Cl - ]. However, C Ag = [Cl - ] = s C Ag = [Ag + ] + [Ag(NH 3 ) + ] + [Ag(NH 3 ) 2 + ] Therefore, [Ag + ] = o s One can then write: AgCl (s) Ag + + Cl - K sp = [Ag + ][Cl - ] K sp = o s * s S = (k sp / o ) 1/2

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Example Find the solubility of AgBr (k sp = 4x ) in pure water and in 0.10 M NH 3 ( o in 0.10 M ammonia = 4.0x10 -6 ). Solution a. In pure water AgBr (s) Ag + + Br - K sp = s * s S = (4.0x ) 1/2, S = 6.3x10 -7 M

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b. In presence of 0.10 M NH 3 K sp = o s * s S = (k sp / o ) 1/2 S = (4.0x /4.0x10 -6 ) 1/2 S = 3.2x10 -4 M % increase in solubility = {(3.2x10 -4 – 6.3x10 -7 )/6.3x10 -7 } x 100 = 5.1X10 4 % Huge increase in solubility is expected as calculation above suggests.

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