# Solubility and Complex Ion Equilibria

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1 CALCITE Solubility and Complex Ion Equilibria a mineral formed by marine organisms through biological precipitation CaCO (s) Ca + (aq)+ CO (aq) K K sp [Ca + ][CO ].8 x 10-9 K sp solubility product constant Chapter 19: 4, 5, 7a-c, 10, 11a,b, 14, 47, 5, 58, 6, 64 K sp describes equilibrium between a solid and dissolved ions. If no solid is present, there is no equilibrium! K sp [Al + ][OH - ] K sp [Ca + ][SO 4 ] K sp [Mg + ] [PO 4 - ] Solubility The concentration of the solid that can dissolve in solution (expressed in moles per L) when in equilibrium with an excess amount of solid. Expressed another way, Solubility is the maximum concentration (in terms of the solid) of a substance that can exist in solution before precipitation begins if sufficiently seeded. CaCO K sp [Ca + ][CO ].8 x 10-9

2 Problems i) We can determine the solubility from K sp. ii) We can determine K sp from solubilities. Example: What is the solubility of AgCl in water at 5 o C? K sp 1.8 x Let us define solubility as S. AgCl (s) Ag + (aq) + Cl - DEFINE S [Ag (aq) ] [Cl - ] I solid 0 0 C x +x +x E solid x S S K sp [Ag + ][Cl - ] (S)(S) S 10 5 S K 1.8X10 1.X10 M sp Example 1 Example: What is the solubility of Mg (AsO 4 ) at 5 o C? K sp.0 x 10-0 Mg (AsO 4 ) Mg + (aq) + AsO 4 - (aq) + [Mg ] [AsO 4 ] DEFINE S K sp [Mg + ] [AsO - 4 ] (S) (S) 7S x 4S 108 S 5 0 Ksp x10 5 S x10 M Molar solubility: 4.5x10-5 moles/l of Mg (AsO 4 ). Mass solubility: (equivalent to molar solubility X molar mass) 4.5x10-5 moles/l x g/mole g/l Example Determine the K sp of Bi S if the solubility is 1.0 x M. Bi S (s) Bi + (aq) + S (aq) K sp [Bi + ] [S ]? + [Bi ] [S ] 15 DEFINE S 1x10 M + [Bi ] S [S ] S K sp [Bi + ] [S ] (S) (S) 108 S (1.0 x ) x 10-7

3 Factors influencing solubility Common ion and salt effects As with other equilibria we ve discussed, adding a common ion will result in a shift of a solubility equilibrium. For example, AgCl (s) Ag + (aq) + Cl - (aq) K sp [Ag + ] [Cl - ] Adding either Ag + or Cl - to our equilibrium system will result in driving it to the left, precipitating out more AgCl, and lowering the solubility of AgCl. Example a) What is the solubility of CaF? K sp 5.x10-9 CaF (s) Ca + (aq) + F - (aq) S [F - ]/ [Ca + ] K sp [Ca + ][F - ] S(S) 4S -9 K sp S M 4 4 b) What is the solubility of CaF in M CaCl? S [F - ]/ [Ca + ] M K sp [Ca + ][F - ] (0.1)(S) (0.1)(4S ) 9 Ksp 5.x10 4 S 1.X10 M 4(0.1) 4(0.1) Notice that the common ion, Ca +, decreases the solubility of CaF by a factor of 15. Uncommon Ion (Salt) Effect Observation: If NaNO salt is added to AgCl precipitate, it s solubility can be increased dramatically. There is no chemical reaction with the NaNO, so what is going on? Thus far we have used molar concentrations in K sp and other equilibrium expressions, but this is an approximation of the exact solution - we should use activities instead! (recall earlier discussion on activities in section on chemical equilibrium) K sp a Ag+ a Cl-

4 Uncommon vs Common ion effect Other experimental evidence of uncommon ion effect on equilibria Conc. Equilib. Cons., K' E-07 1E-06 1E-05 1E-04 1E-0 1E-0 1E-01 NaCl (M) Ksp BaSO4 x 10^10 Ka CHCOH x 10^5 Kw HO x 10^14 What is activity? Activities are effective concentrations. It is the effective concentration of a species when the species behave under ideal conditions. For ionic species, ideal conditions are dilute conditions, where each molecule or ion behaves independently. At higher concentrations, the species in solution do not necessarily behave independently. The ions can be stabilized in solution through electrostatic forces with other ions in solution.

5 Mathematical description of Activity For ions in solution, the effective concentration is lower than the true concentration (usually) due to electrostatic interactions (stabilization). a i γ i [ I ] where a i is the activity of the species γ i is the activity coefficient of species i [ I ] is the true concentration of species i Activities (effective concentrations) are frequently less than true concentrations. γ values are frequently less than 1. Activity coefficients can be predicted using Debye-Huckel theory. Criteria for Precipitation Ion Product - used to decide if something will precipitate Q [Ca + ] exp [CO ] exp If Q > K sp, the solution is supersaturated. In this case, a precipitate will form. Sufficient material will precipitate until solution becomes saturated. If Q < K sp, precipitate will not form at equilibrium, solution is unsaturated. If solid exists, it will dissolve. If Q K sp, solution is saturated and no net change is expected. ph effects Hydrolysis If the anion of a weak acid, or cation of a weak base, is part of a K sp, solubilities will be greater than expected and will also be dependent upon ph. AgCN(s) Ag + (aq) + CN - + (aq) K sp [Ag ][CN ] + H O(l) The The formation of of HCN reduces the the free free [CN [CN - ], - ], thus thus increasing the the solubility of of AgCN. HCN(aq) + OH - (aq) - [HCN][OH ] Kb [CN ]

6 ph effects: dissolution of limestone CaCO (s) Ca + (aq) + CO (aq) H + Dissolution of of limestone, HCO - CaCO, (aq), is is increased through addition of of acid. H + H CO (aq) H O + CO (g) ph effects: quantitative example What is the solubility of CaF in a solution that is buffered at ph 1? K sp (CaF ) 5. x 10-9, K a (HF) 6.6 X 10-4, pk a.18 CaF (s) Ca + (aq) + F - (aq) K K sp F - + H O + HF + H O K 1/K a CaF (s) + H O + Ca + (aq) + HF + H O K K sp /K a K K K sp a + [Ca ][HF] + [H O ] Since the ph is much lower than the pk a, fluoride will exist predominantly in the HF form, we can presume that [F - ] << [HF]. (x)(x) K [H O + ] + Cont d CaF (s) + H O + Ca + (aq) + HF + H O x S I 0.1M 0 0 C -x x x E ~0.1M x x Remember - solution is buffered! 4x 1.X10 [H O ] + K[H O ] (1.X10 )(0.1) x.1x10 M 4 4 The solubility of CaF in a ph1 solution is.1x10 - M. We can compare this to a solubility of 1.7x10 - M in neutral solution determined previously. Solubility goes up by 18x!!

7 Complex Ion Formation The solubility of slightly soluble salts is increased by the formation of complex ions. Example: addition of excess Cl - to solution of AgCl AgCl(s) Ag + (aq) + Cl - (aq) K sp + [Ag ][Cl ] add large + excess of Cl - (aq) [AgCl ] Kf } + chloride [Ag ][Cl ] A AgCl (aq) large excess of of chloride results in in the the formation of of the the complex. More AgCl will will dissolve as as a result. Fractional Precipitation A method used for: a) quantitative analysis using precipitation titrations b) separation and/or purification Textbook example: 0.01M CrO 4 & 0.01 M Br - Both form insoluble precipitates with Ag + Ag CrO 4 (s) Ag + (aq) + CrO 4 (aq) K sp 1.1x10-1 AgBr (s) Ag + (aq) + Br - (aq) K sp 5.0x10-1 Can we separate the chromate and bromide by fractional precipitation with AgNO? example, cont d Which precipitates first, Br - or CrO 4? Calculate [Ag + ] necessary to just start precipitating each - whichever needs the smaller amount, precipitates first. Br - Q sp K sp [Ag + ][Br - ] [Ag + ](0.01M ) 5.0x10-1 [Ag + ] 5.0x10-1 / x10-11 M CrO 4 K sp [Ag + ] [CrO 4 ] [Ag + ] (0.01M ) 1.1x10-1 K + sp 1.1x10 [Ag ] x10-5 M 1

8 cont d Br - precipitates first. As it continues to precipitate, [Br - ] drops and [Ag + ] increases until it reaches a point where Ag CrO 4 just starts to precipitate. When Ag CrO 4 just starts to precipitate, what is [Br - ]? [CrO 4 ] 0.01M, [Ag + ] 1.0x10-5 (calculated previously) K sp [Ag + ][Br - ] (1.0x10-5 M)[Br - ] 5.0x10-1 [Br - ] 5.0x10-8 M So if we stop the addition of AgNO just before Ag CrO 4 starts to precipitate... [Br - ] drops from 0.01M to 5.0x10-8 M. What % of Br - is left? 5.0x10-8 M / 1.0x10 - M *100% % Good separation!!

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