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Handout11-SolubilityEqui

# Handout11-SolubilityEqui - CEM 834 I SOLUBILITY EQUILIBRIA...

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CEM 834 SOLUBILITY EQUILIBRIA FALL 2005 I. Simultaneous Equation Approach without Secondary Equilibria Consider the salt M A m n , which dissociates according to the following overall reaction: M A mM nA m n + We can write the equilibrium constant as m n m n m n m n m n d M A M A M A m n M A a a M A K a M A By rearrangement, m n m n m n d M A m n M A M A K M A If the salt is sparingly soluble, we can define the solubility product constant sp K as m n m n d sp M A K M A K S o Thus, the sp K is the product of the equilibrium constant for dissociation and the molecular or intrinsic solubility S o . If the solubility of the salt expressed in moles per liter is denoted by S, then [M] = mS [A] = nS If all activity coefficients are equal to unity, i.e. for an infinitely dilute solution of M A m n at zero ionic strength, then ( 29 ( 29 K mS nS m n S sp m n m n m n = = + 1. Example: Calculate the molar solubility of ( 29 AgCl K 2.0 x 10 at 25 C sp 10 = ° - in water. 1 1 2 sp 1/2 10 5 AgCl Ag Cl K 1 1 S S 2.0 x 10 1.4 x 10 M II. Simultaneous Equation Approach with Secondary Equilibria

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2 Often the solubility of the salt is much higher than predicted by the calculations above because the metal ion can be complexed with ligands, the anion can be complexed with hydrogen ions, or both of these reactions can occur simultaneously. Each of these cases will be considered in detail below. A. Complexes of the anion A with hydrogen ions f ay y 1 y f 2 a 2 f al K A H H A H K A H H HA K HA H A + - + + The total analytical concentration of the anion is C = nS A The fraction of A present in the unprotonated form ( 29 A 0 α 0 f f 2 f y A a1 a2 ay 1 α [H ] β [H ] ... β [H ] Thus, the equilibrium concentration of A is given by [ ] ( 29 ( 29 A C = nS 0 A A 0 A = α α The equilibrium concentration of the metal ion M is [ ] M mS = Thus, by substitution in the expression for the solubility product constant [ ] [ ] ( 29 ( 29 ( 29 ( 29 K M A mS nS m n S sp m n m 0 A n m n 0 A n m n = = = + α α 1. Example: Calculate the molar solubility of CaF (K = 4.0 x 10 at 25 C) 2 sp 11 - ° at pH 7.00 and at pH 1.00, if the formation constant for the weak acid hydrogen fluoride (HF) is K 1.66 x 10 a f 3 = .
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