6._Chapter_19_solubility_post - Solubility and Complex-Ion...

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Ch. 19 Solubility 1 Solubility and Complex-Ion Equilibria Another equilibrium process Sparingly soluble salts are those that have molar solubilities much less than 1 M The dissolution of AgCl(s) in water is described by the following chemical equation: AgCl(s) Ag + (aq) + Cl¯ The equilibrium constant for this reaction is denoted by K sp and called the solubility product. For a solid such as Ag 3 PO 4 we have: Ag 3 PO 4 (s) 3 Ag + (aq) + PO 4 3- (aq)
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Ch. 19 Solubility 2 Solubility and Complex-Ion Equilibria For sparingly soluble salts, the value of K sp is quite small. If you know the K sp value for a salt you can calculate the molar solubility. Example: Calculate the molar solubility of AgCl in water and the molar solubility of Ag 3 PO 4 in water. AgCl(s) Ag + (aq) + Cl¯ i. excess 0 0 c. –x +x +x e. excess x x K sp = [Ag + ][Cl¯ ] = x 2 x=√K sp = 1.3 x 10 -5 M The molar solibility of AgCl in pure water is 1.3 x 10 -5 M Ag 3 PO 4 (s) 3 Ag + (aq) + PO 4 3- (aq) i. excess 0 0 c. –x + 3x +x e. excess 3x x K sp = [Ag + ] 3 [PO 4 3- ] = (3x) 3 (x) = 27x 4 x= (K sp /27) ¼ = 1.6 x 10 -5 M The molar solubility of Ag 3 PO 4 in pure water is 1.6 x 10-5 M. Note even though the K sp for Ag 3 PO 4 is smaller than the K sp of AgCl, Ag 3 PO 4 is more soluble!!!
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Ch. 19 Solubility 3 Solubility and Complex-Ion Equilibria The solubility product depends on the value of K sp and on the formula of the salt. You cannot predict which salt is more soluble by considering only the K sp value For a 1:1 salt For a 2:1 salt For a 3:1 salt sp K S = 3 4 sp K S = 4 27 sp K S = Common ion Effect Calculate the molar solubility of AgCl(s) in 0.10 M NaCl(aq) (K sp AgCl = 1.8 x 10 -10 )
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Ch. 19 Solubility 4 Precipitation Up to this point, we’ve focused on the dissolution process (i.e. calculating the equilibrium concentrations of ions in solution when excess solid is in equilibrium with the solution avoe it.) Now we turn our attention to the precipitation process, in which we mix together different substances with the aim of predicting whether a precipitate will form. Consider the following equilibrium AgCl(s) Ag + (aq) + Cl¯(aq) At equilibrium, we must have [Ag + ][Cl¯ ] = K sp . The solid is in equilibrium with the solution above it. There is no further dissolution of the solid and the solution is saturated with respect to Ag + and Cl¯.
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