Lec 17 Feb 11 Wed

# Lec 17 Feb 11 Wed - B Calculating solubility from Ksp e.g...

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B. Calculating solubility from Ksp e.g. What is the solubility of Ag 3 PO 4 in water at 25 o C ? Ag 3 PO 4 (s) 3 Ag + (aq) + PO 4 3- (aq) Initial solid 0 0 Change solid + 3x + x Equil. solid 3x x Plug into expression for Ksp and solve for s : Ksp = [Ag + ] 3 [PO 4 3- ]= (3x) 3 (x) 1 × 10 -19 = 27x 4 ; x = 7.8 × 10 -6 mol/L Ksp = 1 x 10 -19 (at 25 o C from data sheet) Since Q = 0, rxn goes to right; Set up i.c.e. table 2.19 D. Precipitation from solution Solutions of “insoluble” salts will only tolerate very low concentrations of these ions before the precipitation of the salt occurs. -- At what concentration will the precipitate form? -- Need to calculate Q for reaction written as: solid cation (aq) + anion (aq) Q > Ksp precipitation occurs (reverse reaction is form solid) Q < Ksp no precipitation, could actually dissolve more solid Q = Ksp solution is saturated and at equilibrium C. Solubility and a common ion e.g. What is the solubility of Ag 3 PO 4 in 0.1 M AgNO 3 (aq) at 25 o C ? Q = 0 since one product is not present, rxn goes to right; Setup i.c.e. table Ag 3 PO 4 (s) 3 Ag + (aq) + PO 4 3- (aq) i solid 0.1 0 c solid + 3x + x e solid 0.1 + 3x x Don’t forget Ag + is common ion Plug into expression for Ksp and solve for x : Ksp = [Ag + ] 3 [PO 4 3- ]= (0.1 + 3x) 3 (x) K is small, so x is small x = 1.0 × 10 -19 /0.1 3 make approx: 0.1+3x~0.1 x = 1.0 × 10 -16 (approx is ok) Solubility is now 1.0 × 10 -16 , almost 100 billion times lower. This is because common ion is a product, equilibrium doesn’t go as far to the right.

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2.20 e.g. Mix a 200 mL solution of 0.10 M Pb(NO 3 ) 2 with a 200 mL solution of 0.10 M NaI. Will a precipitate form? Calculate the equilibrium concentrations of Pb 2+ and I - . Ksp (PbI 2 ) = 1.4 × 10 -8 Are the concentrations of Pb 2+ and I - sufficient for solid formation ?
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