patino_chm2046_chapter16

010 m bano32 b ksp for baso4 11 x 10 10 11 baso4s

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Unformatted text preview: y containing 0.010 M Ba2+ (one common ion. The other is SO42−) 0.010 (one Which way will the “common ion” shift the equilibrium? Will solubility of BaSO4 be less than or greater than in pure water?___ The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. (b) Ksp for BaSO4 = 1.1 x 10-10 1.1 BaSO4(s) ⇄ Ba2+(aq) + SO42-(aq) (s) Solution Solution initial change equilib. equilib. [Ba2+] [Ba 0.010 +y 0.010 + y [SO42-] 0 +y y = the molar solubility the of BaSO4 of The Common Ion Effect The Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. (b) Ksp for BaSO4 = 1.1 x 10-10 1.1 BaSO4(s) ⇄ Ba2+(aq) + SO42-(aq) (s) Solution Solution Ksp = [Ba2+] [SO42-] = (0.010 + y) (y) Because y < 1.1 x 10-5 M (= x, the solubility in pure water), this means 0.010 + y is about equal to 0.010 0.010. Therefore, 0.010 Ksp = 1.1 x 10-10 = (0.010)(y) y = 1.1 x 10-8 M = solubility in presence of 0.010 M 1.1 added Ba2+ ion. y is indeed < 1.1 x 10-5 M added 1.1 The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. Ba(NO Ksp for BaSO4 = 1.1 x 10-10 1.1 BaSO4(s) ⇄ Ba2+(aq) + SO42-(aq) (s) SUMMARY SUMMARY Solubility BaSO4 in pure water = x = 1.1 x 10-5 M Solubility BaSO4 in presence of added (0.010 M) Ba2+ = 1.1 x 10-8 M Solubility = [SO42−] Ba Le Chatelier’s Principle is followed! (solubilty Le dec; or, more BaSO4 ppts. as equil shifts left) dec; Dissolving Precipitates by forming Dissolving Complex Ions Ions Insoluble ionic compounds can dissolve in the presence of a Lewis base. AgCl(s) ⇄ Ag+(aq) + Cl-(aq) Ksp = 1.8 × 10-10 Ag+(aq) + 2NH3(aq)⇄ [Ag(NH3)2]+(aq) Kf = 1.6 × 107 (aq) 2NH (aq) Complex Ion ---------------------------------------------------------------------formation formation + AgCl(s) + 2NH3(aq) ⇄ [Ag(NH3)2] (aq) + Cl (aq) equilibrium (aq) (aq) Cl constant constant Ch. 14: add two equations, Knet = product of the two Ch. net equilibrium constants for the two equations. equilibrium Knet = Ksp • Kf = (1.8× 10-10) (1.6× 107) = 2.9 × 10-3 2.9 this is a moderate value equilibrium constant this...
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