patino_chm2046_chapter16

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online