Mg 3 PO 4 2 s 3 Mg 2 aq 2 PO 4 3 aq 3s 2s K sp Mg 2 3 PO 4 3 2 3s 3 2s 2 3 3 x

# Mg 3 po 4 2 s 3 mg 2 aq 2 po 4 3 aq 3s 2s k sp mg 2 3

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Mg 3 (PO 4 ) 2 (s) 3 Mg 2+ (aq) + 2 PO 4 3 (aq) 3s 2s K sp = [Mg 2+ ] 3 [PO 4 3 ] 2 = [3s] 3 [2s] 2 = 3 3 x 2 2 x s 5 = 108 s 5 = 1.0 x 10 13 4 5 13 13 5 10 9.9 108 10 1.0 108 10 1.0 s Question: Does the compound which has the lower K sp , has the lower solubility? Not necessarily. For Example: a) AgCl K sp = 1.8 x 10 10 b) AgBr K sp = 5.0 x 10 13 a) K sp = [Ag + ][Cl ] = (s)(s) = s 2 = 1.8 x 10 10 s = 1.8 x 10 10 = 1.3 x 10 5 b) K sp = [Ag + ][Br ] = (s)(s) = s 2 = 5.0 x 10 13 s = 5.0 x 10 13 = 7.1 x 10 7 AgBr is less soluble then AgCl BUT!!! a) AgCl K sp = 1.8 x 10 10 b) Ag 2 CrO 4 K sp = 2.4 x 10 12 Question: Which is more soluble? a) K sp = [Ag + ][Cl ] = (s)(s) = s 2 = 1.8 x 10 10 s = 1.8 x 10 10 = 1.3 x 10 5 b) Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2 (aq) 2s s 17-21 K sp = [Ag + ] 2 [CrO 4 2 ] = (2s) 2 (s) = 4s 3 = 2.4 x 10 12 4 10 2.4 s 12 3 5 3 12 10 8.4 4 10 2.4 s Although Ag 2 CrO 4 has the lower K sp , it is the more soluble salt. The Common Ion Effect in Solubility Equilibria 1) Calculate the solubility of PbI 2 (s) in… a) Pure water b) Water containing already dissolved KI which has a concentration of 0.10 M K sp (PbI 2 ) = 7.1 x 10 9 a) PbI 2 (s) Pb 2+ (aq) + 2 I (aq) Initial: 0 M 0 M Equilibrium: s 2 s K SP = [Pb 2+ ][I ] 2 = (s)(2s) 2 = 4s 3 = 7.1 x 10 9 9 7 3 10 8 . 1 4 10 7.1 s M 10 1.2 10 1.8 s 3 3 9 b) Look at the equilibrium PbI 2 (s) Pb 2+ (aq) + 2 I (aq) In this case, the solution already contains some I which comes from another source, which is KI. So one is putting a stress on the system by increasing the concentration of I , therefore the equilibrium will shift to the left. In other words, some of the PbI 2 (s) will precipitate, meaning the solubility will decrease. Quantitatively one can calculate the new solubility, the way the previous calculations were done. 17-22 KI K + (aq) + I (aq) Strong electrolyte, completely dissociates 0.10 M 0.10 M 0.10 M Therefore... [I ] that comes from KI is 0.10 M Next… PbI 2 (s) Pb 2+ (aq) + 2 I (aq) Initial (from KI): 0.10 From PbI 2 : s 2 s Equilibrium: s 0.10 + 2s K sp = [Pb 2+ ][I ] 2 = (s)(0.10 + 2s) 2 = 7.1 x 10 9 Assume… 0.10 + 2s 0.10 In other words… 0.10 >>> 2s K sp = (s)(0.10) 2 = 7.1 x 10 9 7 2 9 10 7.1 (0.1) 10 7.1 s Now Compare… Compound Solublity in Pure H 2 O Solubility in H 2 O containing 0.10 M KI PbI 2 (s) 1.2 x 10 3 M 7.1 x 10 7 M 1.2 x 10 3 M >> 7.1 x 10 7 M Conclusion : Solubility of a sparingly soluble ionic compound decreases in the presence of a common ion. The common ion could be either the cation or the anion. Exercise: Do at home. a) Calculate the solubility of PbI 2 (s) in a solution containing 0.20 M Pb(NO 3 ) 2 . b) Calculate the solubility of Mg 3 (PO 4 ) 2 in pure water. K sp = 1.0 x 10 13 c) Calculate the solubility of Mg 3 (PO 4 ) 2 in 0.10 M (NH 4 ) 3 PO 4 . 17-23 Predicting Precipitation Reactions Question: If a solution of AgNO 3 and a solution of NaCl are mixed would AgCl (s) precipitate? In other words, does one see the formation of a solid AgCl (s)? What is the criterion to observe the formation of a precipitate?  #### You've reached the end of your free preview.

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