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Solubility Lesson 5 Trial Ion Product We have learned that when two ionic solutions are mixed and if one product has low solubility, then there is a reaction where a precipitate will form. Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq) low solubility The solubility chart on page 4 predicts this reaction, but only if the solution is 0.10 M or greater. If the molarity is less than 0.10 M, then the reaction may or may not happen. A trial ion product must be calculated to predict the reaction of all solutions less than 0.10 M. The capacity of a solution to dissolve a solid is described by the Ksp. Pb(NO3)2 NaCl Pb2+ PbCl2(s) ⇌ 2ClPb2+ + 2Cl- The Ksp represents the limit of the solution to dissolve PbCl2. You can add Pb2+ and Cl- until the ion concentrations are equal to the Ksp. The solution is saturated and addition PbCl2 must sit on the bottom not dissolved. 1. 200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur? 1. 200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur? PbCl2(s) ⇌ Pb2+ + 2Cl- Write a dissociation equation for the compound with low solubility. 1. 200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur? PbCl2(s) ⇌ Pb2+ + 2Cl0.10 M 0.20 M Write a dissociation equation for the compound with low solubility. List the initial molarities of each ion. 1. 200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur? PbCl2(s) ⇌ Pb2+ + 2Cl200 500 0.10 M 0.040 M 300 500 0.20 M 0.12 M Write a dissociation equation for the compound with low solubility. List the initial molarities of each ion. Reduce each molarity by the dilution factor: V1/V2. 1. 200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur? PbCl2(s) ⇌ Pb2+ + 2Cl200 500 0.10 M 0.040 M TIP TIP = = = 300 500 0.20 M 0.12 M [Pb2+][Cl-]2 [0.040][0.12] 2 5.8 x 10-4 Write a dissociation equation for the compound with low solubility. List the initial molarities of each ion. Reduce each molarity by the dilution factor: V1/V2. Write the Ksp or TIP (trial ion product) and solve. 1. 200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur? PbCl2(s) ⇌ Pb2+ + 2Cl200 500 0.10 M 0.040 M TIP TIP Ksp = = = = 300 500 0.20 M 0.12 M [Pb2+][Cl-]2 [0.040][0.12] 2 5.8 x 10-4 1.2 x 10-5 TIP > Ksp ppt forms Write a dissociation equation for the compound with low solubility. List the initial molarities of each ion. Reduce each molarity by the dilution factor: V1/V2. Write the Ksp or TIP (trial ion product) and solve. Compare to real Ksp 2. Will a precipitate form if 20.0 mL of 0.010M CaCl2 is mixed with 60.0 mL of 0.0080 M Na2SO4? CaSO4(s) ⇌ Ca2+ 20 80 0.010 M + 0.0025 M Ksp SO42- 60 0.0080 M 80 0.0060 M [Ca2+][SO42-] [0.0025][0.0060] 1.5 x 10-5 TIP TIP = = = = 7.1 x 10-5 TIP < Ksp no ppt forms 3. Will a precipitate form when equal volumes of 0.020 M CaCl2 and 0.040 M AgNO3 are mixed. The Cl- is doubled AgCl(s) ⇌ Ag+ + Cl- 1 0.040 M 2 0.020 M TIP TIP = = = [Ag+][Cl-] [0.020][0.020] 4.0 x 10-4 Ksp = 1.8 x 10-10 TIP > Ksp ppt forms 1 0.040 M 2 0.020 M 4. Consider the two saturated solutions AgCl and Ag2CrO4. Which has the greater Ag+ concentration? AgCl ⇌ s Ksp Ag+ + Cls = s2 s Ag2CrO4 ⇌ 2Ag+ s + 2s Ksp = CrO42s 4s3 1.8 x 10 -10 = s2 1.1 x 10-12 =4s3 s = 1.3 x 10-5 M s = 6.5 x 10-5 M [Ag+] = 1.3 x 10-5 M [Ag+] = 2s = 1.3 x 10-4 M Ag2CrO4 has the greater Ag+ concentration