Solubility Lesson 5 Trial Ion Product

 When two ionic solutions are mixed and if one product has low solubility, 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, if the ions are > 0.10 M. A trial ion product is required if the ion concentrations are < 0.10 M.

The capacity of a solution to dissolve a solid is described by the Ksp. Pb(NO3)2 2Cl- Pb2+ ⇌ NaCl PbCl2(s) ⇌ Pb2+ + 2Cl- The Ksp represents the limit of the solution to dissolve PbCl2. Pb2+ and Cl- will dissolve until the ion concentrations are equal to the Ksp. The solution is saturated- any more ions will form a solid.

1. 200. 0 mL 0. 10 M Pb(NO3)2 is mixed with 300. 0 mL of 0 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- 200 300 0.10 M 0.20 M 500 500 0.040 M 0.12 M TIP = [Pb2+][Cl-]2 TIP = [0.040][0.12] 2 = 5.8 x 10-4 Ksp = 1.2 x 10-5 TIP > Ksp ppt forms

2. Will a precipitate form if 20. 0 mL of 0 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+ + SO42- 20 0.010 M 60 0.0080 M 80 80 0.0025 M 0.0060 M TIP = [Ca2+][SO42-] TIP = [0.0025][0.0060] = 1.5 x 10-5 Ksp = 7.1 x 10-5 TIP < Ksp no ppt forms

3. Will a precipitate form when equal volumes of 0. 020 M AlCl3 and 0 3. Will a precipitate form when equal volumes of 0.020 M AlCl3 and 0.040 M AgNO3 are mixed. The Cl- x 3 AgCl(s) ⇌ Ag+ + Cl- 1 1 0.040 M 0.060 M 2 2 0.020 M 0.030 M TIP = [Ag+][Cl-] TIP = [0.020][0.030] = 6.0 x 10-4 Ksp = 1.8 x 10-10 TIP > Ksp ppt forms

Ag2CrO4 has the greater Ag+ concentration 4. Consider the two saturated solutions AgCl and Ag2CrO4. Which has the greater Ag+ concentration? Ag2CrO4 ⇌ 2Ag+ + CrO42- AgCl ⇌ Ag+ + Cl- s s s s 2s s Ksp = s2 Ksp = 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