Solubility Lesson 5 Trial Ion Product When two

Solubility Lesson 5 Trial Ion Product When two

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 Pb2+ PbCl2(s) NaCl

Pb2+ 2Cl- + 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 0.20 M NaCl, will a precipitate occur? 200 500 PbCl2(s) Pb2+ + 2Cl300 500 0.10 M 0.040 M TIP TIP = Ksp

= = = 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 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) 20 80 Ca2+ + 0.010 M 0.0025 M TIP

= [Ca2+][SO42-] TIP = [0.0025][0.0060] = 1.5 = 7.1 x 10-5 Ksp x 10-5 TIP < Ksp no ppt forms SO4260 0.0080 M 80 0.0060 M 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+ + Cl1 0.060 M 2 0.030 M 1 0.040 M 2 0.020 M TIP = [Ag+][Cl-] TIP = [0.020][0.030] = 6.0 = 1.8 x 10-10 Ksp

TIP > Ksp x 10-4 ppt forms 4. Consider the two saturated solutions AgCl and Ag 2CrO4. Which has the greater Ag+ concentration? AgCl s Ksp 1.8 x 10 -10 s Ag+ s + Cls = s2 = s2 = 1.3 x 10-5 M [Ag+] = 1.3 x 10-5 M Ag2CrO4 s

Ksp 2Ag+ 2s + = 4s3 1.1 x 10-12 = 4s3 s = CrO42s 6.5 x 10-5 M [Ag+] = 2s = 1.3 x 10-4 M Ag2CrO4 has the greater Ag+ concentration

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