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Solubility Lesson 5 Trial Ion Product. When two ionic solutions are mixed and if one product has low solubility , a precipitate will form. Pb(NO 3 ) 2(aq) + 2NaCl (aq) → PbCl 2(s) + 2NaNO 3(aq). low solubility.
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Solubility Lesson 5 Trial Ion Product
When two ionic solutions are mixed and if one producthaslowsolubility, 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 NaCl Pb2+ 2Cl- ⇌ PbCl2(s)⇌ Pb2+ + 2Cl- The Ksp represents the limitof the solution to dissolve PbCl2. Pb2+and Cl- will dissolve until the ion concentrationsare equal to the Ksp. The solution is saturated- anymore 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? 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 0.010M 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 0.040 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
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 1.1 x 10-12 = 4s3 1.8 x 10 -10 = s2 s = 1.3 x 10-5 M s = 6.5 x 10-5 M [Ag+] = 2s = 1.3 x 10-4 M [Ag+] = 1.3 x 10-5 M Ag2CrO4 has the greater Ag+ concentration