1. Solubility Lesson 5 Trial Ion Product

2. 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(NO 3 ) 2(aq) + 2NaCl (aq) → PbCl 2(s) + 2NaNO 3(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.

3. The capacity of a solution to dissolve a solid is described by the Ksp. PbCl 2(s) ⇌ Pb 2+ +2Cl - The Ksp represents the limit of the solution to dissolve PbCl 2. You can add Pb 2+ and Cl - until the ion concentrations are equal to the Ksp. The solution is saturated and addition PbCl 2 must sit on the bottom not dissolved. Pb 2+ 2Cl - Pb(NO 3 ) 2 NaCl

4. 1.200.0 mL 0.10 M Pb(NO 3 ) 2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur?

5. PbCl 2(s) ⇌ Pb 2+ +2Cl - Write a dissociation equation for the compound with low solubility.

6. 1.200.0 mL 0.10 M Pb(NO 3 ) 2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur? PbCl 2(s) ⇌ Pb 2+ +2Cl - 0.10 M0.20 M Write a dissociation equation for the compound with low solubility. List the initial molarities of each ion.

7. 1.200.0 mL 0.10 M Pb(NO 3 ) 2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur? PbCl 2(s) ⇌ Pb 2+ +2Cl - 2000.10 M3000.20 M 500500 0.040 M0.12 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: V 1 /V 2.

8. 1.200.0 mL 0.10 M Pb(NO 3 ) 2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur? PbCl 2(s) ⇌ Pb 2+ +2Cl - 2000.10 M3000.20 M 500500 0.040 M0.12 M TIP=[Pb 2+ ][Cl - ] 2 TIP=[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: V 1 /V 2. Write the Ksp or TIP (trial ion product) and solve.

9. 1.200.0 mL 0.10 M Pb(NO 3 ) 2 is mixed with 300.0 mL of 0.20 M NaCl, will a precipitate occur? PbCl 2(s) ⇌ Pb 2+ +2Cl - 2000.10 M3000.20 M 500500 0.040 M0.12 M TIP=[Pb 2+ ][Cl - ] 2 TIP=[0.040][0.12] 2 =5.8 x 10 -4 Ksp=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: V 1 /V 2. Write the Ksp or TIP (trial ion product) and solve. Compare to real Ksp

10. 2. Will a precipitate form if 20.0 mL of 0.010M CaCl 2 is mixed with 60.0 mL of 0.0080 M Na 2 SO 4 ? CaSO 4(s) ⇌ Ca 2+ +SO 4 2- 200.010 M600.0080 M 8080 0.0025 M0.0060 M TIP=[Ca 2+ ][SO 4 2- ] TIP=[0.0025][0.0060] =1.5 x 10 -5 Ksp=7.1 x 10 -5 TIP < Ksp no ppt forms

11. AgCl (s) ⇌ Ag + +Cl - 10.040 M10.040 M 220.020 M TIP=[Ag + ][Cl - ] TIP=[0.020][0.020] =4.0 x 10 -4 Ksp=1.8 x 10 -10 TIP > Ksp ppt forms The Cl - is doubled 3.Will a precipitate form when equal volumes of 0.020 M CaCl 2 and 0.040 M AgNO 3 are mixed.

12. 4.Consider the two saturated solutions AgCl and Ag 2 CrO 4. Which has the greater Ag + concentration? AgCl ⇌ Ag + + Cl - Ag 2 CrO 4 ⇌ 2Ag + + CrO 4 2- ssss2s s Ksp=s 2 Ksp=4s 3 1.8 x 10 -10 = s 2 1.1 x 10 -12 =4s 3 s = 1.3 x 10 -5 Ms = 6.5 x 10 -5 M [Ag + ] = 1.3 x 10 -5 M[Ag + ] = 2s = 1.3 x 10 -4 M Ag 2 CrO 4 has the greater Ag + concentration