1. Chapter 16 Lesson 1 Solubility and Complex Ion Equilibria

2. Chapter 16 Table of Contents Copyright © Cengage Learning. All rights reserved 2 16.1Solubility Equilibria and the Solubility Product 16.2 Precipitation and Qualitative Analysis 16.3 Equilibria Involving Complex Ions

3. Ksp, the solubility-product constant. An equilibrium can exist between a partially soluble substance and its solution:

4. Section 16.1 Solubility Equilibria and the Solubility Product Return to TOC Copyright © Cengage Learning. All rights reserved 4 Solubility Equilibria Solubility product (K sp ) – equilibrium constant; has only one value for a given solid at a given temperature. Solubility – an equilibrium position. Bi 2 S 3 (s) 2Bi 3+ (aq) + 3S 2– (aq)

5. For example: BaSO 4 (s)  Ba 2+ (aq) + SO 4 2- (aq) When writing the equilibrium constant expression for the dissolution of BaSO 4, we remember that the concentration of a solid is constant. The equilibrium expression is therefore: K = [Ba 2+ ][SO 4 2- ] K = Ksp, the solubility-product constant. Ksp = [Ba 2+ ][SO 4 2- ]

6. The Solubility Expression A a B b (s)  aA b+ (aq) + bB a- (aq) Ksp = [A b+ ] a [B a- ] b Example:PbI 2 (s)  Pb 2+ + 2 I - Ksp = [Pb 2+ ] [I - ] 2

7. Section 16.1 Solubility Equilibria and the Solubility Product Return to TOC Copyright © Cengage Learning. All rights reserved 7 Exercise Calculate the solubility of silver chloride in water. K sp = 1.6 × 10 –10 1.3×10 -5 M Calculate the solubility of silver phosphate in water. K sp = 1.8 × 10 –18 1.6×10 -5 M

8. Solubility and Ksp Three important definitions: 1)solubility: quantity of a substance that dissolves to form a saturated solution 2)molar solubility: the number of moles of the solute that dissolves to form a liter of saturated solution 3)Ksp (solubility product): the equilibrium constant for the equilibrium between an ionic solid and its saturated solution

9. Calculating Molar Solubility Calculate the molar solubility of Ag 2 SO 4 in one liter of water. Ksp = 1.4 x 10 -5 Ag 2 SO 4  2Ag + + SO 4 2- Initial00 Change+2x+x Equilb x2x x Ksp = [Ag + ] 2 [SO 4 2- ] = (2x) 2 (x)= 1.4 x 10 -5 X = 1.5 x 10 -2 mol Ag 2 SO 4 /L (molar solubility)

10. Section 16.1 Solubility Equilibria and the Solubility Product Return to TOC Copyright © Cengage Learning. All rights reserved 10 Concept Check In comparing several salts at a given temperature, does a higher K sp value always mean a higher solubility? Explain. If yes, explain and verify. If no, provide a counter-example. No

11. Common ion Effect PbI 2 (s)  Pb 2+ (aq) + 2I – (aq) Common ion: “The ion in a mixture of ionic substances that is common to the formulas of at least two.” Common ion effect: “The solubility of one salt is reduced by the presence of another having a common ion”

12. PbI 2 (s)Pb 2+ (aq)I – (aq) 12 00.10 x2x 0.10 + 2xx R I C E Ksp = [x] [0.10 + 2x] 2 = 7.9 x 10 –9 Ksp = [Pb 2+ (aq)] [I – (aq)] 2 Example #1 What is the Molar solubility of PbI 2 if the concentration of NaI is 0.10? Ksp = 7.9 x 10 –9 So [I-] = 0.10 M x is small, thus we can ignore 2x in 0.10 + 2x Ksp = [x] [0.10] 2 = 7.9 x 10 –9, x = 7.9 x 10 –7 M

13. Common Ion Effect Thus the solubility of the PbI 2 is reduced by the presence of the NaI. Ksp of PbI 2 = 7.9 x 10 -9, so the molar solubility is 7.9 x 10 -9 = (x)(2x) 2 = 4x 3 X = 1.3 x 10 -3 Which is much greater than 7.9 x 10 -7 when 0.10 M NaI is in solution.  The greater the Ksp the more soluble the solid is in H 2 O (once you have the molar solubility).

14. AgI(s)Ag + (aq)I – (aq) 11 00.20 xx 0.20 + xx R I C E Ksp = [x] [0.20 + x] = 8.3 x 10 –17 Ksp = [Ag + (aq)] [I – (aq)] Example #2 Molar solubility of AgI? Ksp = 8.3 x 10 –17 Concentration of NaI is 0.20, thus [I – ] = 0.20 x is small, thus we can ignore it in 0.20 + x Ksp = [x] [0.20] = 8.3 x 10 –17, x = 4.2 x 10 –16

15. Common Ion Effect When two salt solutions that share a common ion are mixed the salt with the lower Ksp will precipitate first. Example: AgCl ksp = [Ag + ][Cl - ] = 1.6 x10 -10 [Ag + ][Cl - ] = 1.6 x 10 -10 X 2 = 1.6 x 10 -10 X = [Ag + ] = [Cl - ] = 1.3 x 10 -5 M

16. Common Ion Effect Add 0.10 M NaCl to a saturated AgCl solution. [Cl-] = 0.10 (common ion) [Ag + ][Cl - ] = 1.6 x 10 -10 [Ag + ][0.10 + x] = 1.6 x 10 -10 (x is small) [Ag + ]= 1.6 x 10 -10 / 0.10 M [Ag + ]= 1.6 x 10 -9  [Ag+] = 1.3 x 10 -5 from the previous slide So, some AgCl will precipitate when the NaCl is added because the molar solubility of the solution is now less than the that of AgCl alone.

17. Section 16.1 Solubility Equilibria and the Solubility Product Return to TOC Copyright © Cengage Learning. All rights reserved 17 Concept Check How does the solubility of silver chloride in water compare to that of silver chloride in an acidic solution (made by adding nitric acid to the solution)? Explain. The solubilities are the same.

18. Section 16.1 Solubility Equilibria and the Solubility Product Return to TOC Copyright © Cengage Learning. All rights reserved 18 Concept Check How does the solubility of silver phosphate in water compare to that of silver phosphate in an acidic solution (made by adding nitric acid to the solution)? Explain. The silver phosphate is more soluble in an acidic solution.

19. Section 16.1 Solubility Equilibria and the Solubility Product Return to TOC Copyright © Cengage Learning. All rights reserved 19 Concept Check How does the K sp of silver phosphate in water compare to that of silver phosphate in an acidic solution (made by adding nitric acid to the solution)? Explain. The K sp values are the same (temperature dependent)

20. Section 16.1 Solubility Equilibria and the Solubility Product Return to TOC Copyright © Cengage Learning. All rights reserved 20 Exercise Calculate the solubility of AgCl in: K sp = 1.6 × 10 –10 a)100.0 mL of 4.00 x 10 -3 M calcium chloride. 2.0×10 -8 M b)100.0 mL of 4.00 x 10 -3 M calcium nitrate. 1.3×10 -5 M