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Chapter 16 Solubility and Complex Ion Equilibria

Chapter 16 Solubility and Complex Ion Equilibria. 16.1 – Solubility E quilibria and the Solubility P roduct. Solubility Product (Constant), K sp : Take MgF 2 for example, MgF 2(s) ⇌ Mg 2+ ( aq ) + 2F - ( aq ) , K sp = 5.16 x 10 -11

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Chapter 16 Solubility and Complex Ion Equilibria

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  1. Chapter 16Solubility and Complex Ion Equilibria

  2. 16.1 – Solubility Equilibria and the Solubility Product • Solubility Product (Constant), Ksp: Take MgF2 for example, MgF2(s)⇌ Mg2+(aq) + 2F-(aq), Ksp = 5.16 x 10-11 Where Ksp = 5.16 x 10-11 = [Mg2+] x [F-]2

  3. 16.1 – Solubility Equilibria and the Solubility Product • Solubility Product (Constant), Ksp: This K behaves just like any other K. It is influenced by temperature, according to Le Chatelier’s principle. Most dissolution reactions are endothermic, meaning increasing temperature also increases the [M+] and [X-].

  4. 16.1 – Solubility Equilibria and the Solubility Product • Example, what is the concentration of acetate ions when 5g of silver acetate (MW 167g/mol) are dissolved in 150ml of water? Ksp of AgCH3COO is 1.94 x 10-3.

  5. 16.1 – Solubility Equilibria and the Solubility Product • We can also express solubility in terms of moles per litre. Solubility measured in these dimensions can be converted to a Ksp value. • How do you define solubility? • For Ag2S, solubility is 3.4 x 10-17 M We can say that there are 3.4 x 10-17 moles of S2- in 1 litre of solution containing Ag2S. There are 6.8 x 10-17 moles of Ag+ because [Ag+] = 2[S2-] upon dissociation

  6. 16.1 – Solubility Equilibria and the Solubility Product • Example, what is the Ksp for Ag2S? The solubility of silver sulfide is 3.4 x 10-17mol/L

  7. 16.1 – Solubility Equilibria and the Solubility Product • Example, Lead (II) bromide has a Ksp of 4.6 x 10-6. What is its solubility in moles per litre?

  8. 16.1 – Solubility Equilibria and the Solubility Product • Example, Approximately 0.14g of nickel (II) hydroxide dissolve per litre of water at 20oC. Calculate Ksp for Ni(OH)2 at this temperature. MW of Ni (OH)2 = 92.71g/mol

  9. 16.1 – Solubility Equilibria and the Solubility Product • Example, What is the solubility of Ba3(PO4)2 in g/L? Ksp = 6.0 x 10-39, MW = 601.8g/mol

  10. 16.1 – Solubility Equilibria and the Solubility Product • Comparing solubilities • When the salts you’re comparing have the same number of ions (i.e., NaCl, MgSO4, KClO4, etc…) the one with the highest Ksp is most soluble • When the salts you’re comparing have a different number of ions (i.e., MgCl2, ZnF, Ca3(PO4)2, etc…), you must convert each Ksp to molar solubility, then compare solubilities

  11. 16.1 – Solubility Equilibria and the Solubility Product • Comparing solubilities • Example, which has a higher solubility? Zn(OH)2, Ksp = 4.5 x 10-17 BaCO3, Ksp= 1.6 x 10-9 Ag3PO4Ksp= 1.8 x 10-18

  12. 16.1 – Solubility Equilibria and the Solubility Product • Common Ion Effect • We saw this before – adding a common ion to the solution shifts the equilibrium to the reactants. In other words, adding a common ion reduces solubility. • Now, we will look at the quantitative aspect of this:

  13. 16.1 – Solubility Equilibria and the Solubility Product • Example, • Calculate the solubility of SrF2 (Ksp = 7.9 x 10-10) in: • A. Pure water • B. 0.10M Sr(NO3)3 • C. 0.40M NaF

  14. 16.1 – Solubility Equilibria and the Solubility Product • pH and solubility: • If you add an acid/base to a solution, and the H+/OH- can react with the ions in the solution, (taking them out of solution), then solubility increases because Le Chatelier says the equilibrium shifts to the products (i.e., more ions!) • Don’t worry about these calculations

  15. 16.1 – Solubility Equilibria and the Solubility Product • pH and solubility: • Calculations you should worry about are calculating solubility in different pH’s

  16. 16.1 – Solubility Equilibria and the Solubility Product • Example, • Calculate the solubility of Fe(OH)3,Ksp = 4 x 10-38 in the following solutions: • A. Water • B. A solution buffered at pH 5 • C. A solution buffered at pH 11

  17. 16.2 – Precipitation and Qualitative Analysis • We will be doing quantitative work with qualitative analysis now. • This section asks: “If I mix these two solutions, will a precipitate form? If so, what will be the concentration of remaining ion in solution?”

  18. 16.2 – Precipitation and Qualitative Analysis • To do this, we will look at a value similar to Ksp. We will call it the ion product, and denote it with Qsp. • It is calculated the same way as Ksp, but instead of using equilibrium concentrations, Qsp uses initial concentrations.

  19. 16.2 – Precipitation and Qualitative Analysis • Recall:

  20. 16.2 – Precipitation and Qualitative Analysis • When dealing with possible precipitation equilibria, if Qsp > Ksp, then a precipitate will form because the equilibrium shifts towards reactants (i.e., the solid)

  21. 16.2 – Precipitation and Qualitative Analysis • Example, • A 200ml solution of 1.3 x 10-3 M AgNO3 is mixed with 100ml of 4.5 x 10-5 M Na2S solution. Will a precipitation reaction occur?

  22. 16.2 – Precipitation and Qualitative Analysis • Example, • First, find out what your potential ppt is Ag+NO3- S2-Na+ Ag2S appears to be the only ppt that can form Check your table to find Kspof Ag2S is 1.6 x 10-49

  23. 16.2 – Precipitation and Qualitative Analysis • Example, • Next, find the initial concentrations of Ag+ and S2- (upon mixing) [Ag+]i = (0.2L) (0.0013M) = 8.67 x 10-4 M (0.3L) [S2-]i = (0.1L) (0.000045M) = 1.5 x 10-5 M (0.3L)

  24. 16.2 – Precipitation and Qualitative Analysis • Example, • Next, plug these values into the equation for Qsp Ag2S(s)⇌ 2Ag+ + S2- Ksp = [Ag+]2 [S2-] = 1.6 x 10-49 Qsp = (8.67 x 10-4)2 (1.5 x 10-5) = 1.1 x 10-11 Qsp > Ksp, so an Ag2S precipitate will form

  25. 16.2 – Precipitation and Qualitative Analysis • Example, • In the previous question, what will be [Na+], [NO3-], [Ag+], and [S2-] at equilibrium? • [Na+] and [NO3-] are easy, since the original salts Na2S and AgNO3 are 100% soluble: [Na+] = 4.5 x 10-5M and [NO3-] = 1.3 x 10-3M

  26. 16.2 – Precipitation and Qualitative Analysis • Example, • Flip the equation around (inverting Ksp) Ag2S(s)⇌ 2Ag+ + S2-Ksp = 1.6 x 10-49 2Ag+ + S2- ⇌ Ag2S(s) K = 6.25 x 1048

  27. 16.2 – Precipitation and Qualitative Analysis • Example, • Then do a RICE diagram (in moles): 2Ag+ + S2- ⇌ Ag2S(s) K = 6.25 x 1048 0.00026 0.000045 --- -0.000090 -0.000045 --- 0.00017 0 --- 0.00017 moles Ag+, [Ag+] = 0.00017/0.3L = 5.67 x 10-4M

  28. 16.2 – Precipitation and Qualitative Analysis • Example, • At equilibrium, [S2-] is not actually zero because the reaction does not go to completion. A small amount of Ag2S re-dissolves to reach equilibrium. • Now we must ask ourselves: what is the solubility of solid Ag2S in a 5.67 x 10-4M Ag+ solution?

  29. 16.2 – Precipitation and Qualitative Analysis • Example, • To answer this, go back to our solubility product: R Ag2S(s)⇌ 2Ag+ + S2- I 5.67 x 10-4 0 C +2s +s E 5.67 x 10-4 s Ksp = (5.67 x 10-4)2(s) = 1.6 x 10-49… s = [S2-] = 4.98x10-43M

  30. 16.2 – Precipitation and Qualitative Analysis

  31. 16.2 – Precipitation and Qualitative Analysis

  32. 16.3 – Equilibria Involving Complex Ions • A complex ion is a charged species consisting of a metal ion (Lewis Acid) surrounded by multiple ligands (Lewis Base). • Common metal ions: • Ag+, Fe2+/3+, Cu+/2+, Al3+ • Common ligands: • H2O, NH3, CN-, Cl-

  33. 16.3 – Equilibria Involving Complex Ions • The number of ligands attached to the metal ion is called the coordination number. The coordination number is usually 2, 4, or 6, but others are known. • Ligands add one at a time, with each addition having its own equilibrium constant (K1, K2, etc…) or formation/stability constant

  34. 16.3 – Equilibria Involving Complex Ions • For example, in the formation of Cu(NH3)42+, Cu2+ + NH3 ⇌ Cu(NH3)2+ K1 = 1 x 105 Cu(NH3)2++ NH3 ⇌ Cu(NH3)22+K2 = 2 x 105 Cu(NH3)22++ NH3 ⇌ Cu(NH3)32+K3 = 3 x 105 Cu(NH3)32++ NH3 ⇌ Cu(NH3)42+K4 = 4 x 105

  35. 16.3 – Equilibria Involving Complex Ions • For example, • If you were to combine 100ml of 0.001M Cu(NO3)2 with 100ml of 1M NH3 • We know that NH3 will react as a base towards water, but because the equilibrium lies so far to the left (Kb << 1), we can ignore the contribution of the acid-base reaction

  36. 16.3 – Equilibria Involving Complex Ions • For example, • On the other hand, the K’s for the formation of the complex ion are very large, they essentially go to completion • As soon as the two are mixed, but before any reaction occurs, we know the initial concentrations are: [Cu2+]0 = ((0.1L)(0.001M)/(0.2L)) = 0.0005M [NH3]0 = ((0.1L)(1M)/(0.2)) = 0.5M

  37. 16.3 – Equilibria Involving Complex Ions • For example, • The RICE diagram would look something like this: R Cu2+ + 4NH3 Cu(NH3)42+ I 0.0005M 0.5M 0 C -0.0005M -4(0.0005M) +0.0005M E 0 ~0.5M 0.0005M

  38. 16.3 – Equilibria Involving Complex Ions • For example, • Although they will be small, we can calculate [Cu(NH3)2+], [Cu(NH3)22+] and [Cu(NH3)32+] by backtracking to the four formation equations. For example, Cu(NH3)32+ + NH3 ⇌ Cu(NH3)42+ K4 = 4 x 105

  39. 16.3 – Equilibria Involving Complex Ions • For example, Cu(NH3)32+ + NH3 ⇌ Cu(NH3)42+ K4= 4 x 105 K4 = 4 x 105 = [Cu(NH3)42+] [Cu(NH3)32+] [NH3] [Cu(NH3)32+] = (0.0005M)/(4 x 105) = 1.25 x 10-9M

  40. 16.3 – Equilibria Involving Complex Ions • For example, Repeat this procedure with K3, K2, and K1 to get: [Cu(NH3)22+] = 4.2 x 10-15 [Cu(NH3)2+] = 2.1 x 10 [Cu2+] = 1.1 x 10

  41. 16.3 – Equilibria Involving Complex Ions • The solubility of complex ions is important in qualitative analysis. For example, adding HCl to a solution containing Ag+, Pb2+, and Hg22+ will form the precipitates AgCl, PbCl2, and Hg2Cl2. • Now how can we separate them from each other and identify the individual cations?

  42. 16.3 – Equilibria Involving Complex Ions • Sometimes, lowering the pH will cause some precipitates to dissolve. It will not work here because the Cl- is such a weak base, it will have no affinity for H+. AgCl(s)⇌ Ag+(aq) + Cl-(aq) PbCl2(s)⇌Pb+(aq) + 2Cl-(aq) Hg2Cl2(s)⇌ Hg22+(aq) + 2Cl-(aq)

  43. 16.3 – Equilibria Involving Complex Ions • Le Chatelier gives us another choice: lower the amount of Ag+/Pb2+/Hg22+ AgCl(s)⇌ Ag+(aq) + Cl-(aq) PbCl2(s)⇌ Pb2+(aq) + 2Cl-(aq) Hg2Cl2(s)⇌ Hg22+(aq) + 2Cl-(aq)

  44. 16.3 – Equilibria Involving Complex Ions • To do this, we can add excess NH3, which will form a complex ion with Ag+, Ag(NH3)2+ • Adding ammonia will lower [Ag+], which will drive the reaction… AgCl(s)⇌ Ag+(aq) + Cl-(aq) • … forward, dissolving the AgCl(s)

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