Presentation is loading. Please wait.

Presentation is loading. Please wait.

William L Masterton Cecile N. Hurley Edward J. Neth University of Connecticut Chapter 16 Precipitation.

Similar presentations

Presentation on theme: "William L Masterton Cecile N. Hurley Edward J. Neth University of Connecticut Chapter 16 Precipitation."— Presentation transcript:

1 William L Masterton Cecile N. Hurley Edward J. Neth University of Connecticut Chapter 16 Precipitation Equilibria

2 Outline 1. Precipitate formation: the solubility product constant (K sp ) 2. Dissolving precipitates

3 Revisiting Precipitation In Chapter 4 we learned that there are compounds that do not dissolve in water These were called insoluble A reaction that produces an insoluble precipitate was assumed to go to completion In reality, even insoluble compounds dissolve to some extent, usually small An equilibrium is set up between the precipitate and its ions Precipitates can be dissolved by forming complex ions

4 Two Types of Equilibria AgCl (s) Ag + (aq) + Cl - (aq) Solid exists in equilibrium with the ions formed when a small amount of solid dissolves AgCl (s) + 2NH 3 (aq) Ag(NH 3 ) 2 + (aq) + Cl - (aq) Formation of a stable complex ion can cause an otherwise insoluble compound to dissolve There are multiple equilibria at work in this example, in similar fashion to the equilibria underlying the function of a buffer (Chapter 14)

5 Precipitate Formation: Solubility Product Constant, K sp Consider mixing two solutions: Sr(NO 3 ) 2 (aq) K 2 CrO 4 (aq) The following net ionic equation describes the reaction: Sr 2+ (aq) + CrO 4 2- (aq) SrCrO 4 (s)

6 Figure 16.1 – Precipitation of SrCrO 4

7 K sp Expression SrCrO 4 (s) Sr 2+ (aq) + CrO 4 2- (aq) The solid establishes an equilibrium with its ions once it forms We can write an equilibrium expression, leaving out the term for the solid (recall that its concentration does not change as long as some is present) K sp is called the solubility product constant

8 Interpreting the Solubility Expression and K sp K sp has a fixed value at a given temperature For strontium chromate, K sp = 3.6 X at 25 °C The product of the two concentrations at equilibrium must have this value regardless of the direction from which equilibrium is approached

9 Example 16.1

10 K sp and the Equilibrium Concentration of Ions K sp SrCrO 4 = [Sr 2+ ][CrO 4 2- ] = 3.6 X This means that if we know one ion concentration, the other one can easily be calculated If [Sr 2+ ] = 1.0 X M, then If [CrO 4 2- ] = 2.0 X 10 -3, then

11 Example 16.2

12 Example 16.1, (Contd)

13 Table 16.1

14 K sp and Precipitate Formation K sp values can be used to predict whether a precipitate will form when two solutions are mixed Recall the use of Q, the reaction quotient, from Chapter 12 We can calculate Q at any time and compare it to K sp The relative magnitude of Q vs. K sp will indicate whether or not a precipitate will form

15 Q and K sp If Q > K sp, a precipitate will form, decreasing the ion concentrations until equilibrium is established If Q < K sp, the solution is unsaturated; no precipitate will form If Q = K sp, the solution is saturated just to the point of precipitation

16 Figure 16.2

17 Example 16.3

18 Example 16.3, (Contd)


20 K sp and Water Solubility One way to establish a solubility equilibrium Stir a slightly soluble solid with water An equilibrium is established between the solid and its ions BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) If we set the concentration of the ions equal to a variable, s:

21 Precipitation Visualized

22 Example 16.4

23 Example 16.4, (Cont'd)


25 Calculating K sp Given Solubility Instead of calculating solubility from K sp, it is possible to calculate K sp from the solubility Recall that solubility may be given in many different sets of units Convert the solubility to moles per liter for use in the K sp expression

26 Example 16.5

27 Example 16.5, (Cont'd)

28 K sp and the Common Ion Effect BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) How would you expect the solubility of barium sulfate in water to compare to its solubility in 0.10 M Na 2 SO 4 ? Solubility must be less than it is in pure water Recall LeChâteliers Principle The presence of the common ion, SO 4 2-, will drive the equilibrium to the left Common ions reduce solubility

29 Visualizing the Common Ion Effect

30 Example 16.6

31 Example 16.6, (Cont'd)

32 Selective Precipitation Consider a solution of two cations One way to separate the cations is to add an anion that precipitates only one of them This approach is called selective precipitation Related approach Consider a solution of magnesium and barium ions

33 Selective Precipitation, (Cont'd) K sp BaCO 3 = 2.6 X K sp MgCO 3 = 6.8 X Carbonate ion is added Since BaCO 3 is less soluble than MgCO 3, BaCO 3 precipitates first, leaving magnesium ion in solution Differences in solubility can be used to separate cations

34 Figure 16.3

35 Figure 16.4 – Selective Precipitation

36 Example 16.7

37 Example 16.7, (Cont'd)

38 Dissolving Precipitates Bringing water-insoluble compounds into solution Adding a strong acid to react with basic anions Adding an agent that forms a complex ion to react with a metal cation

39 Strong Acid Zn(OH) 2 (s) + 2H + (aq) Zn 2+ (aq) + 2H 2 O This reaction takes place as two equilibria: Zn(OH) 2 (s) Zn 2+ (aq) + 2OH - (aq) 2H + (aq) + 2OH - (aq) 2H 2 O Zn(OH) 2 (s) + 2H + (aq) Zn 2+ (aq) + 2H 2 O Because the equilibrium constant for the neutralization is so large, the reaction goes essentially to completion Note that for the second equilibrium, K = (1/K w ) 2 = 1 X 10 28

40 Example 16.8

41 Example 16.8, (Cont'd)


43 Insoluble Compounds that Dissolve in Strong Acid Virtually all carbonates The product if the reaction is H 2 CO 3, a weak acid that decomposes to carbon dioxide H 2 CO 3 (aq) H 2 O + CO 2 (g) Many sulfides The product of the reaction is H 2 S, a gas that is also a weak acid H 2 S (aq) H + (aq) + HS - (aq)

44 Visualizing Selective Dissolving of Precipitates

45 Example 16.9

46 Complex Formation Ammonia and NaOH can dissolve compounds whose metal cations form complexes with NH 3 and OH - As with the addition of a strong acid, multiple equilibria are at work: Zn(OH) 2 (s) Zn 2+ (aq) + 2OH - (aq)K sp Zn 2+ (aq) + 4NH 3 (aq) Zn(NH 3 ) 4 2+ (aq) K f Net: Zn(OH) 2 (s) + 4NH 3 (aq) Zn(NH 3 ) 4 2+ (aq) + 2OH - (aq) K net = K sp K f = 4 X X 3.6 X 10 8 = 1 X 10 -8

47 Table 16.2

48 Visualizing Dissolving by Complex Formation

49 Example 16.10

50 Example 16.10, (Cont'd)

51 Example 16.11

52 Key Concepts 1. Write the K sp expression for an ionic solid 2. Use the value of K sp to A. Calculate the concentration of one ion, knowing the other B. Determine whether a precipitate will form C. Calculate the water solubility of a compound D. Calculate the solubility of a compound in a solution of a common ion E. Determine which ion will precipitate first

53 Key Concepts 3. Calculate K for A. Dissolving a metal hydroxide in a strong acid B. Dissolving a precipitate in a complexing agent 4. Write balanced, net ionic equations to explain why a precipitate dissolves in A. Strong acid B. Ammonia or hydroxide solution

Download ppt "William L Masterton Cecile N. Hurley Edward J. Neth University of Connecticut Chapter 16 Precipitation."

Similar presentations

Ads by Google