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Introduction and Example 1—Molar Solubility of an AB type Compound. Molar Solubility from K sp.

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Presentation on theme: "Introduction and Example 1—Molar Solubility of an AB type Compound. Molar Solubility from K sp."— Presentation transcript:

1 Introduction and Example 1—Molar Solubility of an AB type Compound. Molar Solubility from K sp

2 Compounds with low solubility always establish an equilibrium when they are placed in water.

3 Compounds with low solubility always establish an equilibrium when they are placed in water. This is called solubility equilibrium.

4 E.g., when lead(II) sulphate is added to water:

5 Compounds with low solubility always establish an equilibrium when they are placed in water. This is called solubility equilibrium. E.g., when lead(II) sulphate is added to water:

6 Compounds with low solubility always establish an equilibrium when they are placed in water. This is called solubility equilibrium. E.g., when lead(II) sulphate is added to water:

7 Compounds with low solubility always establish an equilibrium when they are placed in water. This is called solubility equilibrium. E.g., when lead(II) sulphate is added to water:

8 Compounds with low solubility always establish an equilibrium when they are placed in water. This is called solubility equilibrium. E.g., when lead(II) sulphate is added to water:

9 The equilibrium constant expression, K eq, for this is:

10

11 Solid

12 The equilibrium constant expression, K eq, for this is: For Solubility Equilibrium

13 The equilibrium constant expression, K eq, for this is: For Solubility Equilibrium

14 The equilibrium constant expression, K eq, for this is: The Solubility Product Constant

15 The equilibrium constant expression, K eq, for this is: Product

16 The equilibrium constant expression, K eq, for this is: Product The Solubility Product Constant

17 Solubility depends on the temperature.

18

19 Molar solubility is the number of moles of a substance that will dissolve to form 1.0 L of a saturated solution,

20 Molar solubility is the number of moles of a substance that will dissolve to form 1.0 L of a saturated solution, expressed in moles/L, or molarity (M).

21 A saturated solution is the solution that forms when a compound is at solubility equilibrium.

22 Molar solubility and Concentrations of Ions in a Saturated Solution

23 The concentrations of ions in a saturated solution (one at solubility equilibrium) depend on:

24 The molar solubility of the compound

25 The concentrations of ions in a saturated solution (one at solubility equilibrium) depend on: The molar solubility of the compound The coefficient of each ion in the equilibrium equation

26 For example, consider the compound lead(II) sulphate.

27 For example, consider the compound lead(II) sulphate. It’s solubility equilibrium equation is:

28 Solubility Equilibrium Equation

29 For example, consider the compound lead(II) sulphate. It’s solubility equilibrium equation is:

30

31 × 1/1

32 For example, consider the compound lead(II) sulphate. It’s solubility equilibrium equation is: × 1/1

33 Molar solubility is how much of this dissolves, in moles/L

34 = the molar solubility of PbSO 4 × 1/1 Molar solubility is how much of this dissolves, in moles/L

35 × 1/1 Molar solubility is how much of this dissolves, in moles/L = the molar solubility of PbSO 4 = the molar solubility of PbSO 4

36 × 1/1 One of each ion so PbSO 4 is called an “AB” type compound.

37 Lead(II) bromide

38 Solubility Equilibrium Equation

39 × 1/1 Molar solubility is how much of this dissolves, in moles/L = the molar solubility of PbBr 2

40 × 1/1 × 2/1 Molar solubility is how much of this dissolves, in moles/L = the molar solubility of PbBr 2 = 2 × the molar solubility of PbBr 2

41 × 1/1 × 2/1 1 of one ion and 2 of the other ion so PbBr 2 is called an “AB 2 ” type compound.

42 Finding the Molar Solubility of an AB Type Compound from K sp

43 Find the molar solubility of PbSO 4 at 25°C. Its K sp at 25°C is 1.8 × 10 –8.

44 1 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Solubility Equilibrium Equation

45 1 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Let molar solubility of PbSO 4 = s

46 1 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Let molar solubility of PbSO 4 = s × 1/1 1/1 ratio

47 1 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s The molar solubility of PbSO 4 = 1.3 × 10 –4 M 1/1 ratio s × 1/1

48 1 2 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Ksp Expression

49 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Use this to find the molar solubility

50 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M

51 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M s

52 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M s

53 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M s s

54 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M So Ksp is equal to s time s s s

55 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M

56 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Take the square root of both sides

57 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M

58 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M Substitute 1.8 × 10 –8 in for K sp

59 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M

60 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M

61 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M M is the unit for molar concentration

62 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M 2 significant figures

63 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M round to 2 significant figures

64 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 = 1.3 × 10 –4 M 2 significant figures

65 1 2 3 Find the molar solubility of PbSO 4. Its K sp at 25°C is 1.8 × 10 –8. Let molar solubility of PbSO 4 = s s The molar solubility of PbSO 4 at 25°C = 1.3 × 10 –4 M


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