Chapter 9. Dissolution and Precipitation Equilibria

Transcription

1 Chapter 9 Dissolution and Precipitation Equilibria 9-1 The Nature of Solubility Equilibria 9-2 The Solubility of Ionic Solids 9-3 Precipitation and the Solubility Product 9-4 The Effects of ph on Solubility 9-5 Complex Ions and Solubility 9-6 Controlling Solubility in Qualitative Analysis 2/17/2004 OFB Chapter 9 1

2 Saturated Solution: a solution in equilibrium with excess solute e.g., NaCl solubility in grams per 100 grams water is approximately 36.0 grams = saturated solution Unsaturated Solution: contains less than the equilibrium concentration of the solute Supersaturated Solution: a solution that temporarily contains more of a solute than the equilibrium quantity 2/17/2004 OFB Chapter 9 2

3 Endothermic heat is added to a system Exothermic heat is removed from a system Sharp changes in slope occur if water of crystallization is lost or gained by the solid that is in contact with the solution AgF has two such changes AgF 4H 2 0 AgF 2H 2 0 AgF 2/17/2004 OFB Chapter 9 3

4 9-2 Solubility of Salts This chapter considers only salts which are sparingly soluble or insoluble for which concentrations of saturated salts are [salt] = 0.1 Mol L -1 or less 2/17/2004 OFB Chapter 9 4

5 Solubility Product K sp Describes a chemical equilibrium in which an excess solid salt is in equilibrium with a saturated aqueous solution of its separated ions. General equation AB (s) A + (aq) + B - (aq) 2/17/2004 OFB Chapter 9 5

6 The Solubility of Ionic Solids The Solubility Product AgCl(s) Ag + (aq) + Cl - (aq) K sp = = The solid AgCl, which is in excess, is understood to have a concentration of 1 mole per liter. K sp = at 25 o C 2/17/2004 OFB Chapter 9 6

7 The Solubility of Ionic Solids The Solubility Product Ag 2 SO 4 (s) 2Ag + (aq) + SO 4 2- (aq) K sp = Fe(OH) 3 (s) Fe +3 (aq) + 3OH -1 (aq) K sp = 2/17/2004 OFB Chapter 9 7

8 The Solubility of Ionic Solids Exercise 9-1 The Solubility Product Write the K sp equation for the dissolution of aluminum hydroxide (Al(OH) 3 ) in water. Al(OH) 3 (s) Al 3+ (aq) + 3 OH - (aq) 2/17/2004 OFB Chapter 9 8

9 The Solubility of Ionic Solids The Solubility Product TABLE 9-1contains K sp values at 25C 2/17/2004 OFB Chapter 9 9

10 2/17/2004 OFB Chapter 9 10

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12 2/17/2004 OFB Chapter 9 12

13 Exercise 9-2 The Solubility of Salts Solubility and K sp Determine the mass of lead(ii) iodate dissolved in 2.50 L of a saturated aqueous solution of Pb(IO 3 ) 2 at 25 o C. The K sp of Pb(IO 3 ) 2 is Pb(IO 3 ) 2 (s) Pb 2+ (aq) + 2 IO 3- (aq) [y] [y] [2y] [Pb 2+ ][IO 3- ] 2 = K sp y = [Pb(IO 3 ) 2 ] = [Pb 2+ ] = y = mol L -1 [IO 3- ] = 2y = mol L -1 Gram solubility of Lead (II) iodate = ( mol L -1 ) (557 g mol -1 ) = g L L Pb= /17/2004 OFB Chapter 9 13 I=126.9 O=16 Pb(IO 3 ) 2 = 557g per mole

14 Exercise 9-3 The Solubility of Salts Solubility and K sp Compute the K sp of silver sulfate (Ag 2 SO 4 ) at 25 o C if its mass solubility is 8.3 g L Ag 2 SO 4 (s) 2 Ag + (aq) + 1 SO 4 2- (aq) [y] [2y] [y] 1. Given mass solubility 2. Find molar solubility 3. Find equilibrium expression expressed in molar solubility 4. Find K sp mass moles = Molar Mass mass = moles x Molar Mass 2/17/2004 OFB Chapter 9 14

15 Exercise 9-3 The Solubility of Salts Solubility and K sp Compute the K sp of silver sulfate (Ag 2 SO 4 ) at 25 o C if its mass solubility is 8.3 g L Ag 2 SO 4 (s) 2 Ag + (aq) + 1 SO 4 2- (aq) [y] [2y] [y] [y] = (8.3 g Ag 2 SO 4 L -1 ) (1 mol Ag 2 SO 4 /311.8 g) [y] = [ ] mol Ag 2 SO 4 L -1 [Ag + ] 2 [SO 4 2- ]=K sp 2/17/2004 OFB Chapter 9 15

16 Review The Nature of Solubility Equilibria Dissolution and precipitation are reverse of each other. Dissolution (Solubility) General reaction X 3 Y 2 (s) 3X +2 (aq) + 2Y -3 (aq) [s] [3s] [2s] K sp = [X +2 ] 3 [Y -3 ] 2 = (3s) 3 (2s) 2 s = molar solubility expressed in moles per liter 2/17/2004 OFB Chapter 9 16

17 Review The Nature of Solubility Equilibria Dissolution and precipitation are reverse of each other. Precipation General reaction X 3 Y 2 (s) 3X +2 (aq) + 2Y -3 (aq) Mix [X +2 ] and [Y -3 ] Does a ppt of X 3 Y 2 form? [X +2 ] [Y -3 ] Reaction quotient before mixing occurs: Q (init) = [X +2 ] 3 (init)[y -3 ] 2 (init) Q > K? K sp = [X +2 ] 3 [Y -3 ] 2 2/17/2004 OFB Chapter 9 17

18 Precipitation from Solution: Does a solid ppt form? AgCl(s) Ag + (aq) + Cl - (aq) Q (init) = [Ag + ] (init) [Cl - ] (init)= Reaction quotient Ksp = [Ag + ][Cl - ] If Q > Ksp then the solid precipitates 2/17/2004 OFB Chapter 9 18

19 Precipitation and the Solubility Product Exercise 9-4: Precipitation from Solution The K sp of thallium (I) iodate is at 25 o C. Suppose that 555 ml of a M solution of TlNO 3 is mixed with 445 ml of a M solution of NaIO 3. Does TlIO 3 precipitate at equilibrium? Evaluate : Reaction quotient before mixing occurs: Q (init) = [Tl + ] (init) [IO 3- ] (init) If Q (init) > K sp, solid TlIO 3 precipitates until Q = K sp If Q (init) < K sp, no solid TlIO 3 can appear. [Tl + ] Q > K sp Solid ppt Q < K sp No ppt 2/17/2004 OFB Chapter 9 19 [IO 3- ]

20 Exercise 9-4 The K sp of thallium(i) iodate is at 25 o C. Suppose that 555 ml of a M solution of TlNO 3 is mixed with 445 ml of a M solution of NaIO 3. Does TlIO 3 precipitate at equilibrium? Tl(IO 3 ) (s) Tl + (aq) + IO 3- (aq) [Tl + ] (init) = ( mol L -1 )(555 ml/1000 ml) = mol L -1 [IO 3- ] (init) = ( mol L -1 )(445 ml/1000 ml) Q (init) = [Tl + ] (init) [IO 3- ] (init) = mol L -1 = (0.0012)( ) = Q (init)? K sp = < Because Q (init) < K sp, solid TlIO 3 does NOT precipitate! 2/17/2004 OFB Chapter 9 20

21 Precipitation and the Solubility Product The Common Ion Effect If a solution and a solid salt to be dissolved in it have an ion in common, then the solubility of the salt is depressed. 2/17/2004 OFB Chapter 9 21

22 Exercise 9-6 The Common Ion Effect The K sp of thallium(i) iodate (TlO 3 ) is at 25 o C. Determine the molar solubility of TlIO 3 in mol L -1 KIO 3 at 25 o C. TlIO 3 (s) Tl + (aq) + IO 3- (aq) [Tl + ] (mol L -1 ) [IO 3- ] (mol L -1 ) Initial concentration Change in concentration Equilibrium concentration [Tl + ][IO 3- ] = K sp Assume s is small s = [TlIO 3 ]= mol L -1 which is depressed 2/17/ times OFB relative Chapter 9to the 1.76 x conc. without the common ion

23 The Effects of ph on Solubility Solubility of Hydroxides Many solids dissolve more readily in more acidic solutions Zn(OH) 2 (s) Zn 2+ (aq) + 2 OH - (aq) [Zn 2+ ][OH - ] 2 = K sp = If ph decreases (or made more acidic), the [OH-] decreases. In order to maintain K sp the [Zn 2+ ] must increase and consequently more solid Zn(OH) 2 dissolves. 2/17/2004 OFB Chapter 9 23

24 The Effects of ph on Solubility Solubility of Hydroxides Exercise 9-7 Estimate the molar solubility of Fe(OH) 3 in a solution that is buffered to a ph of 2.9. In pure water: [Fe 3+ ] = y [OH - ] = 3y y(3y) 3 = 27y 4 = K sp = y = mol L -1 = [Fe 3+ ] = [Fe(OH) 3 ]= [OH - ] = 3y = mol L -1 poh = 8.87 (and ph = 5.13) In pure water, Fe(OH) 3 is 5 x 10 6 less soluble than at ph = 2.9 2/17/2004 OFB Chapter 9 24

25 9-7 The Effects of ph on Solubility Solubilities of Hydroxides Solubility of Salts and Weak Bases Selective Precipitation of Ions Metal Sulfides Somewhat more complicated due to other competing reactions. E.g., MS + H 2 O M 2+ + OH - + HS - (Metal Sulfide) But as before solubility of Metal Sulfides increase as ph decreases K sp = [M 2+ ][OH - ][HS - ] As ph decreases (or made more acidic), the [OH-] decreases. In order to maintain K sp the [M 2+ ] must increase and consequently more solid Metal Sulfide 2/17/2004 OFB Chapter 9 25 dissolves.

26 Examples / Exercises 9-1, Ksp calculations 9-2, Ksp calculations 9-3, Ksp calculations 9-4, ppt Q? Ksp 9-5, Equilibrium concentrations 9-6, Common Ion effect 9-7 Effect of ph of on solubility Problems 16, 19, 23, 30, 39, 41, 42 2/17/2004 OFB Chapter 9 26