Download presentation

Presentation is loading. Please wait.

Published byConner Bartell Modified over 8 years ago

1
Solubility Equilibria

2
Write a balanced chemical equation to represent equilibrium in a saturated solution. Write a solubility product expression. Answer questions about Ksp and various missing concentrations using I.C.E. tables.

3
C 6 H 12 O 6(s) C 6 H 12 O 6(aq)

4
There are 3 actions that affect solubility: 1. Nature of the solute and solvent like dissolves like Polar / ionic solute dissolve in polar solvent. Non-polar dissolve in non-polar. Even the most insoluble ionic solids are actually soluble in water to a limited extent

5
2. Temperature Solids in liquids: temperature - solubility. Gases in liquids: in temperature - solubility. 3. Pressure Does not affect the solubility of (s)/(l). (g): pressure solubility.

6
A a B b(s) aA + (aq) + bB¯ (aq) K sp, called the solubility product constant. K sp = [A + ] a [B - ] b Product of ion concentrations in a saturated solution. K c = [A + ] a [B - ] b [A a B b ]

7
Write the dissociation and the product constant equation for the solubility of calcium hydroxide. K sp = [Ca 2+ ][OH - ] 2 Ca(OH) 2 (s ) Pb 3 (PO 4 ) 2(s) 3 Pb 2+ (aq) + 2 PO 4 3- (aq) K sp = [Pb 2+ ] 3 [PO 4 3- ] 2 Write a solubility product expression for Pb 3 (PO 4 ) 2. Ca 2+ (aq) + OH - (aq) 2

8
At equilibrium, the [Ag + ] = 1.3 x 10 -5 M and the [Cl - ] = 1.3 x 10 -5 M, what is the K sp of silver chloride? K sp = [Ag + ][Cl - ] K sp =(1.3 x 10 -5 )(1.3 x 10 -5 ) K sp = 1.7 x 10 -10 AgCl (s ) Ag + (aq) + Cl - (aq) *NOTE: K sp has no units.

9
Solubility And I.C.E. Tables (Yeah!) Solubility - maximum amount of solute that can dissolve in a certain amount of solvent at a certain temperature.

10
Calculate K sp of lead (II) chloride if a 1.0 L saturated solution has of lead ions. I---00 C--- +x+2x E--- K sp = [Pb +2 ][Cl - ] 2 K sp = [1.62 x 10 -2 ][ 3. 24 x 10 -2 ] 2 K sp = 1.70 x 10 -5 PbCl 2(s) Pb 2+ (aq) + 2 Cl - (aq) 1.62 x 10 -2 M 2(1.62 x 10 -2 ) 1.62 x 10 -2 M

11
The solubility of PbF 2 is. What is the value of the solubility product constant? PbF 2(s) Pb 2+ (aq) + 2 F¯ (aq) 0.466 g 1 L 245.2 g 1 mol = 1.90 x 10 -3 M PbF 2 0.466 g/L Pb – 207 + 2 (19) = 245g/mol K sp = [Pb 2+ ][F - ] 2

12
K sp = (1.90 x 10 -3 )(3.80 x 10 -3 ) 2 K sp = 2.74 x 10 -8 PbF 2(s) Pb 2+ (aq) + 2 F¯ (aq) [E][E] 0 1.9 x 10 -3 M [I][I] 1.9 x 10 -3 mol/L 00 [C][C]- x+ x+ 2x 3.8 x 10 -3 M Saturated – all solid reactant dissociates.

13
Calculate K sp if 50.0 mL of a saturated solution was found to contain 0.2207 g of lead (II) chloride. I0.0159 0 0 C-x +x +2x E0 0.0159 M0.0318 M K sp = [Pb 2+ ][Cl - ] 2 0.2207 g 278.1g 1 mol = 0.0159 M PbCl 2 0.05 L PbCl 2(s) Pb 2+ (aq) + 2 Cl - (aq) K sp = [0.0159][0.0318] 2 = 1.61 x 10 -5

14
K sp of magnesium hydroxide is 8.9 x 10 -12. What are the [equilibrium] of ions in saturated solution? I---00 C---+x+2x E--- x 2x Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH - (aq) K sp = [Mg 2+ ][OH - ] 2 8.9 x 10 -12 = [x][2x] 2 8.9 x 10 -12 = [x] 4x 2 8.9 x 10 -12 = 4x 3

15
[Mg 2+ ] = x = 1.3 x 10 -4 mol/L [OH - ] = 2x = 2.6 x 10 -4 mol/L 8.9 x 10 -12 = 4x 3 44 2.23 x 10 -12 = x 3 3 3 1.3 x 10 -4 = x

16
Estimate the solubility in g/L of Ag 2 CrO 4 if the K sp is 1.1 x 10 -12. I---00 C---+2x+x E--- K sp = [Ag + ] 2 [CrO 4 2- ] 1.1 x 10 -12 = [2x] 2 [x] 1.1 x 10 -12 = 4x 3 x = 6.50 x 10 -5 M Ag 2 CrO 4(s) 2 Ag + (aq) + CrO 4 2- (aq) 1.30 x 10 -4 M 6.50 x 10 -5 M

17
[Ag 2 CrO 4 ] i = 6.50 x 10 -5 moles/1L Ag 2 CrO 4(s) 2 Ag + (aq) + CrO 4 2- (aq) 6.5 x 10 -5 mol 330 g 1 mol = 0.022 g/L 1 L E---1.30 x 10 -4 M 6.50 x 10 -5 M 1 1

18
Precipitation

19
Compare value of Q, with given K sp to determine if an aqueous solution is saturated or unsaturated. Q = K sp Saturated solution, no precipitate. Q >K sp Precipitate forms (oversaturated) Q < K sp Solution is unsaturated. Q sp = [A + ] a [B¯] b

20
PbF 2(s) Pb 2+ (aq) + 2 F¯ (aq) Pb – 207 + 2 (19) = 245g/mol K sp of lead (II) fluoride is 1.6 x 10 -5. If 0.57 g are mixed with 1500 mL of water, is solution saturated? 0.57 g 1 L 245.2 g 1 mol = 2.32 x 10 -3 mol/L Q sp = (2.32 x 10 -3 )(4.64 x 10 -3 ) 2 Q sp = 5.0 x 10 -8 Q >K sp Precipitate forms K sp = [Pb 2+ ][F - ] 2

21
0.01 M NaCl0.02 M Pb(NO 3 ) 2 Predict if there is a precipitate of PbCl 2 if 100 mL each of of and are added together. K sp of PbCl 2 = 1.7 x 10 -5 PbF 2(s) Pb 2+ (aq) + 2 Cl¯ (aq) Q sp = [Pb 2+ ][Cl - ] 2 0.01 M NaCl0.02 M Pb(NO 3 ) 2

22
C 1 V 1 = C 2 V 2 (0.01 M)(0.1 L) = (C 2 )(0.2 L) = 0.005 M Cl - (0.02 M)(0.1 L) = (C 2 )(0.2 L) = 0.01 M Pb 2+ Q sp = [0.01][0.005] 2 = 2.5 x 10 -7 Q sp < K sp A precipitate will not form. Because you are adding volumes, you must account for the new diluted concentrations.

23
If 20.0 mL of 0.0010 M silver nitrate is mixed with 20.0 mL of 3.0 x 10 -5 M potassium bromide, does silver bromide (K sp = 5.0 x 10 -13 ) precipitate? Assume the volumes are additive. AgBr (s) Ag + (aq) + Br¯ (aq) K sp = [Ag + ][Br - ] (0.001 M)(0.02 L) = (C 2 )(0.04 L) = 5.0 x 10 -4 M Ag + (2e -5 M)(0.02 L) = (C 2 )(0.04 L) = 1.5 x 10 -5 M Br - Q sp = [5.0 x 10 -4 ][1.5 x 10 -5 ] = 7.5 x 10 -9 Q sp > K sp A precipitate will form.

24
Substances which are insoluble are actually slightly soluble. The solubility product, K sp, describes the product of ion concentrations in saturated solutions. Solubility can be determined from the solubility product.

Similar presentations

© 2022 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google