1 / 18

Solubility Product Constant

Solubility Product Constant. Ksp. is a variation on the equilibrium constant for a solute-solution equilibrium. remember that the solubility equilibrium is based on a saturated solution. Solids are not written in the equilibrium expression. CaF 2 (s) Ca 2+ (aq) + 2F - (aq).

amina
Download Presentation

Solubility Product Constant

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Solubility Product Constant Ksp

  2. is a variation on the equilibrium constant for a solute-solution equilibrium. • remember that the solubility equilibrium is based on a saturated solution. • Solids are not written in the equilibrium expression.

  3. CaF2 (s) Ca 2+ (aq) + 2F -(aq) Ksp = [Ca 2+] [F -]2 Where Ksp is the solubility product constant, or simply the solubility product.

  4. CaCl2 (s) Ca 2+ (aq) + 2 Cl -(aq) Ksp = [Ca 2+] [Cl -]2

  5. Given the solubility, calculate the Ksp of the substance. 1).Write the chemical equation for the substance dissolving and dissociating. 2) Write the Ksp expression. 3) Insert the concentration of each ion and multiply out.

  6. Calculating the solubility product: CuBr has a measured solubility of 2.0 10-4 M at 25 C. Calculate the Ksp value. CuBr (s) Cu + (aq) + Br -(aq) CuBr (s) Cu + (aq) + Br -(aq) Ksp = [Cu +] [Br -] = (2.0 10-4 )(2.0 10-4 ) = 4.0 10-8

  7. x EX: Calculate the Ksp for Bi2 S3 , which has a solubility of 1.36 x 10-15 mol/L at 25°C. Bi2S3 (s) 2 Bi3+ (aq) + 3 S2- (aq) 2(1.36 x 10-15 ) 3(1.36 x 10-15) 2.72 x 10-15 4.08 x 10-15 x 3x 2x

  8. Ksp = [Bi 3+]2 [S 2-]3 = (2.72 x 10-15)2 (4.08 x 10-15)3 = 5.02 x 10-73

  9. Determine the Ksp of calcium fluoride (CaF2 ), given that its molar solubility is 2.14 x 10-2 M. • CaF2 (s) → Ca2+ (aq) + 2 F¯ (aq) • The Ksp expression is: Ksp = [Ca2+ ] [F¯]2 2x x

  10. Ksp = [Ca2+ ] [F¯]2 • Ksp = (2.14 x 10-2 ) (4.28 x 10-2 )2 • Ksp = 3.92 x 10-5

  11. Convert g/l to mol/l The solubility of Fe(OH)2 is found to be 1.4 x 10-3 g/l. What is the Ksp value? Change the g/l to mol/l. Molar mass of Fe(OH)2 is 89.87g/mol Solubility is = 1.56 x 10-5 mol/l Fe(OH)2(s) → Fe2+(aq) + 2OH-(aq) x x 2x Ksp = [Fe2+ ][OH- ]2 =(1.56 x 10-5 ) (2[1.56 x 10-5 ]) 2 =1.5 x 10-14

  12. Calculate the Ksp of silver bromide: ( AgBr ), if the solubility is 1.3 x 10 -4 g / L .

  13. We can also use a known value of Ksp to find out the solubility of a substance. • Write out the chemical equation for the substance • Write the Ksp expression • Solve for x

  14. If the Ksp is 1.8 x 10 -8 of silver chloride, • write Ksp expression for silver chloride. • AgCl (s) ------> Ag+ (aq) + Cl- (aq) • x x

  15. Ksp = [Ag + (aq) ] [Cl - (aq)] • 1.8 x 10 -8 = x2 • 1.3 x 10-8 = x

  16. The Ksp value for Cu(IO3)2 is 1.4 x 10-7 at 25°C. Calculate its solubility at this temperature. Cu(IO3)2 (s) Cu2+ (aq) + 2 IO31- (aq) x x 2x

  17. Ksp = [Cu 2+] [IO31-]2 • 1.47 x 10-7 = (x) (2x)2 • 1.47 x 10-7 = 4x3 • 1.47 x 10-7 = 4x3 4 4 3.3 x 10-3 = x Solubility of Cu(IO3)2

  18. If Ksp of calcium hydroxide is 5.5 x 10-6, find the solubility. Ca(OH)2 (s)→ Ca+2 (aq) + 2OH-(aq) x=0.011 M

More Related