1 / 30

Solubility Equilibrium

Solubility Equilibrium. Chapter 7. The Solubility Equilibrium. Remember from SPH3U: Solubility is the amount of solute that dissolves in a given amount of solvent (usually water) at a particular temperature. Measured in g/100 mL or mol/L (molar concentration)

ewa
Download Presentation

Solubility Equilibrium

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 Equilibrium Chapter 7

  2. The Solubility Equilibrium Remember from SPH3U: • Solubility is the amount of solute that dissolves in a given amount of solvent (usually water) at a particular temperature. • Measured in g/100 mL or mol/L (molar concentration) • Very soluble - > 10 g/100 mLSoluble – 1 – 10 g/100 mLSlightly soluble – 0.1 – 1 g/100 mLInsoluble - < 0.1 g/100 mL

  3. The Solubility Equilibrium • Produced solubility curves • Most solids increased in solubility as the temperature increased. While, most gases decreased in solubility as the temperature increased. • Terminology – Saturated, Unsaturated, Supersaturated

  4. The Solubility Process Supersaturated solution Saturated solution Solubility(gm/100 ml) Unsaturated solution Temperature(oC)

  5. The Solubility Equilibrium • Dissociation of ionic compounds into ions • Ionic equations, net ionic equations, spectator ions • Electrolytes, non-electrolytes • Solubility charts • Predict precipitates in ionic solution combinations (see periodic table given)

  6. Solubility Product Constant – Ksp The Ksp (solubility product constant) is the rate constant between a salt and its dissolved ions in a solution. The value indicated the degree to which a compound dissociated in water. The higher the Ksp, the more soluble the compound.

  7. The Solubility Product Constant A solution equilibrium can be treated as a chemical equilibrium. • The solubility equilibrium must be saturated and have excess solute (salt), therefore they occur with low solubility compounds. • The balanced chemical reaction (dissociation equation) can be used to generate an equilibrium expression. • Remember, solids have a constant concentration and are not used in the equilibrium law expression. The equilibrium constant is simplified to the product of the ion concentration raised to their proper exponents. • The equilibrium constant (Keq) can be considered a solubility product constant (Ksp) in an solubility equilibrium.

  8. The Solubility Product Constant A solution equilibrium can be treated as a chemical equilibrium. PbI2 (s) Pb 2+(aq) + 2 I 1-(aq) (saturated sol’n) PbI2 is a solid so its concentration does not alter and is therefore removed from the equilibrium expression.

  9. The Solubility Product Constant A general application of the solubility equilibria results in the following Ksp expression. BC(s)Db B+(aq) + c C-(aq) Where BC(s) is a slightly soluble salt, and B+(aq) and C-(aq) are dissociated aqueous ions. Ksp values are reported in reference tables such as those in the appendix. (Pg 802)

  10. The Solubility Product Constant usage • Calculating solubility constants from solubility data. (i.e.- solubility of 0.0003 g/100 mL @ 25oC) • Predicting whether a precipitate will form in a solubility equilibrium. • Trial ion product, Q– the reaction quotient applied to the ion concentrations. • Q > Ksp (supersaturated) precipitate will formQ = Ksp (saturated) precipitate will not formQ < Ksp (unsaturated) precipitate will not form

  11. Calculating solubility constants from solubility data Determine the solubility of AgCl(s) at SATP given that its Ksp is 1.8X 10-10. Strategy Balanced chemical equation for the system. Equilibrium Law expression. Initial, Change in & Equilibrium Concentration table. Sub-n-solve.

  12. Calculating solubility constants from solubility data Determine the solubility of AgCl(s) at SATP given that its Ksp is 1.8X 10-10. Strategy Balanced chemical equation for the system. AgCl(s)D Ag +1(aq) + Cl-1(aq)

  13. Calculating solubility constants from solubility data Determine the solubility of AgCl(s) at SATP given that its Ksp is 1.8X 10-10. Equilibrium Law expression. Ksp = [Ag+1] [Cl-1] = 1.8 x 10-10

  14. Calculating solubility constants from solubility data Determine the solubility of AgCl(s) at SATP given that its Ksp is 1.8X 10-10. Initial, Change in & Equilibrium Concentration table.

  15. Calculating solubility constants from solubility data Determine the solubility of AgCl(s) at SATP given that its Ksp is 1.8X 10-10. Sub-n-solve.

  16. Calculating solubility constants from solubility data Determine the solubility of AgCl(s) at SATP given that its Ksp is 1.8X 10-10. Check assumptions

  17. The Solubility Product Constant usage • The common ion effect • The reduction in the solubility of a salt caused by the addition of a second salt with a common ion. • Eg. The solubility of NaCl is decreased if Cl- is already present in the solution • Le Chatelier’s principle is applied and results in a shift of the solubility equilibrium. • Use equilibrium (ICE) table to solve for new concentrations.

  18. Example Calculate the molar solubility of PbSO4 in 0.100 M Na2SO4 at SATP. Ksp= 1.96 x 10-8 Split this problem into two steps. Initially, NaSO4(aq) has already dissociated in water to a concentration of 0.100 mol/L Na2SO4 (s) 2Na+(aq) + SO42- (aq) The [SO42- (aq)] = [Na2SO4 (s)] = 0.100M PbSO4 is then added to this solution. PbSO4 D Pb2+(aq) + SO42- (aq)

  19. Because there are already SO42- ions present in the solution, the PbSO4cannot completely dissociate in this solution. We cannot use the assumption that the concentration of ions will be equal to the concentration of PbSO4. Let molar solubility of PbSO4 be x. Ksp = [Pb2+] [SO42-] = 1.96´ 10-8 PbSO4(s)® Pb2+(aq) + SO42-(aq) Initial -- 0 0.100 Change -- x + x Equil. -- x 0.100 + x

  20. Ksp = [Pb2+] [SO42-] (x)(0.100 + x) = 1.96´ 10-8 x2 + 0.100 x -1.96´ 10-8 = 0 Using the quadratic equation, x = 2.00 x 10-7 M

  21. Molar solubility of PbSO4 in 0.100 M Na2SO4 = 2.0 ´ 10-7 M Molar solubility of PbSO4 in pure water = 1.40´10-4 M Note the solubility is very much reduced in 0.100 M Na2SO4because of the common ion effect.

  22. Predicting Precipitate • Reaction Quotient (Q): • Used to determine if a solubility reaction has reached saturation. It tells you how far a reaction has proceeded.

  23. Using Q • To do so, the concentration of ions in solution can be used to calculate a trial ion product and then compare it to the listed value of Ksp • Given a reaction AB(s) D A+(aq) + bB-(aq) Trial Ion Product = [A+][B-]b

  24. Determining if a precipitate will form Strategy Balanced chemical equation for the system. Identify the concentrations to be considered. Use the equilibrium law expression to generate the “Trial Ion Product (Qsp)” expression. Use given concentration and the expression to solve for Qsp.

  25. Determining if a precipitate will form Strategy • Compare the Qsp value to the known Ksp value to determine if the situation is at equilibrium. • Qsp = Ksp  the solution is saturated & at equilibrium so no precipitate will form. • Qsp < Ksp  the solution is unsaturated so no precipitate will form. • Qsp > Ksp  the solution is “supersaturated” so a precipitate will form.

  26. Comparing Q to Ksp

  27. Determining if a precipitate will form If 50 mL of 0.200 mol/L NiCl2 (aq) and 50 mL of 0.0500 mol/L Na2CO3 (aq) combined at 25oC will a precipitate form? Ksp for NiCO3 (s) is 8.2 x 10-4. NiCl2 (aq) + Na2CO3 (aq)→ NiCO3 (s) + 2 NaCl (aq) (Use the solubility rules to identify which of the products is most likely to form a solid.) NiCO3 (s)⇌ Ni 2+(aq) + 2 Cl-(aq)

  28. Determining if a precipitate will form If 50 mL of 0.200 mol/L NiCl2 (aq) and 50 mL of 0.0500 mol/L Na2CO3 (aq) combined at 25oC will a precipitate form? Ksp for NiCO3 (s) is 8.2 x 10-4. NiCl2 (aq)→ Ni 2+(aq) + 2 Cl- (aq) [Ni2+]= 0.200 mol/L(50 mL)/(100 mL) = 0.100 mol/L Na2CO3 (aq)→ 2 Na+(aq) + CO32- (aq) [CO32-]= 0.0500 mol/L(50 mL)/(100 mL) = 0.0250 mol/L

  29. Determining if a precipitate will form If 50 mL of 0.200 mol/L NiCl2 (aq) and 50 mL of 0.0500 mol/L Na2CO3 (aq) combined at 25oC will a precipitate form? Ksp for NiCO3 (s) is 8.2 x 10-4. NiCO3 (s)⇌ Ni 2+(aq) + CO32- (aq)

  30. Determining if a precipitate will form If 50 mL of 0.200 mol/L NiCl2 (aq) and 50 mL of 0.0500 mol/L Na2CO3 (aq) combined at 25oC will a precipitate form? Ksp for NiCO3 (s) is 8.2 x 10-4. Qsp > Ksp (0.0025 > 8.2 x 10-4 ) In this case the solution is either supersaturated or a precipitate will be formed.

More Related