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Chapter 15

Chapter 15. Applications of Aqueous Equilibria. The Common Ion Effect. A common ion is an ion that is produced by multiple species in solution (other than H) It comes from having a salt dissolved in the same solution as an acid.

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Chapter 15

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  1. Chapter 15 Applications of Aqueous Equilibria

  2. The Common Ion Effect • A common ion is an ion that is produced by multiple species in solution (other than H) • It comes from having a salt dissolved in the same solution as an acid. • If the acid is a weak acid, additional ions of the conjugate base may affect the pH of the solution.

  3. Consider the Following… HF   H+ + F- and NaF  Na+ + F- What happens to the equilibrium expression when the NaF dissociates (because it is ionic and that’s what ionic compounds do?) Equilibrium Shifts To Left

  4. Check This Out Calculate [H+] and percent dissociation of HF in a solution of 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF. Compare those values to a 1.0 M HF solution with no NaF…

  5. BUFFERS!! • Solutions that resist pH change when small amounts of acid or base are added. HA + H2O H3O+ + A- Based on the Common Ion Effect Weak Acid + Its Salt Two Types Weak Base + Its Salt

  6. Buffers Try Me • A buffered solution contains 1.00 M acetic acid (Ka= 1.80 x 10-5) and 1.00 M sodium acetate. Calculate the pH of the solution.

  7. Buffy Buff Buffers • Calculate the change in pH that occurs when 0.03 mole of solid NaOH is added to 1.0 Liter of the buffered solution from the problem we just finished.

  8. Henderson Hasselbalch • Useful for calculating the pH of solutions when the ratios [HA]/[A-] is known. pH = pKa + log ([A-]/[HA])

  9. Note: When a solution contains equal concentrations of an acid and its conjugate base, [H+] = Ka the pH of the buffer is equal to the log of the Ka

  10. What if I graphed it all? buffered unbuffered

  11. Try This Buffer Problem Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 x 10-4) and 0.25 M sodium lactate. Lactic acid (HC3H5O3) is a common constituent of biologic systems. For example, it is found in milk and is present in human muscle tissue during exertion. • Use Henderson Hasselbach OR B) Use an I.C.E. Chart

  12. Solubility Equilibria • Ion dissociation is an equilibrium process • You can write an equilibrium expression for it. • Ksp = solubility product constant • Solubility product ≠ solubility

  13. Calculating Ksp • Copper (I) bromide has a measured solubility of 2.0 x 10-4 mol/L at 25oC. Calculate its Ksp value.

  14. Calculating Solubility • The Ksp value for copper (II) iodate is 1.4 x 10-7 at 25oC. Calculate its solubility at this temperature.

  15. Comparing Solubilities • # of ions produced matters: • If the number is the same AgCl AgOH AgI • If the number is different CuS Ag2S Bi2S3 chart

  16. pH and Solubility • Example of Mg(OH)2 solubility • Common ion effect causes milk of magnesia to dissolve in stomach fluids but it is relatively insoluble in non-acidic solutions because of the presence of other OH- ions. If X- is an effective base, (HX is a weak acid) The salt MX will be more soluble in acid. Precipitation occurs as a result of decreased solubility due to the Common Ion Effect

  17. Precipitation Reactions • To calculate the likelihood of the formation of precipitate formation upon mixing two solutions, we use Only this time we call it the ion product. Q Q > Ksp Precipitation! Q < Ksp No Precipitation!

  18. Try It Out • A solution is prepared by adding 750.0 mL of 4.00 x 10-3 M Ce(NO3)3 to 300.0 mL of 2.00 x 10-2 M KIO3. Will Ce(IO3)3 precipitate from this solution (Ksp = 1.9 x 10-10)

  19. Precipitation Try ME • A solution is prepared by mixing 150.0 mL of 1.00 x 10-2 M Mg(NO3)2 and 250.0 mL of 1.00 x 10-1 M NaF. Calculate the concentrations of Mg2+ and F- at equilibrium with solid MgF2 (Ksp = 6.4 x 10-9)

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