1. Solubility Equilibria (Sec 6-4) K sp = solubility product AgCl(s) = Ag + (aq) + Cl - (aq) K sp = CaF 2 (s) = Ca 2+ (aq) + 2F - (aq) K sp = in general A m B n = mA n+ + nB m- K sp = [A n+ ] m [B m- ] n We use K sp to calculate the equilibrium solubility of a compound.

2. Calculating the Solubility of an Ionic Compound (p.136) e.g. PbI 2

3. in general for A m B n = mA n+ + nB m-

4. The Common Ion Effect (p. 137) What happens to the solubility of PbI 2 if we add a second source of I - (e.g. the PbI 2 is being dissolved in a solution of 0.030 M NaI)? The common ion =

5. Ch 12: A Deeper Look at Chemical Equilibrium Up to now we've ignored two points- 1. 2.

6. PbI 2 (s) = Pb 2+ (aq) + 2I - (aq) K sp = 7.9 x 10 -9 (ignoring PbOH +, PbI 3 -, etc) K' sp =

7. Activity Coefficients - concentrations are replaced by "activities" aA + bB = cC + dD We can calculate the activity coefficients if we know what the ionic strength of the solution is.

8. Charge Effects - an ion with a +2 charge affects activity of a given electrolyte more than an ion with a +1 charge  = ionic strength, a measure of the magnitude of the electrostatic environment C i = concentration Z i = charge e.g. calculate the ionic strength of an aqueous soln of 0.50M NaCl and 0.75M MgCl 2

9. The Extended Huckel-Debye Equation  A = activity coefficient Z = ion charge  = ionic strength (M)  = hydrated radius (pm) works well for   0.10M

14. Example (p. 247) - Find the activity coefficient in a solution of 3.3 mM Mg(NO 3 ) 2 Easy solution using Table 12-1:

15. Harder solution using Huckel-Debye:

16. Example - What is the concentration of Ba 2+ in a solution saturated with BaSO 4 in (a) pure H 2 O, (b) 0.050 M KCl

17. The Real Definition of pH What is the concentration of H + in (a) pure H2O and (b) 0.10M NaCl?

18. Systematic Treatment of Equilibria (Sec 12-3 and 12-4) A procedure for solving any equilibrium problem no matter how complicated. Charge Balance - the sum of the positive charges in solution must equal the sum of negative charges. e.g. sulfate ion C SO42- = 0.0167 M

19. General charge balance equation - n 1 [C 1 ] + n 2 [C 2 ] +…. = m 1 [A 1 ] + m 2 [A 2 ] + … whereC = cation concentration n = cation charge A = anion concentration m = anion charge e.g. write the charge balance equation for a soln of Na 2 SO 4 and NaCl in water.

20. Mass Balance The sum of all substances in solution containing a particular atom (or group of atoms) must equal the quantity added to solution. e.g. solution of 0.050 M Hac HAc HAc HAc HAc HAc

21. e.g. solution of 0.025 M H 3 PO 4 H 3 PO 4

22. Mass balance for a sparingly soluble salt follows this approach: e.g. PbCl 2 initially:[I-] = 2 [Pb 2+ ]

23. General Procedure 1. Write down all the relevant chemical equations 2. Write the charge blance 3. Write the mass balance 4. Write down the equilibrium constant expressions (only step where activities may be used) 5. Make sure that the number of unknowns equals the number of equations 6. Solve the system of equations -make approximations -use a computer

24. Coupled Equilibria: Solubility of CaF 2 1.Relevant equations CaF 2 (s) Ca 2+ + 2 F - F - + H 2 O HF + OH - H 2 O H + + OH - 2.Charge Balance

25. 3.Mass Balance 4.Equilibrium Expressions

26. 5.Number of equations = number of unknowns [H + ], [OH - ], [Ca 2+ ], [F - ], [HF] = unknowns 6.Simplifying Assumptions and Solution fix the pH using a buffer { [H + ] = C H+ }, removes one unknown adding a buffer and associated ions nullifies the charge balance equation so now we have 4 equations and 4 unknowns

27. After buffering to pH = 3.0, [H + ] = 1.0 x 10 -3 M [OH - ] = K w /[H + ] = 1.0 x 10 -11 M and now subst into K b [HF] = 1.5[F - ] and now subst 1.5[F-] for [HF] in the mass balance equation [F - ] + [HF] = 2[Ca 2+ ] [F - ] + 1.5[F - ] = 2[Ca 2+ ] [F - ] = 0.80[Ca 2+ ] and finally subst 0.80[Ca 2+ ] for [F - ] into K sp [Ca 2+ ][F - ] 2 = K sp [Ca 2+ ](0.80[Ca 2+ ]) 2 = K sp [Ca 2+ ] = (K sp /0.80 2 ) 1/3 = 3.9 x 10 -4 M