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Solubility Equilibria (Sec 6-4)

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.

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Solubility Equilibria (Sec 6-4)

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  1. Solubility Equilibria (Sec 6-4) Ksp = solubility product AgCl(s) = Ag+(aq) + Cl-(aq) Ksp = CaF2(s) = Ca2+(aq) + 2F-(aq) Ksp = in general AmBn = mAn+ + nBm- Ksp = [An+]m[Bm-]n We use Ksp to calculate the equilibrium solubility of a compound.

  2. Calculating the Solubility of an Ionic Compound (p.131) PbI2 = Pb2+ + 2I-

  3. in general for AmBn = mAn+ + nBm-

  4. The Common Ion Effect (p. 132) What happens to the solubility of PbI2 if we add a second source of I- (e.g. the PbI2 is being dissolved in a solution of 0.030 M NaI)? The common ion = PbI2 = Pb2+ + 2I-

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

  6. PbI2(s) = Pb2+(aq) + 2I-(aq) Ksp = 7.9 x 10-9 (ignoring PbOH+, PbI3-, 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 Ci = concentration Zi = charge e.g. calculate the ionic strength of an aqueous soln of 0.50M NaCl and 0.75M MgCl2

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

  10. Example (p. 262) - Find the activity coefficient in a solution of 3.3 mM Mg(NO3)2 Data from Table 12-1:

  11. Example (p. 264) –A Better Estimate of the Solubility of PbI2 PbI2 = Pb2+ + 2I-

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

  13. 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 CSO42- = 0.0167 M

  14. Charge balance equation, p. 266 Solution containing H+, OH-, K+, H2PO4-, HPO42-, and PO43-

  15. General charge balance equation - n1[C1] + n2[C2] +…. = m1[A1] + m2[A2] + … where C = cation concentration n = cation charge A = anion concentration m = anion charge e.g. write the charge balance equation for a soln of Na2SO4 and NaCl in water.

  16. 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 HAcHAcHAcHAcHAcHAcHAcHAcHAcHAc

  17. Mass Balance e.g. solution of 0.025 M H3PO4 H3PO4 H3PO4 H3PO4 H3PO4 H3PO4 H3PO4 H3PO4 H3PO4 H3PO4 H3PO4

  18. Mass balance for a sparingly soluble salt is different: e.g. CaF2

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

  20. Coupled Equilibria: Solubility of CaF2 • Relevant equations • CaF2(s) Ca2+ + 2 F- • F- + H2O HF + OH- • H2O H+ + OH- • Charge Balance

  21. Coupled Equilibria: Solubility of CaF2 • Mass Balance • Equilibrium Expressions

  22. Coupled Equilibria: Solubility of CaF2 • Number of equations = number of unknowns • [H+], [OH-], [Ca2+], [F-], [HF] = unknowns • Simplifying Assumptions and Solution • fix the pH using a buffer { [H+] = CH+ } , removes one unknown • adding a buffer and associated ions nullifies the charge balance equation • so now we have 4 equations and 4 unknowns

  23. After buffering to pH = 3.0, [H+] = 1.0 x 10-3 M [OH-] = Kw/[H+] = 1.0 x 10-11 M and now subst into Kb [HF] = 1.5[F-] and now subst 1.5[F-] for [HF] in the mass balance equation [F-] + [HF] = 2[Ca2+] [F-] + 1.5[F-] = 2[Ca2+] [F-] = 0.80[Ca2+] and finally subst 0.80[Ca2+] for [F-] into Ksp [Ca2+][F-]2 = Ksp [Ca2+](0.80[Ca2+])2 = Ksp [Ca2+] = (Ksp/0.802)1/3 = 3.9 x 10-4 M

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