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Solubility Equilibria (Sec 6-4)

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Solubility Equilibria (Sec 6-4) Solubility Equilibria (Sec 6-4) Ksp = solubility product AgCl(s) = Ag+(aq) + Cl-(aq) Ksp = 2+ - CaF2(s) = Ca (aq) + 2F (aq) Ksp = n+ m- in general AmBn = mA + nB n+ m m- n Ksp = [A ] [B ] We use Ksp to calculate the equilibrium solubility of a compound. Calculating the Solubility of an Ionic Compound (p.131) 2+ - PbI2 = Pb + 2I 1 n+ m- in general for AmBn = mA + nB 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 = 2+ - PbI2 = Pb + 2I 2 Ch 12: A Deeper Look at Chemical Equilibrium Up to now we've ignored two points- 1. 2. 2+ - PbI2(s) = Pb (aq) + 2I (aq) -9 + - Ksp = 7.9 x 10 (ignoring PbOH , PbI3 , etc) K'sp = 3 Activity Coefficients - concentrations are replaced by "activities" aA + bB = cC + dD [C]c [D]d K old definition [A]a [B]b γc [C]c γd [D]d K C D new definition using activity γa [A]a γb [B]b A B We can calculate the activity coefficients if we know what the ionic strength of the solution is. 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 1 2 μ C Z Ci = concentration 2 i i Zi = charge e.g. calculate the ionic strength of an aqueous soln of 0.50M NaCl and 0.75M MgCl2 4 The Extended Huckel-Debye Equation 0.51z 2 log A 1 / 305 A = activity coefficient Z = ion charge = ionic strength (M) = hydrated radius (pm) works well for 0.10M 5 6 Example (p. 262) - Find the activity coefficient in a solution of 3.3 mM Mg(NO3)2 Data from Table 12-1: 7 Example (p. 264) – A Better Estimate of the Solubility of PbI2 2+ - PbI2 = Pb + 2I The Real Definition of pH pH logΑH logγH [H ] + What is the concentration of H in (a) pure H2O and (b) 0.10M NaCl? 8 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 Charge balance equation, p. 266 + - + - Solution containing H , OH , K , H2PO4 , 2- 3- HPO4 , and PO4 9 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. 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 HAc HAc HAc HAc HAc 10 Mass Balance e.g. solution of 0.025 M H3PO4 H3PO4 H3PO4 H3PO4 H3PO4 H3PO4 H3PO4 H3PO4 H3PO4 H3PO4 H3PO4 Mass balance for a sparingly soluble salt is different: e.g. CaF2 11 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 Coupled Equilibria: Solubility of CaF2 1. Relevant equations 2+ - CaF2(s) Ca + 2 F - - F + H2O HF + OH + - H2O H + OH 2. Charge Balance 12 Coupled Equilibria: Solubility of CaF2 3. Mass Balance 4. Equilibrium Expressions Coupled Equilibria: Solubility of CaF2 5.Number of equations = number of unknowns [H+], [OH-], [Ca2+], [F-], [HF] = unknowns 6.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 13 After buffering to pH = 3.0, [H+] = 1.0 x 10-3 M - + -11 [OH ] = Kw/[H ] = 1.0 x 10 M and now subst into Kb [HF] K 1.5 x 1011 b 1.5 [F ] [OH ] 1.0 x 1011 [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+] - 2+ 2+ - [F ] = 0.80[Ca ] and finally subst 0.80[Ca ] for [F ] into Ksp 2+ - 2 [Ca ][F ] = Ksp 2+ 2+ 2 [Ca ](0.80[Ca ]) = Ksp 2+ 2 1/3 -4 [Ca ] = (Ksp/0.80 ) = 3.9 x 10 M 14.
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