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CHEMISTRY 59-320 ANALYTICAL CHEMISTRY Fall - 2010

CHEMISTRY 59-320 ANALYTICAL CHEMISTRY Fall - 2010. Chapter 8: Activity and the systematic treatment of equilibrium. Chemical Equilibrium Electrolyte Effects.

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CHEMISTRY 59-320 ANALYTICAL CHEMISTRY Fall - 2010

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  1. CHEMISTRY 59-320ANALYTICAL CHEMISTRYFall - 2010 Chapter 8: Activity and the systematic treatment of equilibrium

  2. Chemical Equilibrium Electrolyte Effects • The right hand side figure shows that the equilibrium constant of (8-1) decreases as electrolyte that does not participate the reaction is added (Why?). • Electrolyte: Substances producing ions in solutions. • Can electrolytes affect chemical equilibria? • (A) “Common Ion Effect”  Yes • Decreases solubility of BaF2 with NaF • F- is the “common ion” • (B) No common ion: “inert electrolyte effect” • or “diverse ion effect”: Add Na2SO4 to saturated solution of AgCl Increases solubility of AgCl

  3. 8-1 The effect of ionic strength on solubility of salts • Why does the solubility increase when salts are added to the solution? • The formation of ionic atmosphere. The greater the ionic strength of a solution, the higher the charge in the ionic atmosphere. • The ionic atmosphere attenuates the attraction between ions since each ion-plus-atmosphere contains less net charge. • Increasing ionic strength therefore reduces the attraction between any particular Ag+ and any Cl-, relative to the case in distilled water.

  4. What is ionic strength? • Ionic strength, μ, is a measure of the total concentration of ions in solution. • m = ½SCiZi2 • where Ci is the concentration of the ion, actually a ratio of Ci/1 M, Zi = charge on each individual ion. Example: Find the ionic strength of (a) 0.10 M NaCl; (b) 0.020 M KBr plus 0.01 M Na2SO4. Solution: (a) the soultion contains 0.10 M Na+ and 0.10 M Cl- m = ½[(0.10M/1M)*(+1)2 + (0.10M/1M)*(-1)2] = ½ [0.10 + 0.10] = 0.10 (b) the solution contains 0.020 M K+, 0.020 M Br-, 0.02 M Na+ and 0.01 M SO42-. • m = ½[(0.020M/1M)*(+1)2 + (0.020M/1M)*(-1)2 + (0.020M/1M)*(+1)2 + (0.010M/1M)*(-2)2 = ½ [0.020 + 0.020 + 0.020 + 0.040] = 0.050. Ions with a larger charge number have greater contribution on the ionic strength.

  5. 8-2 Activity Coefficients • To account for the effect of ionic strength, concentrations in the calculation of equilibrium constant show be replaced by activities: ẲC = [C]*γC • The activity of species j is its concentration multiplied by its activity coefficient, ai = Ciƒi ƒi= activity coefficient. • At low ionic strength, activity coefficients approach unity.

  6. Activity and Activity Coefficients • Calculation of Activity Coefficients • Extended Debye-Huckel Equation: • ai = ion size parameter in angstrom (Å) 1 Å = 100 picometers (pm, 10-10 meters) • Limitations: singly charged ions = 3 Å • log ƒi = - 0.51Zi2(m)½/(1+(m)½) • On page 144: “the equation works fairly well for μ≤0.1 M “ (??)

  7. Effect of ionic strength, ion charge, and ion size on the activity coefficient • The ion size αin eq 8-6 is an empirical parameter that provides good agreement between measured activity coefficients. • α is the diameter of the hydrated ion. • As ionic strength increases, the activity coefficient decreases. • The activity coefficient approaches unity as the ionic strength approaches 0. • As the magnitude of the charge of the ion increases, the departure of its activity coefficient from unity increases. • The smaller the ion size (α), the more important activity effects become.

  8. Activity coefficients for differently charged ions with a constant ionic size

  9. Activity coefficient under high ionic strength

  10. Activity coefficient for non-ionic compounds • Case 1: neutral molecules in solution phase, the activity coefficient is unity and thus the activity is numerically the same as their concentration. • Case 2: Gases. The fugacity (i.e. activity) is calculated as the product of pressure and the fugacity coefficient (i.e. activity coefficient). When the pressure is below 1 bar, the fugacity coefficient is close to unity.

  11. Diverse Ion (Inert Electrolyte) Effect: • Kspo = aAg+ . aCl- = 1.75 x 10-10 • Adding Na2SO4 to saturated solution of AgCl, at high concentration of diverse (inert) electrolyte: higher ionic strength, m • aAg+< [Ag+] ; aCl-< [Cl-] • aAg+ . aCl- < [Ag+] [Cl-] • Kspo < [Ag+] [Cl-] ;  Kspo < [Ag+] = solubility • Solubility = [Ag+] >  Kspo

  12. Equilibrium calculations using activities • Solubility of PbI2 in 0.1M KNO3 • m = 0.1 = {0.1(1+)2 + 0.1(1-)2}/2 (ignore Pb2+,I-) • ƒPb = 0.35 ƒI = 0.76 • Kspo= (aPb)1(aI)2 = ([Pb2+]Pb )1([I-]I )2 • Kspo = ([Pb2+] [I-]2)(Pb I2) = Ksp (Pb I2) • Ksp = Kspo / (Pb I ) • Ksp = 7.1 x 10-9 /((0.35)(0.76)2) = 3.5 x 10-8 • (s)(2s)2 = Ksp s = (Ksp/4)1/3 s =2.1 x 10-3 M • Note: If s = (Kspo/4)1/3 then s =1.2 x 10-3M • The solubility is increased by approx. 43%

  13. 8-3 pH revisited Addition example: Calculate the pH of water containing 0.010 M KCl at 25 oC. (see in class discussion).

  14. 8-4 Systematic treatment of equilibrium • Step 1: Write the pertinent reactions. • Step 2: Write the charge balance equation. • Step 3: Write mass balance equations. • Step 4: Write equilibrium constant expression for each chemical reaction. • Step 5: Count the equations and unknowns. The numbers should be the same. • Step 6: Solve those equations.

  15. 8-5 Applying the systematic treatment of equilibrium Solubility of calcium sulfate Step 1 Step 2 Step 3 Step 4 Step 5: counting Step 6

  16. Solubility of magnesium hydroxide Step 6

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