Presentation on theme: "CHEMISTRY 59-320 ANALYTICAL CHEMISTRY Fall - 2010 Chapter 8: Activity and the systematic treatment of equilibrium."— Presentation transcript:
CHEMISTRY 59-320 ANALYTICAL CHEMISTRY Fall - 2010 Chapter 8: Activity and the systematic treatment of equilibrium
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 BaF 2 with NaF F - is the “common ion” (B) No common ion: “inert electrolyte effect” or “diverse ion effect”: Add Na 2 SO 4 to saturated solution of AgCl Increases solubility of AgCl
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.
What is ionic strength? Ionic strength, μ, is a measure of the total concentration of ions in solution. = ½ C i Z i 2 where C i is the concentration of the ion, actually a ratio of C i /1 M, Z i = 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 Na 2 SO 4. Solution: (a) the soultion contains 0.10 M Na + and 0.10 M Cl - = 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 SO 4 2-. = 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.
8-2 Activity Coefficients To account for the effect of ionic strength, concentrations in the calculation of equilibrium constant show be replaced by activities: Ẳ C = [C]*γ C The activity of species j is its concentration multiplied by its activity coefficient, a i = C i ƒ i ƒ i = activity coefficient. At low ionic strength, activity coefficients approach unity.
Activity and Activity Coefficients Calculation of Activity Coefficients Extended Debye-Huckel Equation: i = ion size parameter in angstrom ( Å ) 1 Å = 100 picometers (pm, 10 -10 meters) Limitations: singly charged ions = 3 Å log ƒ i = - 0.51Z i 2 ½ ½ On page 144: “the equation works fairly well for μ≤0.1 M “ (??)
Effect of ionic strength, ion charge, and ion size on the activity coefficient The ion size α in 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.
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.
Diverse Ion (Inert Electrolyte) Effect: K sp o = a Ag +. a Cl - = 1.75 x 10 -10 Adding Na 2 SO 4 to saturated solution of AgCl, at high concentration of diverse (inert) electrolyte: higher ionic strength, a Ag + < [Ag + ] ; a Cl - < [Cl - ] a Ag +. a Cl - < [Ag + ] [Cl - ] K sp o < [Ag + ] [Cl - ] ; K sp o < [Ag + ] = solubility Solubility = [Ag + ] > K sp o
Equilibrium calculations using activities Solubility of PbI 2 in 0.1M KNO 3 2 (ignore Pb 2+,I - ) ƒ Pb = 0.35 ƒ I = 0.76 K sp o = (a Pb ) 1 (a I ) 2 = ([Pb 2+ ] Pb ) 1 ([I - ] I ) 2 K sp o = ([Pb 2+ ] [I - ] 2 )( Pb I 2 ) = K sp ( Pb I 2 ) K sp = K sp o / ( Pb I ) K sp = 7.1 x 10 -9 /((0.35)(0.76) 2 ) = 3.5 x 10 -8 (s)(2s) 2 = K sp s = (Ksp/4) 1/3 s =2.1 x 10 -3 M Note: If s = (K sp o /4) 1/3 thens =1.2 x 10 -3 M The solubility is increased by approx. 43%
8-3 pH revisited Addition example: Calculate the pH of water containing 0.010 M KCl at 25 o C. (see in class discussion).
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.
8-5 Applying the systematic treatment of equilibrium Solubility of calcium sulfate Step 1 Step 2 Step 3 Step 4 Step 6 Step 5: counting