Lecture # 9. Chapter 8. Activity and Systematic Treatment of Equilibrium. K = [Fe(SCN) 2+ ] [Fe 3+ ] [SCN - ] Fe 3+ + SCN - Fe(SCN) 2+

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1 Lecture # 9 Chapter 8 Activity and Systematic Treatment of Equilibrium K = [Fe(SCN) ] [Fe 3 ] [SCN ] Fe 3 SCN Fe(SCN) CaCO 3 (s) Ca (aq) CO K sp =.5 x 10 9 CaCO 3 (s) CO H O Ca HCO 3 CaCO 3 (s) Ca (aq) CO CO (g) CO (aq) CO (aq) H O H CO H CO H HCO 3 HCO H CO 3 H O H OH 1

2 a way to deal with all types of chemical equilibria, regardless of their complexity Step 1: Write all pertinent equations. Step : Write the charge balance equation. Step 3: Write the mass balance equation(s). Step : Write the equilibrium constant expression for each reaction. Step 5: Count the equations and unknowns. They should be equal. Step 6: Solve for all unknowns. Dissociation of Ions in Water Dissociation of Ions in Water CaCO 3 solubility Effects of Ions on Solubility CaCO 3 (s) Ca (aq) CO Common Ion Effect CaCl Concentration of Added Substance (M) CaCO 3 (s) Common Ion Effect Ca (aq) CO 3 (aq) Common Ion Effect CaCO 3 (s) Ca (aq) CO

3 CaCO 3 solubility Effects of Ions on Solubility CaCO 3 (s) Ca (aq) CO KNO 3 CaCl Concentration of Added Substance (M) Dissociation of Ions in Water Ionic Atmosphere Ionic Atmosphere Ionic Strength measure of the total concentration of ions in solution = 1 (c 1 z 1 c z c 3 z 3.) = 1 (c i z i ) c = concentration (M) z = charge (e.g. 1, 1,,, etc.) Activity and Equilibrium A X = [X] c = Activity Where c = Activity Coefficient Activity Coefficients Extended DebyeHückel Equation: log = 0.51z (at 5 C) 1 ( /305) aa bb cc dd K = [C] c [D] d [A] a [B] b Where ± z = ion charge, = ionic strength and = ion diameter (pm) K = A c C A d D A a A A b B = [C] c c c [D] d d d [A] a a a [B] b b b Note: 1 for neutral compounds and gases Works well for <

4 log γ = 0.51z μ 1 (α μ /305) Activity Coefficient (γ) Activity Coefficients 1 M M M M Ionic Strength (μ) 1. As ionic strength (μ) increases, the activity coefficient (γ) decreases.. As magnitude of charge (z) increases, increased departure from unity (1). 3. The smaller the ion size, the more important activity. Activity Coefficients Interpolation of Activity Coefficients Unknown yinterval Δy = Known xinterval Δx? y?? x Example: What is the γ for (CH3H7)N at μ = 0.015?

5 Interpolation of Activity Coefficients Unknown yinterval = Known xinterval y x y?? Example: What is the for (CH 3 H 7 ) N at = 0.015? (0.91 ) = ( ) ( ) ( ) = (0.035/0.0) x (0.06) 0.91 = 0.858? a way to deal with all types of chemical equilibria, regardless of their complexity Step 1: Write all pertinent equations. Step : Write the charge balance equation. Step 3: Write the mass balance equation(s). Step : Write the equilibrium constant expression for each reaction. Step 5: Count the equations and unknowns. They should be equal. Step 6: Solve for all unknowns. Charge Balance The sum of the positive charges in solution equal the sum of the negative charges in solution (c i z i ) = 0 n 1 [C 1 ] n [C ] = m 1 [A 1 ] m [A ] Example: CaCO 3 (s) CO H O Ca HCO 3 CaCO 3 (s) CO (g) CO (aq) H O Ca (aq) CO 3 (aq) CO (aq) H CO H CO H HCO 3 HCO 3 (aq) H CO 3 H O H OH [H ] [Ca ] = [CO 3 ] [HCO 3 ] [OH ] Mass Balance the quantity of all species in a solution containing a particular atom (or group of atoms) must equal the amount of that atom (or group) delivered to the solution Example: CaCO 3 (s) CO H O Ca HCO 3 CaCO 3 (s) CO (g) CO (aq) H O Ca (aq) CO 3 (aq) CO (aq) H CO H CO H HCO 3 HCO 3 (aq) H CO 3 H O H OH [Ca ] = [CO 3 ] [Ca ] = [CaCO 3 ] [Ca ] [CO 3 ] = [CO 3 ] [HCO 3 ] [H CO 3 ] [CaCO 3 ] [Ca ] = [CO 3 ] [HCO 3 ] [H CO 3 ] a way to deal with all types of chemical equilibria, regardless of their complexity Step 1: Write all pertinent equations. Step : Write the charge balance equation. Step 3: Write the mass balance equation(s). Step : Write the equilibrium constant expression for each reaction. Step 5: Count the equations and unknowns. They should be equal. Step 6: Solve for all unknowns. Equilibrium Constants K sp = [Ca ] Ca [CO 3 ] CO3 = 6 x 10 9 [CO (aq)] CO = K CO P(CO ), K CO = 3. x 10 K a1 = [H ] H [HCO 3 ] HCO3 =.5 x 10 7 [H CO 3 ] HCO3 K a = [H ] H [CO 3 ] CO3 =.69 x [HCO 3 ] HCO3 K w = [H ] H [OH ] OH = 1 x

6 Equations (TOTAL) K sp = [Ca ] Ca [CO 3 ] CO3 = 6 x 10 9 [CO (aq)] CO = K CO P(CO ), K CO = 3. x 10 K a1 = [H ] H [HCO 3 ] HCO3 =.5 x 10 7 [H CO 3 ] HCO3 K a = [H ] H [CO 3 ] CO3 =.69 x [HCO 3 ] HCO3 K w = [H ] H [OH ] OH = 1 x 10 1 [H ] [Ca ] = [CO 3 ] [HCO 3 ] [OH ] [CaCO 3 ] [Ca ] = { [CO 3 ] [HCO 3 ] [H CO 3 ] } Unknowns: [Ca ], [CO 3 ], [H CO ], [HCO 3 ], [CO (aq)], [H ], [OH ] 7 7 a way to deal with all types of chemical equilibria, regardless of their complexity Step 1: Write all pertinent equations. Step : Write the charge balance equation. Step 3: Write the mass balance equation(s). Step : Write the equilibrium constant expression for each reaction. Step 5: Count the equations and unknowns. They should be equal.??? Step 6: Solve for all unknowns. Example: Ca (s) Ca Ca (s) Ca (aq) Ca H O CaOH H H O H OH H O H OH Example: Ca (s) Ca Ca (s) Ca (aq) Ca H O CaOH H H O H OH H O H OH Unknowns: [Ca ], [ ], [Ca (aq)], [CaOH ], [H ]. [H ], [OH ] = 7 unknowns Charge Balance: [Ca ] [CaOH ] [H ] = [ ] [H ] [OH ] Mass Balance: [Total Calcium] = [Total Sulfate] [Ca ] [Ca (aq)] [CaOH ] = [ ] [H ] [Ca (aq)] Example: Ca (s) Ca Ca (s) Ca (aq) Ca H O CaOH H H O H OH H O H OH K sp = [Ca ] Ca [ ] =. x 10 5 K ion pair = [Ca (aq)] = 5.0 x 10 3 K a = [H ] H [CaOH] CaOH =.0 x [Ca ] Ca K b = [H ] H [OH ] OH = 9.8 x [ ] K W = [H ] H [OH ] OH = 1.0 x 10 1 = 7 equilibrium expression [Ca ] [CaOH ] [H ] = [ ] [H ] [OH ] [Ca ] [Ca (aq)] [CaOH ] = [ ] [H ] [Ca (aq)] Example: Ca (s) Ca Ca (s) Ca (aq) Ca H O CaOH H H O H OH H O H OH K sp = [Ca ][ ] =. x 10 5 K ion pair = [Ca (aq)] = 5.0 x 10 3 K a = [H ][CaOH] =.0 x [Ca ] K b = [H ][OH ] = 9.8 x [ ] K W = [H ][OH ] = 1.0 x 10 1 [Ca ] [CaOH ] [H ] = [ ] [H ] [OH ] [Ca ] [Ca (aq)] [CaOH ] = [ ] [H ] [Ca (aq)] 6

7 Example: Ca (s) Ca Ca (s) Ca (aq) Ca H O CaOH H H O H OH H O H OH K sp = [Ca ][ ] =. x 10 5 K ion pair = [Ca (aq)] = 5.0 x 10 3 [Ca (aq)] = 5.0 x 10 3 Example: Ca (s) Ca Ca (s) Ca (aq) Ca H O CaOH H H O H OH H O H OH K sp = [Ca ][ ] =. x 10 5 [Ca ] 1 = [ ] 1 =. x 10 5 =.9 x 10 3 [Ca ] [CaOH ] [H ] = [ ] [H ] [OH ] [Ca ] [Ca (aq)] [CaOH ] = [ ] [H ] [Ca (aq)] = 1/ ([Ca ] 1 () [ ] 1 () ) = 1/ (8 x.9 x 10 3 ) = (.9 x 10 3 ) = 0.00 M Example: Ca (s) Ca Ca (s) Ca (aq) Ca H O CaOH H H O H OH H O H OH K sp = [Ca ][ ] =. x 10 5 Ca = 0.68 = K sp = [Ca ] Ca [ ] =. x 10 5 [Ca ] (0.68) [Ca ] (0.606) =. x 105 [Ca ] = 7.9 x 10 3 K sp = [Ca ] Ca [ ] =. x 10 5 Solving by Iteration: Iteration Ca [Ca ] (M) (M) Are CaOH and H negligible? K a = [H ][CaOH] =.0 x [Ca ] K b = [H ][OH ] = 9.8 x [ ] K W = [H ][OH ] = 1.0 x 10 1 [CaOH ] = (.0 x )(0.009) (1.0 x 10 7 ) = x 10 8 [H ] = (9.8 x )(0.009) (1.0 x 10 7 ) = 9 x

8 [Ca ] = M [ ] = M [Ca (aq)] = 5.0 x 10 3 M [CaOH ] = x 10 8 m [H ] = 9 x 10 8 M [H ] = 1 x 10 7 M [OH ] = 1 x 10 7 M 8