Updates

1. Updates • Assignment 06 is due today (in class) • Midterm 2 is THIS Thurs., March 15 and will cover Chapters 16 & 17 • Huggins 10, 7-8pm • For conflicts: ELL 221, 6-7pm (must arrange at least one week in advance)

2. Acid-Base Equilibria andSolubility Equilibria Chapter 17

3. Precipitation and separation of ions • Predicting what precipitate might form from a mixture of ions (solubility rules, pg. 97) • Using quantitative means to decide whether a precipitate will form • Predicting selective precipitation of an ion from a mixture of ions

4. Will a Precipitate Form? Remember Q is the reaction quotient, which is obtained by substituting the initial concentrations into the equilibrium expression. • In a solution, • If Q = Ksp, the system is at equilibrium and the solution is saturated. • If Q < Ksp, more solid will dissolve until Q = Ksp. • If Q > Ksp, the salt will precipitate until Q = Ksp.

5. 2 Q = [Ca2+]0[OH-]0 If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a precipitate form? • Note what ions present in solution: Na+, OH-, Ca2+, Cl-. • Note that the only possible precipitate is Ca(OH)2 (solubility rules). • Is Q > Kspfor Ca(OH)2? [Ca2+]0 = 0.100 M [OH-]0 = 4.0 x 10-4M = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8 Ksp = [Ca2+][OH-]2 = 8.0 x 10-6 Q < Ksp No precipitate will form 17.6

6. What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br- and Cl- at a concentration of 0.02 M? • Use Ksp for AgBr to figure out the solubility of Ag+ when [Br-] is 0.02 M • Use Ksp for AgCl to figure out the solubility of Ag+ when [Cl-] is 0.02 M • Is it possible to choose a concentration between the two solubility values where only AgBr precipitates? This will only work if the Ksp for AgCl is greater than that for AgBr since the concentrations of the counterions are identical in the problem. 17.7

7. AgCl (s) Ag+(aq) + Cl-(aq) AgBr (s) Ag+(aq) + Br-(aq) Ksp= 7.7 x 10-13 Ksp= 1.6 x 10-10 = 8.0 x 10-9M = 3.9 x 10-11M [Ag+] = [Ag+] = 7.7 x 10-13 1.6 x 10-10 = = 0.020 0.020 Ksp Ksp [Cl-] [Br-] What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br- and Cl- at a concentration of 0.02 M? AgBr is less soluble than AgCl so a selective precipitation should be possible When [Ag+] is greater than 3.9 x 10-11M, AgBr will precipitate; when [Ag+] is greater than 8.0 x 10-9, AgCl will precipitate. 3.9 x 10-11M < [Ag+] < 8.0 x 10-9M 17.7

8. Factors that affect solubility • We have considered the solubility of ionic compounds in pure water; we noted that temperature and ionic strength has an effect, but we did not discuss these influences further • We will now examine three factors that affect the solubility of ionic compounds in water • Presence of common ions • pH of solution • Presence of complexing agents

9. The presence of a common ion suppresses the ionization of a weak acid or a weak base. CH3COONa (s) Na+(aq) + CH3COO-(aq) common ion CH3COOH (aq) H+(aq) + CH3COO-(aq) The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid). 17.2

10. HCOOH (aq) H+(aq) + HCOO-(aq) Initial (M) Change (M) Equilibrium (M) [HCOO-] pH = pKa + log [HCOOH] [0.52] pH = 3.77 + log [0.30] What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Mixture of weak acid and conjugate base! 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x Common ion effect 0.30 – x 0.30 = 4.01 0.52 + x 0.52 HCOOH pKa = 3.77 17.2

11. Common ions affect solubility • CaF2 Ca2+ + F- • The presence of either Ca2+ or F- (from another source) reduces the solubility of CaF2, shifting the solubility equilibrium of CaF2 to the left

12. AgBr (s) Ag+(aq) + Br-(aq) The presence of a common ion decreases the solubility of the salt. What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr? The Common Ion Effect and Solubility Ksp = 7.7 x 10-13 The solubility of Ag and Br ions equals (Ksp)1/2 only when AgBr is the only source of Ag and Br ions (a)! s2 = Ksp s = 8.8 x 10-7 17.8

13. AgBr (s) Ag+(aq) + Br-(aq) The presence of a common ion decreases the solubility of the salt. What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr? AgBr (s) Ag+(aq) + Br-(aq) Ksp = 7.7 x 10-13 s2 = Ksp NaBr (s) Na+ (aq) + Br-(aq) s = 8.8 x 10-7 The Common Ion Effect and Solubility [Br-] = 0.0010 M [Ag+] = s [Br-] = 0.0010 + s 0.0010 Ksp = 0.0010 x s s = 7.7 x 10-10 17.8

15. The presence of a common ion decreases the solubility. • Insoluble bases dissolve in acidic solutions • Insoluble acids dissolve in basic solutions OH-(aq) + H+(aq) H2O (l) remove add Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) pH and Solubility At pH less than 10.45 Ksp= [Mg2+][OH-]2 = 1.2 x 10-11 Lower [OH-] Ksp = (s)(2s)2 = 4s3 4s3 = 1.2 x 10-11 Increase solubility of Mg(OH)2 s = 1.4 x 10-4M At pH greater than 10.45 [OH-] = 2s = 2.8 x 10-4M pOH = 3.55 pH = 10.45 Raise [OH-] Decrease solubility of Mg(OH)2 17.9

16. Complex Ions Affect Solubility • Complex Ions • The formation of these complex ions increases the solubility of these salts.

17. 2- Co2+(aq) + 4Cl-(aq) CoCl4(aq) 2- stability of complex Kf 2+ Co(H2O)6 2- CoCl4 [CoCl4 ] Kf = [Co2+][Cl-]4 Complex Ion Equilibria and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. The formation constant or stability constant (Kf) is the equilibrium constant for the complex ion formation. 17.10

18. 17.10

19. 17.11

20. Selective Precipitation of Ions • Common cations can be divided into five groups • Insoluble chlorides • Acid-insoluble sulfides • Base-insoluble sulfides and hydroxides • Insoluble phosphates • Alkali metal ions and NH4+ remain in solution; each ion can be tested for individually using a flame test

21. Qualitative Analysis of Cations 17.11

22. Flame Test for Cations sodium lithium potassium copper 17.11

23. Ca2+ (aq) + CO32- (aq) CaCO3 (s) carbonic CO2 (g) + H2O (l) H2CO3 (aq) anhydrase H2CO3 (aq) H+ (aq) + HCO3- (aq) HCO3- (aq) H+ (aq) + CO32- (aq) Chemistry In Action: How an Eggshell is Formed

24. …explain why “as the polarity of H-X bonding increases, the acid strength increases”? …especially the section involving oxoacids

25. Binary acids (HX, H2X, H3X, H4X) • Bond strength determines acidity within the same group (column), size • Bond polarity determines acidity within the same period (row), electronegativity

26. Oxyacids Central atoms derived from same group (same oxidation state) • More electronegative central atom polarizes the OH bond more, facilitating ionization (effect is weakening the O-H bond) • More electronegative central atom better able to stablize resulting negative charge following ionization, making a happier (more stable) conjugate base

27. 16.41) Calculate the concentrations of all the species (HCN, H+, CN- and OH-) in a 0.15 M HCN solution. If [H+] is 8.6 x 10-6, then [OH-]: If [H+] = [CN-] = 8.6 x 10-6, then we have lost this amount of HCN, so: [HCN] = 0.15 – (8.6 x 10-6) = 0.15 M

28. 16.97) Henry’s law constant for CO2 at 38oC is 2.28 x 10-3 mol/L.atm. Calculate the pH of a solution of CO2 at 38oC in equilibrium with the gas at a partial pressure of 3.20 atm. Remember that Henry’s law describes the effect of pressure on the solubility of gases. The solubility of CO2 can be calculated from Henry’s law: 2.28 x 10-3 mol/L.atm x 3.20 atm = 7.30 x 10-3 mol/L. Remember that CO2 dissolves in water to form H2CO3. Therefore, the pH will depend on the extent of ionization of H2CO3, which can be found from Ka (4.2 x 10-7): 4.2 x 10-7 = x2/(7.30 x 10-3M) = 5.54 x 10-5M; pH = -log x, pH = 4.26.

29. 17.47) insert