Chapter 15

1. Chapter 15 Applications of Aqueous Equilibria

2. The Common Ion Effect • When the salt with the same anion as weak acid is added to that acid, • It reverses the dissociation of the acid. • Lowers the percent dissociation of the acid. • The same principle applies to salts with the cation of a weak base. • Common Ion Effect: Shift in equilibrium position that occurs when ion is added that is already part of equilibrium reaction. • Example: NaF is added to an HF solution. • The calculations are the same as last chapter.

3. The Common Ion Effect HF(aq) ↔ H+(aq) + F-(aq) HF(aq)H+(aq) + F-(aq) NaF(s) Na+(aq) + F-(aq) Expected shift single reaction Equilibrium shift from added common ion

4. Acidic Solutions Containing Common Ions Calculate [H+] and the percent dissociation of HF in a solution containing 1.0M HF (Ka=7.2x10-4) and 1.0M NaF • Write the major species • Consider chemical properties of each • Write equilibrium reaction for [H+] • Write equilibrium expression for [H+] • Create ICE Table • Determine Equilibrium concentration in terms of x • Notice initial [F-]o comes from NaF • Since x is small, it has little change on [F-] & [HF] and solve for x in terms of [H+] • Calc. percent dissociation = [H+]/[HF] x 100 Do #23 p. 740

5. pH of Buffered Solution • A solution that resists a change in pH when either an H+ or OH- is added. • Buffers are: • weak acid + salt Ex. HF & NaF • weak base + salt Ex. NH3 & NH4Cl • Salt has “Common Ion Effect” • Make a buffer by varying the concentrations • Use same steps to solve as before. Do #33 p. 741

6. pH Changes in Buffered Solutions • Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0L of buffered solution of 0.50 M acetic acid (ka=1.8x10-5) and 0.50M sodium acetate. • Calculate pH of original solution • Write the major species • Consider chemical properties of each • Write equilibrium reaction for [H+] • Write equilibrium expression for [H+] • Create ICE Table • Determine Equilibrium concentration in terms of x • Calculate pH

7. pH Changes in Buffered Solutions • A strong base will grab protons from the weak acid reducing [HA]0 • OH- is very strong base • Write the major species • Consider chemical properties of each • Write equilibrium reaction OH- + HC2H3O2 H2O + C2H3O2- • Assume the reaction goes to completion until all OH- are consumed • Do Stoichiometry problem first • Then do equilibrium calculation.

8. pH Changes in Buffered Solutions • Find the difference between the pH of the buffered solution and the buffered solution plus the NaOH. • Compare this with what happens when 0.01 mol solid NaOH is added to 1.0L water to give 0.01M NaOH solution. Water starts at pH 7.00 • 12.00 (new sol.) – 7.00 (water) = 5.00 change in pH #35 p. 741

9. Derivation of Henderson-Hasselbalch Equation The stability of pH can be understood by examining equilibrium expression of HA p.687 Or rearranged The [H+] depends on the ratio [HA]/[A-] Another useful form of this equation is arrived at by taking the negative log of both sides -log [H+] = -log Ka –log [HA]/[A-] Substuting pH & pKa and inverting the final log term you get Henderson-Hasselbalch equation

10. Henderson-Hasselbalch equation Also in this form • Use Henderson-Hasselbalch to calculate the pH of the following 2 buffered solutions: • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate Ka=1.4x10-4 • 0.25 M NH3 and 0.40 M NH4Cl Kb = 1.8 x 10-5

11. Adding Strong Acid to a Buffered Solution • A buffered solution of 0.25 M NH3 and 0.40 M NH4Cl Kb = 1.8 x 10-5 has a pH=9.05 • Calculate the pH of the solution that results when 0.10 mol of gaseous HCl is added to a 1.0L of the buffered solution listed above • Do the stoichiometry first • List major species • Calculate concentrations • Use Henderson-Hasselbach • What is the change in pH after the strong acid is added?

12. Summary of Characteristics of Buffered Solutions Buffered Solutions contain large conc. of weak acid or weak base and their conjugates Resists change in pH when [H+] or [OH-] are added pH in solution is determined by Henderson-Hasselbalch pH = pKa + log(base/acid) p stands for potential H stands for Hydrogen But the jury is still out on the origin of the term

13. Buffer capacity • Buffering capacity is a measure of the amount of protons or hydroxide ions a buffer can absorb without altering the pH. • The pH of a buffered solution is determined by the ratio [A-]/[HA]. • Capacity of buffer is determined by magnitude (how big the numbers are) of [HA] and [A-].

14. Buffer Capacity Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following solutions. Ka (HC2H3O2)= 1.8x10-5 • 5.00 M HC2H3O2 and 5.00 M NaC2H3O2 • 0.050 M HC2H3O2 and 0.050 M NaC2H3O2

15. Buffer capacity Change in ratio of [A-]/[HA] for 2 solutions when 0.01 mol H+ is added to 1.0L of each solution • The best buffers have a ratio [A-]/[HA] = 1 • This is most resistant to change • Optimal buffering occurs when [A-] = [HA]

16. Titrations and pH Curves • Titration is adding a solution with a known concentration (titrant) into a solution of unknown concentration until the substance being analyzed is consumed. • Example: Put an acid of known concentration into a base of unknown concentration until all the base is neutralized by the acid. To see that point we use indicators which turn colors at the stoichiometric equivalence point.

17. Titrations • Millimole (mmol) = 1/1000 mol • Molarity = mmol/mL = mol/L • Makes calculations easier because we will rarely add Liters of solution. • Adding a solution of known concentration until the substance being tested is consumed. • This is called the equivalence point. • Graph of pH vs. mL is a titration curve.

18. Strong acid-Strong Base Titration • Do the stoichiometry • There is no equilibrium • SA & SB dissociate completely • Calc. pH at certain points where specific volumes of 0.100 M NaOH has been added Don’t forget this formula

19. Case Study The titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH

20. Weak acid with Strong base • Similar to series of buffer problems • Calc. the pH curve for weak acid-Strong base involves a 2 step procedure • Do stoichiometry to determine concentrations • Do equilibrium to determine the weak acid equilibrium and pH • Titrate 50.0 mL of 0.10M H(C2H3O2) (Ka = 1.5 x 10-5) with 0.10 M NaOH

21. Titration of a Weak Acid Hydrogen cyanide gas (HCN) is a very weak acid (Ka=6.2x10-10) when dissolved in water. If a 50.0mL sample of 0.100M HCN is titrated with 0.10M NaOH, calculate the pH of the solution • After 8.00 mL of 0.100M NaOH has been added • At the halfway point of the titration • At the equivalence point of the titration

22. Strong acid with strong Base • Equivalence at pH 7 7 pH mL of Base added

23. Weak acid with strong Base Equivalence at pH >7 >7 pH mL of Base added

24. Strong base with strong acid Equivalence at pH 7 7 pH mL of Acid added

25. Weak base with strong acid Equivalence at pH <7 <7 pH mL of Acid added

26. Indicators • Weak acids that change color when they become bases. • Weak acid written HIn • Weak base In- • HIn  H+ + In- For phenolphthalein fuchsia • Equilibrium is controlled by pH • End Point - when the indicator changes color and stays fuchsia ( in the case of phenolphthalein) clear

27. Indicators • Since it is an equilibrium the color change is gradual. • In the titration of acid with base, color change will occur at a pH where • Choose the indicator with a pKa 1 less than the pH at equivalence point if you are titrating with base. • Choose the indicator with a pKa 1 greater than the pH at equivalence point if you are titrating with acid.

28. Solubility Equilibria • All dissolving is an equilibrium problem • If there is not much solid it will all dissolve • As more solid is added the solution will become saturated • The solid will precipitate as fast as it dissolves • Resulting in an equilibrium which can be used to set up an equilibrium constant expression

29. Solubility-Product Constant Ksp • Example: • The salt will dissolve and an equilibrium is set up • Can write an equilibrium expression for the reaction • Called the solubility productfor each compound.

30. Solubility Equilibria • Solubility Product = product of the concentration of ions in equation, raised to power of coefficients • Solubility Product is an equilibriumconstant • It doesn’t change except with temperature • Solubilityis an equilibrium position (can change) • It’s the amount of solid that will dissolve under a given set of conditions • Solubilityis not the same as solubility product • A common ion can change this • See Table 15.4 for values of Ksp

31. Calculating Ksp • Copper (I) bromide has a measured solubility of 2.0 X 10-4 mol/L at 25oC. Calculate it’s Ksp • Start by writing the reaction • Write the equilibrium expression • Build an ICE Table • Plug the concentrations into the eq. expr.

32. Calculating Ksp • The solubility of Bismuth (III) Iodide is 7.78x10-3 g/L Calculate it’s Ksp • Need to convert g/L to mol/L - use the molar mass of the Bismuth (III) Iodide - (1 mol = 589.7g) • Build ICE table

33. Calculating Ksp • Plug into Ksp expression • Notice the 3 exponent on the I- shows up in 2 places, as an exponent above and as a multiplier in the ICE table (that is the stoichiometric usage), in the expression it is the exponent.

34. Calculate Solubility • The Ksp of Mg(OH)2 is 8.9x10-12 at 25oC. Calculate the solubility. • Plug it in • =(x)(2x)2=4x3=8.9x10-12 • X=1.3x10-4 mol/L

35. Relative solubilities • Ksp will only allow us to compare the solubility of solids that fall apart into the same total number of ions. • The bigger the Ksp the more soluble. • Example 1 • AgI Ksp=1.5x10-16 • Ba(SO4) Ksp=1.5x10-9 • Ni(CO3) Ksp=1.4x10-7 • Example 2 • CuS Ksp=8.5x10-45 • Ba3(PO4)2 Ksp=6x10-39 • Fe(OH)3 Ksp=4.0x10-38

36. 3 FACTORS AFFECT SOLUBILITY OF IONIC SUBSTANCES: • COMMON ION • pH • COMPLEX IONS Common Ion Effect • If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve, equilibrium will shift to the left

37. pH and solubility • pH greatly affects salt’s solubility • Adding OH-or increasing pHdecreases solubility • Removing OH-or decreasing pHincreasessolubility • For other anions if they come from a weak acid they are more soluble in a acidic solution than in water • Brown LeMay #34

38. Precipitation and Ions • Ion Product, Q =[M+]a[Nm-]b use initial concentrations • If Q>Ksp precipitation forms until Q = Ksp • If Q<Ksp solid dissolves until Q = Ksp • If Q = Ksp equilibrium. • A solution of 750.0 mL of 4.00 x 10-3M Ce(NO3)3 is added to 300.0 mL of 2.00 x 10-2M KIO3. Will Ce(IO3)3 (Ksp= 1.9 x 10-10M)precipitate and if so, what is the concentration of the ions?

39. Selective Precipitations • Used to separate mixtures of metal ions in solutions. • Add anions that will only precipitate certain metals at a time. • Used to purify mixtures. • Often use H2S because in acidic solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate.

40. Complex Ion Equilibria • A charged species consisting of metal ion surrounded by ligands • Ligand is Lewis base using their lone pair to stabilize the charged metal ions • Common ligands are NH3, H2O, Cl-, CN- • Coordination number is the number of attached ligands • Cu(NH3)4+2 has a coordination # of 4

41. Complex Ion Equilibria • Usually the ligand is in large excess. • Ligands are added 1 step at a time: Example Ag+ + NH3 Ag(NH3)+ K1 = 2.1 x 103 Ag(NH3)++ NH3 Ag(NH3)2+ K2 = 8.2 x 103 • The formation constant or stability constant is Kx where x = step number • Individual K’s will be large so treat them as if reactions goes to equilibrium. • The complex ion will be the biggest ion in solution.