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Chapter 15

Chapter 15. Applications of Equilibrium- Titrations & Solubility. Section 15.4-15.5: Recall Titrations. What are titrations used for? Find an unknown concentration. End point When the indicator changes color. Equivalence point Equal stoichiometric amounts of acid and base are present.

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Chapter 15

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  1. Chapter 15 Applications of Equilibrium- Titrations & Solubility

  2. Section 15.4-15.5: Recall Titrations • What are titrations used for? • Find an unknown concentration. • End point • When the indicator changes color. • Equivalence point • Equal stoichiometric amounts of acid and base are present. • Titrant vs. Analyte • Substance whose concentration is known vs. substance whose concentration is unknown.

  3. Titration/pH Curves As a titration proceeds, the pH can be monitored. No matter what is titrated with what (strong acid, strong base; weak acid, strong base; etc.), they all have characteristic components.

  4. Strong acid-Strong base *Notice the steep drop around the equivalence point. *Acid reacts with added base and is in excess (pH still acidic) UNTIL equivalence point. *Equivalence point pH = 7; no acid or base present, and ions present are neutral.

  5. Strong base – Strong acid *Notice the similarities between the curves. *Base reacts with added acid and is in excess (pH still basic) UNTIL equivalence point. *Equivalence point pH = 7; no acid or base present, and ions present are neutral.

  6. Weak acid – Strong base *Still looks similar to previous curves, but initial pH is higher than with the strong acid. *Notice the initial steep increase of the pH, until buffer kicks in. *Equivalence point pH = 9. Why? *Conjugate base present is weak!

  7. Weak base – Strong acid *Initial pH is lower than with the strong base. *Again, notice initial steep decrease of the pH, until buffer kicks in. *Equivalence point pH = 5. Why? *Conjugate acid present is weak!

  8. pH of Titrations Consider a weak acid that is titrated with a strong base. Titration can be broken down into the following parts when studying pH changes: Initial pH: find pH of a weak acid. Acid & base react completely- LR problem. If weak acid remains, a bufferdevelops. At equivalence point, weak acid has all reacted into the conjugate base. Calculate pH of the weak base. After equivalence point, find pH of a strong base.

  9. Example of Titration Components Na+ has been left out- spectator! Consider a titration where nitrous acid is titrated with sodium hydroxide. Only weak acid: HNO2 NO2- + H+ Some NaOH added: HNO2 + OH-  NO2- + H2O *The HNO2 completely reacts with OH-. This is a stoichiometry problem at first! Find LR. *Say OH- is LR, so some HNO2 still present: HNO2  H+ + NO2- (HNO2 & NO2- present) *Buffer problem to find pH!

  10. Example of Titration Components Consider a titration where nitrous acid is titrated with sodium hydroxide. More NaOH added: *HNO2 continues to completely react with NaOH. Calculations are the same as step 2 UNTIL equivalence point is reached. *At equivalence point, no HNO2 & no NaOH. *Only conjugate base NO2- left: NO2- + H2O  HNO2 + OH- *Weak base problem to find pH!

  11. Example of Titration Components Consider a titration where nitrous acid is titrated with sodium hydroxide. Even more NaOH added: *All acid has been used up. Now you have a strong and weak base mixed together. *Strong base only needs to be considered. *Find concentration of OH-, then find pOH and pH.

  12. Homework Pg. 741 #51

  13. Example Calculations To demonstrate the components described, various amounts of strong base (NaOH) will be added to the weak acid (HNO2). Notice that, with the exception of the initial pH, the same approach is used initially. Depending on what remains, various equations may be used to find the pH.

  14. Example Calculations (1) Initially, 100.0mL of 0.150M nitrous acid is present. Ka = 4.5 x 10-4 • Set up ICE table and solve! • Note: cannot ignore x here! • x = 0.0080 = [H+] • pH = 2.10 to start with

  15. Example Calculation (2) 25.00mL of 0.300M NaOH is added. • Stoichiometry problem- IF table! • Acid and base react to completion. • Determine which reactant is limiting: NaOH + HNO2  NO2- + H2O I 0.00750mol 0.0150mol 0 ----- F 0mol0.00750mol 0.00750mol • Use new molar amounts to find new concentrations. You now have a buffer!

  16. Example Calculation (2) 25.00mL of 0.300M NaOH is added. • Buffer: HNO2 H+ + NO2- • Henderson-Hasselbalch to find pH. • New concentrations (total V = 125.00mL): [HNO2] = [NO2-] = 0.0600M • pH = 3.35 • This buffer will continue until the equivalence point is reached!

  17. Example Calculation (3) 50.00mL of 0.300M NaOH is added. • Stoichiometry problem- IF table! • Acid and base react to completion. • Determine which reactant is limiting: NaOH + HNO2  NO2- + H2O I 0.0150mol 0.0150mol 0 ----- F 0mol0mol 0.0150mol • No buffer this time! Only weak conjugate base. Solve for pH of a weak base.

  18. Example Calculation Equivalence point pH! (3) 50.00mL of 0.300M NaOH is added. Weak base (new [NO2-] = 0.100M): NO2- + H2O  HNO2 + OH- I 0.100M 0M 0M C -x +x +x E 0.100 – x x x Solve for x: x = 1.50 x 10-6 = [OH-] pOH = 5.82; pH = 8.18 

  19. Example Calculation Past equivalence point! (4) 75.00mL of 0.300M NaOH is added. • Stoichiometry problem- IF table! • Acid and base react to completion. • Determine which reactant is limiting: NaOH + HNO2  NO2- + H2O I 0.0225mol 0.0150mol 0 ----- F 0.0075mol 0 0.0150mol • Strong and weak base remain. Strong base determines pH- solve for pH of a strong base. • pOH = 1.37; pH = 12.63 

  20. Strategy • When asked to find the pH at some point in a titration: • Unless it’s the initial pH, ALWAYS set up and use an IF table. The acid/base reaction will go to completion based on the LR. • Determine what species remain and how much. • Determine if you have a buffer, are at the equivalence point (only conjugate species present), or are past the equivalence point (strong species & conjugate species present). • Then calculate the pH appropriately.

  21. Homework Pg. 742 #53

  22. Section 15.6- Solubility Equilibria Recall that salts typically dissociate in water. But do all salts dissolve? No! Some are only slightly soluble, and therefore are in equilibrium between the solid salt and dissolved ions. Ex: PbSO4 (s)  Pb+2 (aq) + SO4-2 (aq) It is important in these problems to take into account how many moles of each ion are in one mole of the salt!

  23. Solubility Product Constant, Ksp • Ksp is used for the equilibrium expression. It is known as the solubility product constant. • Still found by taking [products]/[reactants] in equilibrium. • However, the reactant will always be a solid salt, so it is not included. • Thus, Ksp only depends on the concentration of ions. • From previous ex: Ksp = [Pb+2][SO4-2]

  24. Sample Calculation Ksp = 8 x 10-8 = 4x3 x = 3 x 10-3 • The Ksp of magnesium fluoride in water is 8 x 10-8. How many grams of magnesium fluoride will dissolve in 0.250L of water? First solve for x: MgF2 (s)  Mg+2 + 2F- I ----- 0 0 C ----- +x +2x E ----- x 2x

  25. Sample Calculation • All we need to solve this problem is one of the concentrations of the ions. • Once we have this, we can use stoichiometry/mole ratios to find moles of the salt from moles of the ion. x = 3 x 10-3M= [Mg+2]  use this with the given volume to find moles: 3 x 10-3mol x 0.250L = 8 x 10-4mol Mg+2 L

  26. Sample Calculation • Recall: MgF2(s)  Mg+2 (aq) + 2F- (aq) 8 x 10-4 mol Mg+2 x 1mol MgF2 x 62.31g MgF2 1mol Mg+2 1mol MgF2 = 0.05g MgF2

  27. Practice If the Ksp of lead (II) sulfate is 1.6 x 10-8, how many grams will dissolve in 0.500L of water? PbSO4 (s)  Pb+2 (aq) + SO4-2 (aq) Set up ICE table, solve for x: 1.6 x 10-8 = x2 x = 1.3 x 10-4M = [Pb+2] = [SO4-2] Use stoichiometry: 0.020g will dissolve

  28. More Practice • What is the value of Ksp if the solubility of CaC2O4 is 6.1 x 10-3g/L? • Set up ICE table CaC2O4 (s)  Ca+2 (aq) + C2O2-2 (aq) I ----- 0 0 C ----- +x +x E ----- x x • Convert solubility to mol/L: 4.8 x 10-5 • Ksp = (4.8 x 10-5M)2 Ksp = 2.3 x 10-9

  29. Homework Pg. 743 # 75, 77(b), 79, 81(a)

  30. Recall LeChatelier’s Principle • Remember what this said? • Any stress added to a system at equilibrium will cause the equilibrium to respond and be re-established. • Ex: N2 (g) + 3H2 (g)  2NH3 (g) • If more H2 is added, what is the effect? • Equilibrium shifts right to use up extra H2, so N2 decreases and NH3 increases. • This principle can be applied to solubility: common-ion effect.

  31. Common-Ion Effect • Adding a solution with an ion also present in the slightly soluble salt will decrease the salt’s solubility. • Ex: PbSO4 (s)  Pb+2 (aq) + SO4-2 (aq) • If Na2SO4 is added, what is the effect? • Adds more SO4-2, so equilibrium shifts left. This means more of the salt remains as a solid, making it less soluble. • Calculations now involve taking into account additional ion added.

  32. Common-Ion Calculation • What is the concentration of silver ions if a 1.00M Na2CrO4 solution is added to solid Ag2CrO4? Ksp = 1.9 x 10-12 • Steps are the same as previously practiced, except now the concentration of CrO4- comes from the solution AND the salt. • Note that for every mole of Na2CrO4 that dissolves, one mole of CrO4- ions form: Na2CrO4 (s)  2Na+ (aq) + CrO4-2 (aq) • [Na2CrO4] = [CrO4-2] = 1.00M add to ICE table

  33. Common-Ion Calculation • Ag2CrO4 (s)  2Ag+ (aq) + CrO4-2 (aq) I ----- 0 1.00 C ----- +2x +x E ----- 2x 1.00 + x • 1.9 x 10-12 = (2x)2(1.00 + x)  x is negligible! • x = 6.9 x 10-7 • [Ag+] = 2x  [Ag+] = 1.4 x 10-6M

  34. AP Practice Question No calculators! • A student wishes to reduce the zinc ion concentration in a saturated zinc iodate solution to 1 x 10-6 M. How many moles of solid KIO3 must be added to 1.00L of solution? Ksp for Zn(IO3)2 = 4 x 10-6 • As usual, write down the equilibrium reaction: Zn(IO3)2 (s)  Zn+2 (aq) + 2IO3- (aq) • Now, think about what you know! • Know Ksp and the desired equilibrium [Zn+2] • Solve for [IO3-], then convert to moles!

  35. AP Practice Question A student wishes to reduce the zinc ion concentration in a saturated zinc iodate solution to 1 x 10-6 M. How many moles of solid KIO3 must be added to 1.00L of solution? Ksp for Zn(IO3)2 = 4 x 10-6 1mol 0.5mol 2mol 4mol

  36. Homework Pg. 743 #90

  37. Remember Q? • Recall Q from the beginning of the equilibrium section. • Q = reaction quotient • Tells you where the reaction is with respect to reaching equilibrium • Smaller = before equilibrium; equal = at equilibrium; larger = past equilibrium • This same idea can be used with Ksp!

  38. Ion-product • Technical name for Q with respect to Ksp. • Found the same way Ksp is calculated, then compared to Ksp value. • Before (smaller than Ksp) = no ppt • After (larger than Ksp) = ppt • Calculation strategy: ONLY difference is the new [ ]’s must be found (changing V!), then calculate Q.

  39. Example Calculation • If 10.0mL of 0.100M BaCl2 is added to 40.0mL of 0.0250M Na2SO4 solution, will BaSO4 precipitate? Ksp for BaSO4 = 1.1 x 10-10. • Write down equilibrium reaction: BaSO4 (s)  Ba+2 (aq) + SO4-2 (aq) • Only concentrations you need to worry about are for Ba+2 and SO4-2! Nothing else is needed to find Q! • Find new concentrations after mixing: [Ba+2] = 0.0200M [SO4-2] = 0.0200M

  40. Example Calculation with Q • If 10.0mL of 0.100M BaCl2 is added to 40.0mL of 0.0250M Na2SO4 solution, will BaSO4 precipitate? Ksp for BaSO4 = 1.1 x 10-10. • Now find Q: Q = (0.0200)(0.0200) = 0.000400 • Compare Q to Ksp: Q > Ksp, therefore BaSO4 will ppt. (past equilibrium, so excess ions will form a ppt.)

  41. Warm Up Problem Will a precipitate form if 100.0mL of 4.0 x 10-4M Mg(NO3)2 is mixed with 100.0mL of 2.0 x 10-4M NaOH? Ksp for Mg(OH)2 = 8.9 x 10-12. 1) Mg(OH)2 (s)  Mg+2 (aq) + 2OH- (aq) 2) Find new concentrations: [Mg+2] = 2.0 x 10-4M [OH-] = 1.0 x 10-4M 3) Find Q: Q = 2.0 x 10-12; Q < K, so no ppt.

  42. Homework Pg. 743 #97 (look on pg. 718 for Ksp values!)

  43. pH and Solubility Decreases solubility! Increases solubility! • pH can impact the solubility of salts • This means that adding an acid or base to a salt can increase or decrease solubility. • Consider the following example: Mg(OH)2 (s)  Mg+2(aq) + 2OH-(aq) • What if a base (OH-) is added? • What if an acid is added? • Why is this seen? • Consider possible reactions and then apply Le Chatelier’s Principle/Common Ion Effect

  44. pH and Solubility Cont. • Consider another example: Ag3PO4(s)  3Ag+(aq) + PO4-3(aq) • What if an acid is added? • PO4-3 reacts with H+ to form HPO4-2, which is a weak acid • As we learned, weak acids don’t dissociate much. Therefore, some PO4-3 is removed from solution to form HPO4-2 • Reaction shifts right, making Ag3PO4 more soluble.

  45. pH and Solubility Cont. • Here’s another example: AgCl(s)  Ag+(aq) + Cl-(aq) • What if an acid is added? • Cl- would react with H+, but it forms a strong acid (HCl). • As we learned, strong acids dissociate completely. Therefore, Cl- is removed from solution. • Reaction is not affected, and the salt has the same solubility in acid as it does in water.

  46. pH and Solubility Cont. Br-, I-, NO3-, SO4-2, and ClO4- • From the two examples previously considered, a general rule can be used: • Any salt, MX, that forms a weak acid, HX, will be more soluble in acid than in water. • The H+ ions in the acid will bond with and remove anion X- from solution to make the salt more soluble. • Common anions that are more soluble in acid: OH-, S-2, CO3-2, C2O4-2, and CrO4-2. • Can you think of anions, besides Cl-, that are not more soluble in acid?

  47. AP Practice Question The addition of nitric acid increases the solubility of which of the following compounds? KCl(s) Pb(CN)2(s) Cu(NO3)2(s) NH4NO3(s)

  48. Complex Ion Equilibria Complex ion: a metal ion surrounded by other molecules or ions. Equilibrium exists for complex ion formation- Kf is used, called formation constant or stability constant. Example: Ag+ + 2NH3 Ag(NH3)2+ Kf = [Ag(NH3)2+]/[Ag+][NH3]2

  49. Complex Ion Equilibria • Complex ions form in a series of steps. • Continuing with the previous ex: Ag+ + 2NH3 Ag(NH3)2+ • Steps are as follows: Ag+ + NH3  Ag(NH3)+ K1 = 2.1 x 103 Ag(NH3)+ + NH3  Ag(NH3)2+ K2 = 8.2 x 103 • Notice an additional NH3 molecule is added to the silver ion each time! • Overall Kf value? • Kf = K1 x K2  Kf = (2.1 x 103) x (8.2 x 103) = 1.7 x 107

  50. Complex Ion Equilibria • To determine the steps involved in forming complex ions, simply look at the final complex ion formed. • Look to see how many ions/molecules are bonded to the metal. • The number of ions/molecules is how many steps there are, with each step adding an additional ion/molecule to the metal. • We will not worry about calculations at this time!

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