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A.P. Chemistry

A.P. Chemistry. Chapter 15 Applications of Aqueous Equuilibria. 15.1 Solutions of Acids or Bases Containing a Common Ion Common ion - the presence of an ion which appears in both the acid (or base) and a salt in the solution

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A.P. Chemistry

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  1. A.P. Chemistry Chapter 15 Applications of Aqueous Equuilibria

  2. 15.1 Solutions of Acids or Bases Containing a Common Ion Common ion- the presence of an ion which appears in both the acid (or base) and a salt in the solution Common ion effect- the shift in equilibrium position that occurs because of the addition of anion already involved in the equilibrium reaction For example: if a solution of NaF and HF are mixed together, F- would be the common ion. If we look at the dissociation of each compound- NaF Na+(aq) + F-(aq) HF(aq)  H+(aq) + F-(aq) According to LeChatelier’s Principle, an increase in concentration of an ion will shift the equilibrium away from the ion added. HF(aq)  H+(aq) + F-(aq) Equilibrium shifts away added F- ions from From added component; NaF Fewer H+ ions present This makes the solution less acidic than a solution of HF alone.

  3. Example: 41 g of sodium acetate, CH3COONa have been added to 500.0 mL of a 0.50 M CH3COOH solution. Assume no change in volume. Determine the pH of the solution. (Ka for CH3COOH is 1.8 x 10-5)(5.05)

  4. 15.2 Buffered Solutions Buffered Solution- a solution that resists a change in pH when either hydroxide ions or protons (hydrogen ions) are added. A buffered solution is composed of a weak acid and its salt (or a weak base and its salt) They are simply solutions of weak acids or bases containing a common ion. The pH calculations on buffered solutions require exactly the same procedures introduced in ch. 14. When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium calculations.

  5. How does a buffer work? Suppose in a buffered solution we have large amounts of a weak acid HA and its conjugate base A- If OH- ions are added to this system (the reaction will proceed in the forward direction) HA + OH- A- + H2O The net result is that OH- are not allowed to accumulate but are replaced by A- ions. So pH changes only slightly. If H+ ions are added to this system (the reaction will proceed in the reverse direction) H+ + A- HA The net result is that H+ ions are not allowed to accumulate but are replaced by HA molecules, so the pH changes only slightly.

  6. Thus the stability of the pH under these conditions can be understood by examining the equilibrium expression for the dissociation of HA: Ka = [H+][A-] [HA] Rearrange to [H+] = Ka [HA] [A-] In other words, the equilibrium concentration of H+ and thus pH, is determined by the ratio [HA]/[A-]. If the amounts of HA and A- are very large to start with compared to the amount of H+ (or OH-) ions added, the change in [HA]/[A-] will be small.

  7. What if the buffered solution is made from a weak base and the salt of the conjugate acid? If H+ ions are added to this system (the reaction will proceed in the forward direction) B + H+ BH+ If OH- ions are added to this system(the reaction will proceed in the reverse direction) BH+ + OH- B + H2O Henderson-Hasselbalch equation- using the above relationship, if we were to take the log form of the Ka expression we would get an equation which is useful for calculating the pH of solutions when the [HA]/[A-] is known. pH = pKa + log( [A-]/[HA]) = pKa + log([base]/[acid]) For a particular buffering system (acid conjugate base pair) all solutions that have the same ratio [A-]/[HA] will have the same pH.

  8. Example: Determine which of the following solutions form buffers and identify each solution as acidic (pH < 7.00), basic (pH > 7.00), or neutral (pH = 7.00) • 50.0 mL of 1.0 M HCl & 50.0 mL of NaCl (not buffer; pH < 7.00) • 100.0 mL of 1.0 M HF & 100.0 mL of 1.0 M NaF (acidic buffer; pH < 7.00) • 100.0 mL of 1.0 M KOH & 50.0 mL of 2.0 M HBr (not buffer; pH = 7.00) • 30.0 mL of 1.0 M NH3 & 15.0 mL of 1.0 M HCl (basic buffer; pH > 7.00) • 30.0 mL of 1.0 M NH3 & 45.0 mL of 1.0 M HCl (not buffer; pH < 7.00)

  9. Example: Calculate the pH of a buffer solution that is 0.25 M HF and 0.50 M NaF. (3.45)

  10. Example: What ratio of [F-] to [HF] would you use to make a buffer of pH = 2.85?

  11. 15.3 Buffer capacity Buffer capacity-represents the amount of protons (hydrogen ions) or hydroxide ions the buffer can absorb without a significant change in pH. The capacity of a buffered solution is determined by the magnitude of [HA] and [A-] Example: Adding Strong Acid to a Buffer Suppose 3.0 mL of 2.0 M HCl is added to exactly 100. mL of the buffer described in the above example; what is the new pH of the buffer after the HCl is neutralized?

  12. 15.4 Titrations and pH curves (Examples on in-class worksheet) pH curve (titration curve)- the progress of an acid-base titration monitored by plotting the pH of the solution being analyzed as a function of the amount of titrant added. Millimole- since most titrations involve very small quantities, the mole is inconveniently large. Since most volumes will be measured in milliliters, it would also be convenient to measure in a quantity equal to a thousandth of a mole. Molarity = mol solute = mol solute/1000 = mmol L solution L solution/1000 mL solution

  13. Strong acid-strong base titration- since both the acid and the base are strong, each will dissociate completely. Therefore it is important to determine the number of moles of H+ and OH- produced. Three situations could occur: If you have equal moles of H+ and OH-, they will cancel out each other and you will be left with a neutral solution. If you have more H+ ions than OH- ions, the extra H+ will make the solution acidic (the other H+ ions will combine with the OH- ions to make water) If you have more OH- ions than H+ ions, the extra OH- will make the solution basic (or alkaline)(the other OH- ions will combine with the H+ ions to make water).

  14. Stoichiometric Point (or equivalence point)- the point in the titration where an amount of base has been added to exactly react with all the acid originally present. Weak acid- strong base titration- this type of titration actually has two components: the stoichiometry of the reaction and the equilibrium which will occur. Each should be dealt with separately. The stoichiometry- the reaction of hydroxide ion with the weak acid is assumed to run to completion and the concentrations of the acid remaining and the conjugate base formed are determined. The equilibrium- the position of the weak acid equilibrium is determined and the pH is calculated. The pH of the equivalence point of a titration of a weak acid and a strong base is always greater than 7.

  15. Weak base- strong acid titration-(see above) The pH at the equivalence point of a titration of a weak base with a strong acid is always less than 7. (Make sure you know the general shapes of acid-base titration curves.)

  16. Example: Calculate the pH when the 10.0 mL of a 0.15 M HNO3 is added to 50.0 mL of 0.10 M NaOH.

  17. 15.5 Acid-Base Indicators Acid-base indicator: marks the end point of a titration by changing color. It is important to note that the end point is defined by the change in color of the indicator. The equivalence point is defined by the reaction stoichiometry. Example: Choose an indicator for the titration of 50.0 mL of a 0.10 M HI solution with 0.10 M NH3.

  18. Second Half of the Chapter!!

  19. 15.6 Solubility Equilibria and the Solubility Product Solubility product constant (Ksp)- expresses the equilibrium position that occurs for the solubility of solids. It also expresses at what point solution would be considered to be saturated. For the following reaction: AB(s) nA+(aq) + mB-(aq) The equilibrium expression would be Ksp = [A+]n[B-]m Example: The solubility of magnesium fluoride in water is 7.3 x 10-3 g per 100. mL of solution. What is the solubility product constant?

  20. Example: The Ksp of Zn(OH)2 at 25oC is 1.8 x 10-14. Determine the molar solubility of zinc hydroxide in pure water at 25oC. Determine the molar solubility of zinc hydroxide if 0.10 mol of Zn(NO3)2 are added to 1.0 L of a saturated solution of zinc hydroxide at 25oC. (assume no volume change) Determine the molar solubility of zinc hydroxide if the solution has an adjusted pH of 2.0 at 25oC.

  21. 15.7 Precipitation and Qualitative Analysis Ion product (Q) – just like the expression for Ksp for a given solid except that initial concentrations are used instead of equilibrium concentrations. If Q > Ksp, precipitation occurs, and will continue until the concentrations are reduced to the point that they satisfy Ksp. Is Q < Ksp, no precipitation will occur. Selective precipitation: uses a reagent whose anion forms a precipitate with only one or a few of the metal ions in a mixture.

  22. Example:10.0 mL of 0.010 M Pb(NO3)2 are mixed with 10.0 mL of 0.01 M NaI solution. Determine whether a precipitate will occur and justify your answer.

  23. Example: Determine all ion concentrations at equilibrium when 10.0 mL of a 0.050 M Pb(NO3)2 solution are mixed with 20.0 mL of a 0.010 M Na2CrO4.

  24. Example: What is the solubility of PbCl2 in 0.50 M NaCl solution?

  25. 15.8 Equilibria Involving Complex ions (read this section) Take note: The complex ions that appear most often on the AP exam are: Ag(NH3)2+, Ag(CN)2-, Al(OH)4-, AlF63-, Cu(NH3)42+, Fe(SCN)2+, Fe(SCN)63-, Ni(NH3)x2+, Ni(OH)42-, Zn(OH)42-, Zn(NH3)62+, Zn(NH3)42+, CoCl42- See Eqx Writing on Complex Ion. Example: Write equations for the formation reaction of each of the following complex ions: Cu(NH3)42+ Cu(CN)42- Ag(CN)2-

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