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Chemistry, The Central Science , 10th edition AP edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten.
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Chemistry, The Central Science, 10th edition AP edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 17: Additional Aspects of Aqueous EquilibriaJohn D. Bookstaver, St. Charles Community College, St. Peters, MO, 2006, Prentice Hall, Inc.(ppt modified for our requirements)
Ch. 16 Acids-Base Equilibria • s • Ch. 17 Additional Aspects of Aqueous Equilibria • Common Ion effect, • Buffered solutions, • Acid-Base Titrations, • Solubility Equilibria, • Factors that affect solubility, • Precipitation and Separation, • Qualitative Analysis for Metallic Elements • Resources and Activities • Textbook - chapter 17 & ppt file • Online practice quiz • Lab activities • POGIL activities • Chem Guy video lecture series on Acids-Bases (many) • http://www.cosmolearning.com/video-lectures/acids-and-bases-i-properties-of-molecular-neutral-ionic-solutions/ • Chemtour videos from Norton • http://www.wwnorton.com/college/chemistry/gilbert2/contents/ch16/studyplan.asp • For titrations and indicator choices: • http://www.avogadro.co.uk/chemeqm/acidbase/titration/phcurves.htm
Activities and Problem set for chapter 17 (due date_______) Lab activities: Titration of Weak Acid (wet lab) Virtual labs from http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animationsindex.htm POGILS (5) : Introduction to Acids and Bases, pH scale, Acid-Base Titrations, Weak Acid-Base Equilibria, Buffer Solutions. Online practice quiz ch 17 due by_____ Chapter 17 reading guide and practice problems packet Independent work - students to view animations & interactive activities (8 in total from Norton) and write summary notes on each. These summaries are to be included in your portfolio. Some of these will be previewed in class. Norton Animations: (Acid rain, acid-base ionization pH scale, self-ionization of water, acid strength and molecular structure, buffers, strong acid-strong base titration, titrations of weak acids) http://www.wwnorton.com/college/chemistry/gilbert2/contents/ch16/studyplan.asp
Outline Common ion effect • Effect on % ionization • Effect on pH Buffered solutions • Henderson-Hasselbalch equation • Addition of a strong acid or a strong base to a buffered solution Acid-Base Titrations • Strong Acid and Strong base • Weak Acid- Strong Base • Strong Acid-Weak base • Weak Acid-Weak base • Titration of Polyprotic Acid Solubility Equilibria Factors that affect solubility Precipitation and Separation Qualitative Analysis for Metallic Elements
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) The Common-Ion Effect • Consider a solution of acetic acid: If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left. • What is the pH of a solution prepared by adding 0.30 mol of ethanoic acid (HC2H3O2) and 0.30 mol of sodium ethanoate to enough water to make a 1.0 L solution?
The Common-Ion Effect “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.” What is the pH of a solution prepared by adding 0.30 mol of ethanoic acid (HC2H3O2) and 0.30 mol of sodium ethanoate to enough water to make a 1.0 L solution?
Analyze and Plan • Identify major species in solution & consider acidity and basicity • Identify the impt. Equilibrium that determines pH • Use ICE or RICE chart to tabulate concentrations • Use equilibrium constant expression to calculate [H3O+]
Ka = [H3O+] [F−] [HF] = 6.8 10-4 The Common-Ion Effect Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Ka for HF is 6.8 10−4.
HF(aq) + H2O(l) H3O+(aq) + F−(aq) The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.
(0.10) (x) (0.20) 6.8 10−4 = (0.20) (6.8 10−4) (0.10) The Common-Ion Effect = x 1.4 10−3 = x
The Common-Ion Effect • Therefore, [F−] = x = 1.4 10−3 [H3O+] = 0.10 + x = 0.10 + 1.4 10−3 = 0.10 M • So, pH = −log (0.10) pH = 1.00
Buffers: • Solutions of a weak conjugate acid-base pair. • They are particularly resistant to pH changes, even when strong acid or base is added.
Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.
Buffers If acid is added, the F− reacts to form HF and water.
Ka = [H3O+] [A−] [HA] HA + H2O H3O+ + A− Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA:
[A−] [HA] [H3O+] Ka = base [A−] [HA] −log [H3O+] + −log −log Ka = pKa acid pH Buffer Calculations Rearranging slightly, this becomes Taking the negative log of both side, we get
[base] [acid] pKa = pH − log [base] [acid] pH = pKa + log Buffer Calculations • So • Rearranging, this becomes • This is the Henderson–Hasselbalchequation.
Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4 10−4.
[base] [acid] (0.10) (0.12) pH = pKa + log pH = −log (1.4 10−4) + log Henderson–Hasselbalch Equation pH = 3.85 + (−0.08) pH = 3.77
pH Range • The pH range is the range of pH values over which a buffer system works effectively. • It is best to choose an acid with a pKa close to the desired pH.
When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base is consumed in the reaction.
Addition of Strong Acid or Base to a Buffer • Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. • Use the Henderson–Hasselbalch equation to determine the new pH of the solution.
Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.
Calculating pH Changes in Buffers Before the reaction, since mol HC2H3O2 = mol C2H3O2− pH = pKa = −log (1.8 10−5) = 4.74
Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC2H3O2(aq) + OH−(aq) C2H3O2−(aq) + H2O(l)
(0.320) (0. 200) pH = 4.74 + log Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + 0.06 pH = 4.80
Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base).
Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.
Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly.
Titration of a Strong Acid with a Strong Base Just before and after the equivalence point, the pH increases rapidly.
Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.
Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off.
Titration of a Weak Acid with a Strong Base • Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. • The pH at the equivalence point will be >7. • Phenolphthalein is commonly used as an indicator in these titrations.
Titration of a Weak Acid with a Strong Base At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.
Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.
Titration of a Weak Base with a Strong Acid • The pH at the equivalence point in these titrations is < 7. • Methyl red is the indicator of choice.
Titrations of Polyprotic Acids In these cases there is an equivalence point for each dissociation.
BaSO4(s) Ba2+(aq) + SO42−(aq) Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO4 in water:
Solubility Products The equilibrium constant expression for this equilibrium is Ksp = [Ba2+] [SO42−] where the equilibrium constant, Ksp, is called the solubility product.
Solubility Products • Ksp is not the same as solubility. • Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).
AgCl (s) Ag+(aq) + Cl-(aq) Initial (M) Change (M) Equilibrium (M) s = Ksp 1.3 x 10-5 mol AgCl 1 L soln x 143.35g AgCl 1 mol AgCl What is the solubility of silver chloride in g/L ? Ksp= 1.6 x 10-10 0.00 0.00 Ksp= [Ag+][Cl-] Ksp= s2 +s +s s s s = 1.3 x 10-5 [Cl-] = 1.3 x 10-5M [Ag+] = 1.3 x 10-5M = 1.9 x 10-3g/L Solubility of AgCl = 16.6
BaSO4(s) Ba2+(aq) + SO42−(aq) Factors Affecting Solubility • The Common-Ion Effect • If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease.
Factors Affecting Solubility • pH • If a substance has a basic anion, it will be more soluble in an acidic solution. • Substances with acidic cations are more soluble in basic solutions.
Factors Affecting Solubility • Complex Ions • Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.
Factors Affecting Solubility • Complex Ions • The formation of these complex ions increases the solubility of these salts.
Factors Affecting Solubility • Amphoterism • Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. • Examples of such cations are Al3+, Zn2+, and Sn2+.
Will a Precipitate Form? • 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.
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? The ions present in solution are Na+, OH-, Ca2+, Cl-. Only possible precipitate is Ca(OH)2 (solubility rules). Is Q > Kspfor Ca(OH)2? [OH-]0 = 4.0 x 10-4M [Ca2+]0 = 0.100 M = 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 16.6
Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture.