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Applications of Aqueous Equilibria

Applications of Aqueous Equilibria. Chapter 15. Common Ion Effect. The shift in equilibrium that occurs because of the addition of an ion already involved in the equilibrium reaction. AgCl( s )  Ag + ( aq ) + Cl  ( aq ) The amount of Ag + ion decreases. Common Ion Effect Calculations.

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Applications of Aqueous Equilibria

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  1. Applications of Aqueous Equilibria Chapter 15

  2. Common Ion Effect • The shift in equilibrium that occurs because of the addition of an ion already involved in the equilibrium reaction. • AgCl(s)  Ag+(aq) + Cl(aq) • The amount of Ag+ ion decreases.

  3. Common Ion Effect Calculations • A 1.0 M HF solution has an [H+] of 2.7 x 10-2 M and a 2.7 % dissociation. What is the [H+] and the % dissociation for a solution of 1.0 M HF and 1.0 M NaF? • Major Species: HF, F-, Na+, & HOH • HF(aq) <---> H+(aq) + F-(aq) • Ka = 7.2 x 10-4 = [H+][F-]/[HF]

  4. Common Ion Effect CalculationsContinued RICE [HF] [F-] [H+] Initial (mol/L) 1.0 1.0 0 Change (mol/L) - x + x + x Equil. (mol/L) 1.0 - x 1.0 + x x

  5. Common Ion Effect CalculationsContinued • Ka = 7.2 x 10-4 = [H+][F-]/[HF] • 7.2 x 10-4 = [x][1.0 + x]/[1.0 - x] • Ka is more than 100 x smaller than concentration, x is neglected in denominator and numerator. • Ka = 7.2 x 10-4 = [x][1.0]/[1.0] • x= 7.2 x 10-4 M

  6. Common Ion Effect CalculationsContinued • % Dissociation =( [H+]/[HF]o)(100%) • % Dissociation = ([7.2 x 10-4]/[1.0]) (100%) • % Dissociation = 0.072% The 1.0 M HF solution was 2.7 % dissociated while in the mixture of 1.0 M HF and 1.0 M NaF, the HF is only dissociated 0.072 %.

  7. A Buffered Solution • . . . resists change in its pH when either H+ or OH are added. 1.0 L of 0.50 M H3CCOOH + 0.50 M H3CCOONa • pH = 4.74 • Adding 0.010 mol solid NaOH raises the pH of the solution to 4.76, a very minor change.

  8. Preparation of Buffered Solutions • Buffered solution can be made from: • a weak acid and its salt (e.g. HC2H3O2 & NaC2H3O2). • a weak base and its salt (e.g. NH3 & NH4Cl). • Other examples of buffered pairs are: • H2CO3 & NaHCO3 • H3PO4 & NaH2PO4 • NaH2PO4 & Na2HPO4 • Na2HPO4 & Na3PO4

  9. Buffer Calculations • A buffered solution contains 0.50 M acetic acid and 0.50 M sodium acetate. Calculate the pH of this solution. See Sample Exercise 15.2 on pages 717-718 for the long solution. The short solution is: • pH = pKa + log([A-]/[HA]) • pH= -log (1.8 x 10-5) + log ([.50]/[.50]) • pH = 4.74 + 0 = 4.74

  10. Henderson-Hasselbalch Equation • Useful for calculating pH when the [A]/[HA] ratios are known.

  11. Key Points on Buffered Solutions • 1. They are weak acids or bases containing a common ion. • 2. After addition of strong acid or base, deal with stoichiometry first, then equilibrium.

  12. NaOH Added to Buffered Solution • Calculate the change in pH that occurs when 0.010 mol of solid NaOH is added to 1.0 L of buffered solution from the previous example (0.5 M). • Stoichiometry Problem HC2H3O2(aq) + OH-(aq) ---> C2H3O2-(aq) + HOH(l) • The stoichiometry of the neutralization reaction must be done first, then the equilibrium calculation.

  13. NaOH Added to Buffered SolutionContinued ICE(Stoichiometry) • [HC2H3O2] [OH-] [C2H3O2-] Initial (mol) 0.50 0.010 0.50 Change (mol) - 0.010 -0.010 +0.010 End (mol) 0.49 0.00 0.51

  14. NaOH Added to Buffered SolutionContinued • HC2H3O2(aq) <---> H+(aq) + C2H3O2-(aq) ICE(Equilibrium) • [HC2H3O2] [H+] [C2H3O2-] • Initial (mol) 0.49 0 0.51 • Change (mol) - x + x + x • Equil. (mol) 0.49 - x x 0.51 + x Initial concentrations are defined after the reaction with OH- is complete but before the system adjusts to equilibrium Now follow use HH Eqn to find pH change

  15. NaOH Added to Buffered SolutionContinued • pH = pKa + log([A-]/[HA]) • pH= -log (1.8 x 10-5) + log ([.51]/[.49]) • pH = 4.74 + 0.017 • pH = 4.76 • Note: The pH only changed 0.02 pH units. The addition of the same amount of NaOH to water would have changed the pH by 5.00 units.

  16. Buffered Solution Characteristics • Buffers contain relatively large amounts of weak acid and corresponding base. • Added H+ reacts to completion with the weak base. • Added OH reacts to completion with the weak acid. • The pH is determined by the ratio of the concentrations of the weak acid and weak base.

  17. Buffered Solution Characteristics • For a particular buffering system (acid-conjugate base pair), all solutions that have the same ratio [A-]/[HA] will have the same pH. • Henderson-Hasselbach Equation.

  18. Buffering Capacity • . . . represents the amount of H+ or OH the buffer can absorb without a significant change in pH.

  19. Buffering Capacity • The pH of a buffered solution is determined by the ratio [A-]/[HA] • The buffering capacity of a buffered solution is determined by the magnitudes of [HA] and [A-]. • When the ratio [A-]/[HA] equals 1, then the system is said to have optimal buffering. pH = pKa

  20. Buffering Capacity • The pKa of a weak acid in the buffer should be as close as possible to the desired pH. • A chemist needs a solution buffered at pH 4.30. Which of the following acids and their sodium salts would be best? • chloroacetic acid Ka = 1.35 x 10-3 • propanoic acid Ka = 1.4 x 10-5 • benzoic acid Ka = 6.4 X 10-5 • hypochlorous acid Ka = 3.5 x 10-8

  21. Titration (pH) Curve • A plot of pH of the solution being analyzed as a function of the amount of titrant added. • Equivalence (stoichiometric) point: Enough titrant has been added to react exactly with the solution being analyzed.

  22. Titration curve for a strong base added to a strong acid -- the equivalence point has a pH of 7.

  23. Titration curve for the addition of strong acid to a strong base -- pH at equivalence point is 7.00.

  24. Strong Acid - Strong Base Titration • Before equivalence point, [H+] is determined by dividing number of millimoles of H+ remaining by total volume of solution in mL. • At equivalence point, pH is 7.00. • After equivalence point [OH-] is calculated by dividing number of millimoles of excess OH- by total volume of solution in mL. • See Example on pages 730-733.

  25. Weak Acid - Strong Base Titration • Know & understand example on pages 734-737. • Step 1 - A stoichiometry problem - reaction is assumed to run to completion - then determine remaining species. • Step 2 - An equilibrium problem - determine position of weak acid equilibrium and calculate pH. • At halfway to the equivalence point the concentration of A- & HA are equal –[H+] = Ka & pH = pKa.

  26. Sample Exercise 15.9: Titration of a weak acid • HCN is a weak acid (Ka=6.2 x 10-10) when dissolved in water. If a 50.0 mL sample of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH of the solution • After 8.00 mL of 0.1 M NaOH has been added • At the halfway point of the titration • At the equivalence point of the titration

  27. a. After 8.00 mL of 0.1 M NaOH has been added • Stoichiometry problem HCN + OH  CN- + H2O Before (50 mLx0.1 M) (8.00 mLx.1 M) =5.0 mmol =0.8 mmol =0mmol After (5.0 -.8) (.8-.8) (+.8) 5.8 mmol 0 mmol .8 mmol

  28. After 8.00 mL of 0.1 M NaOH has been added • Snice the solution contains the species HCN, CN-, Na+, H2O, the position of the acid dissociation equilibrium is: HCN  H+ + CN- This will determine the pH • Equilibrium problem HCN  H+ + CN- Initial [ ] = (4.2 mmol)/(58 mL) 0 (.8 mmol/58 mL)

  29. RICE TABLE SET UP RICE TABLE LIKE BEFORE Ka=6.2 x10-10 = [H+][CN-] / [HCN] =6.2 x10-10 =[(4.2/58) – x ][x] / [(.8/58)+ x] ≈ [(4.2/58) ][x] / [(.8/58)] = 6.2 x 10-10 = x (.8/4.2) x = 3.3 x 10-9 M = [H+], pH=8.49

  30. b. At the halfway point of the titration • The amount of HCN originally present can be found from the original volume and molarity 50.0 mL x 0.100 M = 5.00 mmol • Thus,the halfway point will be when 2.50 mmol of OH- has been added: mL Volume OH- x 0.100 M= 2.50 mmol OH- Volume of NaOH=25 mL • OR the halfway point [HCN] is equal to the [CN-] and pH is equal to pKa, thus after 25.0 mL of 0.100 M NaOH has been added: pH= pKa= -log(6.2 x 10-10)=9.21

  31. c. At the equivalence point of the titration • Eq. Point will occur when a total of 5.00 mmol of NaOH has beed added. Since the NaOH solution is 0.100 M, the equivalence point occurs when 50 mL of NaOH has been added. This amount will form 5.00 mmol CN-. • The reaction that will control pH is: CN- + H2O  HCN(aq) + OH-(aq) • Kb= Kw/Ka=1.0x10-14/6.2x10-10 = 1.6 x 10-5 • Kb= [HCN][OH-] / [CN-]

  32. RICE TABLE SET UP RICE TABLE LIKE BEFORE Kb=1.6 x10-5 = [HCN][OH-] / [CN-] Kb=1.6 x10-5 = [x][x] / [.050 –x] ≈ x2 / .050 1.6 x10-5 = (x2)/.05 x = 8.9 x 10-4 M = [OH-], pOH=3.05, pH=10.95

  33. Titration curve for the addition of a strong base to a weak acid-- pH is above 7.00.

  34. The equivalence point is defined by the stoichiometry, not the pH.

  35. The pH curves for the titrations of 50.0 mL samples of 0.10 M acids with various Ka values with 0.10 M NaOH.

  36. Titration curve for the addition of a strong acid to a weak base -- the pH at equivalence is below 7.00.

  37. Determining the End Point in a Titration • Two methods are used: • pH meter • acid-base indicator

  38. Acid-Base Indicator • . . . marks the end point of a titration by changing color. The color change will be sharp, occurring with the addition of a single drop of titrant. • The equivalence point is not necessarily the same as the end point. • Indicators give a visible color change will occur at a pH where: •  pH = pKa  1

  39. 15_333 OH – HO O O C C OH – C O – C O O O – (Colorless acid form, HIn) ( Pink base form, In ) The acid and base forms of the indicator phenolphthalein.

  40. The useful pH ranges of several common indicators -- the useful range is usually pKa 1. Why do some indicators have two pH ranges?

  41. On the left is the pH curve for the titration of a strong acid and a strong base. On the right is the curve for a weak acid and a strong base.

  42. Solubility • Allows us to flavor foods -- salt & sugar. • Solubility of tooth enamel in acids. • Allows use of toxic barium sulfate for intestinal x-rays.

  43. Solubility – a physical property • This measures the extent to which a solute dissolves in a solvent. http://www.elmhurst.edu/~chm/vchembook/171solublesalts.html

  44. Saturation • Every solvent has a maximum amount of solute it can dissolve. • Expressed in g/designated volume • e.g. The solubility of NaCl is 36g/100mL

  45. http://www.kirksville.k12.mo.us/khs/Teacher_Web/alternative/saturated.gifhttp://www.kirksville.k12.mo.us/khs/Teacher_Web/alternative/saturated.gif

  46. Solubility Product Constant (Ksp) • “Insoluble” = a tiny, tiny bit dissolves • Equilibrium between solid and dissolved ions AgCl (s) ⇄ Ag+ (aq) + Cl- (aq) • Ksp is usually a very small value (~10-10 or smaller); equilibrium lies WAY to the left

  47. Ksp Equation • Precipitate (solid) = reactant • Ions (aqueous) = products • BALANCE (welcome back, coefficients) • Write Ksp the same way we’ve written all other K’s (prod/react) – make sure to OMIT SOLIDS! • Coefficients become exponents!

  48. Examples: • Saturated solution of AgCl • Saturated solution of Ag2CO3 AgCl (s) ⇄ Ag+ (aq) + Cl- (aq) Ksp = [Ag+][Cl-] Ag2CO3 (s) ⇄ 2Ag+ (aq) + CO32- (aq) Ksp = [Ag+]2[CO32-]

  49. Example Calc of Ksp • Copper (I) bromide has a measured solubility of 2.0 x 10-4 mol/L at 25 C. Calculate Ksp value CuBr(s)  Cu+ (aq) + Br- (aq) Ksp = [Cu+][Br-] • Initially, the solution contains no Cu+ or Br-, so initial concentrations are 0 • The equilibrium concentrations can be found from the measured solubility of CuBr: CuBr(s)  Cu+ (aq) + Br- (aq) • Ksp = [2.0x 10-4][2.0x10-4] = 4.0 x 10-8

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