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Applications of Aqueous Equilibria

Applications of Aqueous Equilibria. “Common Ions”. When we dissolve acetic acid in water, the following equilibrium is established: CH 3 COOH  CH 3 COO - + H + If sodium acetate were dissolved in solution, which way would this equilibrium shift?. “Common Ions”.

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Applications of Aqueous Equilibria

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  1. Applications of Aqueous Equilibria

  2. “Common Ions” • When we dissolve acetic acid in water, the following equilibrium is established: • CH3COOH  CH3COO- + H+ • If sodium acetate were dissolved in solution, which way would this equilibrium shift?

  3. “Common Ions” • When we dissolve acetic acid in water, the following equilibrium is established: • CH3COOH  CH3COO- + H+ • If sodium acetate were dissolved in solution, which way would this equilibrium shift? • Adding acetate ions from a strong electrolyte would shift this equilibrium left (Le Chatelier’s principle).

  4. Common Ion Effect • Whenever a weak electrolyte (acetic acid) and a strong electrolyte (sodium acetate) share a common ion, the weak electrolyte ionizes less than it would if it were alone. (Le Chatelier’s) • This is called the common-ion effect.

  5. Steps for Common-Ion Problems • 1. Consider which solutes are strong electrolyte and weak electrolytes. • 2. Identify the important equilibrium (weak) that is the source of H+ and therefore determines pH. • 3. Create an ICE chart using the equilibrium and strong electrolyte concentrations. • 4. Use the equilibrium constant expression to calculuate [H+] and pH.

  6. What are the sources of Ag+1? • What is the concentration of silver and chromate in a solution with silver chromate in 0.1 M silver nitrate Ag2CrO4(s)  2Ag+1 + CrO4-2 I 0.1 0 C +2x +x E 0.1 + 2x x Ksp = 9.0 x 10-12 = [Ag+1]2• [CrO4-2] = 9.0 x 10-12 = [.1 +2x]2• [x] 9.0 x 10-12 = 0.12 • x [CrO4-2] = x = 9.0 x 10-10 M [Ag+1] = .1 + 2x = .1M [Ag2CrO4] = 9.0 x 10-10 Why is this a plus sign?

  7. Sample Problem • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution? • Ka for acetic acid = 1.8 x 10-5 pH = 4.74

  8. Sample Problem • Calculate the fluoride concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl • Ka for HF = 6.8 x 10-4. • [F-] = 1.2x10-3M • pH = 1.00

  9. Buffers Solutions that resist pH change when small amounts of acid or base are added. Two types weak acid and its salt weak base and its salt HX(aq) + H2O(l) H+(aq) + X-(aq) Add OH- Add H+ shift to right shift to left Based on the common ion effect.

  10. Buffers buffered unbuffered pH ml HCl added

  11. Buffers and blood Control of blood pH Oxygen is transported primarily by hemoglobin in the red blood cells. CO2 is transported both in plasma and the red blood cells. CO2 (aq) + H2O H2CO3 (aq) H+(aq) + HCO3-(aq)

  12. Buffers • Composition and Action of Buffered Solutions • The Ka expression is • A buffer resists a change in pH when a small amount of OH- or H+ is added.

  13. Buffered Solutions • Addition of Strong Acids or Bases to Buffers • With the concentrations of HX and X- (note the change in volume of solution) we can calculate the pH from the Henderson-Hasselbalch equation Equation not necessary, since you know Ka = [X-] [H+] [HX]

  14. Buffers • Action of Buffered Solutions

  15. Buffers Buffer Capacity and pH Buffer capacity is the amount of acid or base neutralized by the buffer before there is a significant change in pH. The greater the amounts(molarity) of the conjugate acid-base pair, the greater the buffer capacity. The pH of the buffer depends on Ka

  16. Buffered Solutions Addition of Strong Acids or Bases to Buffers

  17. Buffer Addition of Strong Acids or Bases to Buffers Break the calculation into two parts: stoichiometric and equilibrium. The amount of strong acid or base added results in a neutralization reaction: X- + H+ HX + H2O HX + OH- X- + H2O. By knowing how much H+ or OH- was added (stoichiometry) we know how much HX or X- is formed.

  18. Buffers The final [HX] and [X-] after the neutralization reaction are used as the initial concentrations for the equilibrium reaction. HX H+ X- Initial conc., M Change, DM Eq. Conc., M Then the equilibrium constant expression is used to find [H+] and pH = - log [H+] Ka = [H+] [X-] [HX]

  19. Buffer Example Determine the initial pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 oC. Ka = 6.5 x 10-5 HBz(aq) + H2O(l) H+(aq) + Bz-(aq) HBz H+ Bz- Initial conc., M 0.10 0.00 0.20 Change, DM -x+x +x Eq. Conc., M 0.10 - x x 0. 20 + x

  20. x (0. 20 + x) 0.10 - x Buffer Example Solve the equilibrium equation in terms of x Ka = 6.5 x 10-5 = x = (6.5 x 10-5 )(0.10) / (0. 20) (assuming x<<0.10) = 3.2 x 10-5 M H+ pH = - log (3.2 x 10-5 M) = 4.5 initial pH

  21. Buffer Example Determine the pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 oC after 0.05 moles of HCl is added. First, find the concentrations of the HBz and Bz- after HCl is added. HBz H+ Bz- Initial conc., M 0.15 0.00 0.15 Change, DM -x+x +x Eq. Conc., M 0.15 - x x 0. 15 + x The 0.05 mol HCl reacts completely with 0.05 mol Bz-(aq) to form 0.05 mol HBz(aq) Then equilibrium will be re-established based on the new initial concentrations of 0.15 M HBz(aq) and 0.15 M Bz-(aq).

  22. x (0. 15 + x) 0.15 - x Buffer Example Solve the equilibrium equation in terms of x Ka = 6.5 x 10-5 = x = (6.5 x 10-5 )(0.15) / (0. 15) (assuming x<<0.15) = 6.5 x 10-5 M H+ pH = - log (6.5 x 10-5 M) = 4.2 after 0.05 mole HCl added

  23. Buffer Example First, find the concentrations of HBz and Bz- after NaOH is added. Determine the pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 oC after 0.05 moles of NaOH is added. HBz H+ Bz- Initial conc., M 0.05 0.00 0.25 Change, DM -x+x +x Eq. Conc., M 0.05 - x x 0. 25 + x The 0.05 mol NaOH reacts completely with 0.05 mol HBz(aq) to form 0.05 mol Bz-(aq) Then equilibrium will be re-established based on the new initial concentrations of 0.05 M HBz(aq) and 0.25 M Bz-(aq).

  24. x (0. 25 + x) 0.05 - x Buffer Example Solve the equilibrium equation in terms of x Ka = 6.5 x 10-5 = x = (6.5 x 10-5 )(0.05) / (0. 25) (assuming x<<0.05) = 1.3 x 10-5 M H+ pH = - log (1.3 x 10-5 M) = 4.9 after 0.05 mole NaOH added

  25. Acid-Base Titrations Strong Acid-Base Titrations • The plot of pH versus volume during a titration is a titration curve.

  26. Remember the Acid-Base Titration Curves? Buffer Zone HC2H3O2 HCl

  27. Acetic Acid/Acetate Ion Buffer Lab For this experiment, you will prepare a buffer that contains acetic acid and its conjugate base, the acetate ions. The equilibrium equation for the reaction is shown below: HC2H3O2(aq) + H2O(l) <=> H+ (aq)+ C2H3O2- (aq) The equilibrium expression for this reaction, Ka, has a value of 1.8 x 10-5 at 25ºC.

  28. Acetic Acid/Acetate Ion Buffer Lab • The ratio between the molarity of the acetate ions to the molarity of the acetic acid in your buffer must equal the ratio between the Ka value and 10- assigned pH. • This ratio should be reduced , so that either the [HC2H3O2] or [C2H3O2- ] has a concentration of 0.10 M, and the concentration of the other component must fall within a range from 0.10 M to 1.00 M. • Complete the calculations only that are needed to prepare 100.0 mL of your assigned buffer solution that has these specific concentrations. Can you predict the final pH when a strong acid or base is added to the buffer solution?

  29. Acetic Acid/Acetate Ion Buffer Lab

  30. Acetic Acid/Acetate Ion Buffer Lab

  31. Making a Buffer Calculations You want to prepare 500.0 mL of a buffer with a pH = 10.00, with both the acid and conjugate base having molarities between 0.10 M to 1.00 M. You may choose from any of the acids listed below: You must select an acid with a Ka value close to 10- assigned pH. The only two options are ammonium or the hydrogen carbonate ions.

  32. Making a Buffer Calculations

  33. Making a Buffer Calculations Place ~250 mL of distilled water in a 500 mL volumetric flask. Add 2.68 g NH4Cl and dissolve. Then mix in 18.9 mL of 14.8 M NH3. Fill with distilled water to the 500 mL mark on the flask. If there is no concentrated NH3 available, the NH3 can be produced by neutralizing additional NH4Cl with 1.00 M NaOH. Dissolve 19.06 g NH4Cl (2.68 g + 16.38 g) in 280. mL of 1.00 M NaOH. Then dilute with distilled water and fill to the 500 mL mark on the flask.

  34. Making a Buffer Calculations Place ~250 mL of distilled water in a 500 mL volumetric flask. Add 2.68 g NH4Cl and dissolve. Then mix in 18.9 mL of 14.8 M NH3. Fill with distilled water to the 500 mL mark on the flask. If there is no NH4Cl available, the NH4+ can be produced by neutralizing additional NH3 with 1.00 M HCl. Place ~250 mL distilled water in volumetric flask. Add 50.0 mL of 1.00 M HCl and mix. Then add 22.3 mL of concentrated NH3 (18.9 mL + 3.4 mL). Mix and fill with distilled water to the 500 mL mark on the flask.

  35. Making a Buffer Calculations Prepare 500. mL of the buffer that has [CO32-] = 0.100 M and [HCO31-] = 0.179 M.

  36. Making a Buffer Calculations Prepare 500. mL of the buffer that has [CO32-] = 0.100 M and [HCO31-] = 0.179 M. Place ~250 mL distilled water in a 500 mL volumetric flask. Add 5.30 g Na2CO3 and 7.52 g NaHCO3 and dissolve. Fill with distilled water to the 500 mL mark on the flask.

  37. Acid-Base Titrations Strong Acid-Base Titrations • The plot of pH versus volume during a titration is a titration curve.

  38. Indicator examples • Acid-base indicators are weak acids that undergo a color change at a known pH. pH phenolphthalein

  39. Indicator examples Select the indicator that undergoes a color change closest to the pH at the equivalence point, where all of the acid has been neutralized by the base. bromthymol blue methyl red

  40. Acid-Base Titrations • Strong Acid-Base Titrations • Consider adding a strong base (e.g. NaOH) to a solution of a strong acid (e.g. HCl). • Before any base is added, the pH is given by the strong acid solution. Therefore, pH < 7. • When base is added, before the equivalence point, the pH is given by the amount of strong acid in excess. Therefore, pH < 7. • At equivalence point, the amount of base added is stoichiometrically equivalent to the amount of acid originally present. Therefore, the pH is determined by the salt solution. Therefore, pH = 7.

  41. Acid-Base Titrations • Strong Acid-Base Titrations • Consider adding a strong base (e.g. NaOH) to a solution of a strong acid (e.g. HCl).

  42. Acid-Base Titrations • Strong Acid-Base Titrations • We know the pH at equivalent point is 7.00. • To detect the equivalent point, we use an indicator that changes color somewhere near 7.00. • Usually, we use phenolphthalein that changes color between pH 8.3 to 10.0. • In acid, phenolphthalein is colorless. • As NaOH is added, there is a slight pink color at the addition point. • When the flask is swirled and the reagents mixed, the pink color disappears. • At the end point, the solution is light pink. • If more base is added, the solution turns darker pink.

  43. Acid-Base Titrations • Strong Acid-Base Titrations • The equivalence point in a titration is the point at which the acid and base are present in stoichiometric quantities. • The end point in a titration is the observed point. • The difference between equivalence point and end point is called the titration error. • The shape of a strong base-strong acid titration curve is very similar to a strong acid-strong base titration curve.

  44. Acid-Base Titrations Strong Acid-Base Titrations

  45. Acid-Base Titrations • Strong Acid-Base Titrations • Initially, the strong base is in excess, so the pH > 7. • As acid is added, the pH decreases but is still greater than 7. • At equivalence point, the pH is given by the salt solution (i.e. pH = 7). • After equivalence point, the pH is given by the strong acid in excess, so pH < 7.

  46. Acid-Base Titrations • Weak Acid-Strong Base Titrations • Consider the titration of acetic acid, HC2H3O2 and NaOH. • Before any base is added, the solution contains only weak acid. Therefore, pH is given by the equilibrium calculation. • As strong base is added, the strong base consumes a stoichiometric quantity of weak acid: • HC2H3O2(aq) + NaOH(aq)  C2H3O2-(aq) + H2O(l)

  47. Acid-Base Titrations Weak Acid-Strong Base Titrations

  48. Acid-Base Titrations • Weak Acid-Strong Base Titrations • There is an excess of acetic acid before the equivalence point. • Therefore, we have a mixture of weak acid and its conjugate base. • The pH is given by the buffer calculation. • First the amount of C2H3O2- generated is calculated, as well as the amount of HC2H3O2consumed. (Stoichiometry.) • Then the pH is calculated using equilibrium conditions. (Henderson-Hasselbalch.)

  49. Acid-Base Titrations • Weak Acid-Strong Base Titrations • At the equivalence point, all the acetic acid has been consumed and all the NaOH has been consumed. However, C2H3O2- has been generated. • Therefore, the pH is given by the C2H3O2- solution. • This means pH > 7. • More importantly, pH  7 for a weak acid-strong base titration. • After the equivalence point, the pH is given by the strong base in excess.

  50. Acid-Base Titrations • Weak Acid-Strong Base Titrations • For a strong acid-strong base titration, the pH begins at less than 7 and gradually increases as base is added. • Near the equivalence point, the pH increases dramatically. • For a weak acid-strong base titration, the initial pH rise is more steep than the strong acid-strong base case. • However, then there is a leveling off due to buffer effects.

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