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Chemistry 100 - Chapter 17

Chemistry 100 - Chapter 17 . Applications of Aqueous Equilibria. The Common Ion Effect. Add sodium format e (HCOONa) to a solution of formic acid (HCOOH) that has already established equilibrium? HCOOH (aq) ⇄ H + (aq) + HCOO - (aq) weak acid

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Chemistry 100 - Chapter 17

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  1. Chemistry 100 - Chapter 17 Applications of Aqueous Equilibria

  2. The Common Ion Effect Add sodium formate (HCOONa) to a solution of formic acid (HCOOH) that has already established equilibrium? HCOOH (aq) ⇄ H+ (aq) + HCOO- (aq) weak acid HCOONa (aq)  HCOO- (aq) + Na+ (aq) strong electrolyte

  3. Another Example of the Common Ion Effect What would happen if we added HCOOH to a solution of a strong acid? HCl (aq)  H+ (aq) + Cl- (aq) strong acid HCOOH (aq) ⇄ H+ (aq) + HCOO- (aq) The ionization of the weak acid would be supressed in the presence of the strong acid! By Le Chatelier’s Principle, the  value of the weak acid is decreased!

  4. How would we calculate the pH of these solutions?

  5. Define pKa = -log (Ka ) note pH = -log [H+]

  6. The Buffer Equation • Substituting and rearranging

  7. The Generalized Buffer Equation The solution pH is determined by the ratio of the weak acid to the conjugate base at equilibrium. Henderson-Hasselbalch equation

  8. The Definition of a Buffer Buffer  a reasonably concentrated solution of a weak acid and its conjugate base. Buffer solutions resists pH changes when additional strong acid or strong base are added to the solutions. Note: The Henderson-Hasselbalch equation is really only valid for pH ranges near the pKa of the weak acid!

  9. How Do We Use the Buffer (the H-H) Equation? The pH of the buffer is determined by the concentration ratio of weak acid to conjugate base at equilibrium. How different are the equilibrium concentrations of weak acid/conjugate base from the initial concentrations?

  10. Buffer  CH3COONa (aq) and CH3COOH (aq)) CH3COOH (aq) ⇄ CH3COO- (aq) + H+ (aq) The Equilibrium Data Table

  11. What if the original concentrations of acid and base ([A] and [B], respectively) are much larger than x (i.e., the  value of the weak acid is very small)? According to the Henderson-Hasselbalch Equation, the pH of the solution is calculated as follows

  12. The pH of the solution changes very little after adding strong acid or base (it is buffered) The pH of the solution will be almost entirely due to the original concentrations of acid and base!!

  13. Examples of Buffer Calculations • How would we prepare a buffer solution of a specified pH? How do we calculate the pH of a buffer solution?

  14. The pH of a Buffer Solution • Major task • obtain the ratio of the concentrations of conjugate base to weak acid! • Using the Ka of the appropriate acid, the pH of the solution is obtained from the Henderson-Hasselbalch equation.

  15. Preparing a Buffer Solution of a Specific pH The first step in the process is to choose a suitable weak acid. The H-H equation is really only valid in a pH range near the pKa of the weak acid; Second Step  Calculate the required ratio of conjugate base to weak acid from the Henderson-Hasselbalch Equation

  16. From the previous problem, there are a number of concentrations where the ratio of the conjugate base to the weak acid will be acceptable [CH3CH2COONa] = 0.13 M; [CH3CH2COOH] = 0.10 M [CH3CH2COONa] = 0.065 M;[CH3CH2COOH] = 0.050 M [CH3CH2COONa] = 0.39 M; [CH3CH2COOH] = 0.30 M [CH3CH2COONa] = 0.65 M; [CH3CH2COOH] = 0.50 M [CH3CH2COONa] = 1.3 M; [CH3CH2COOH] = 1.0 M

  17. Buffer Capacity Buffer Capacity  refers to the amount of strong acid/base that can be added to the buffer solution Buffer capacity is directly related to the concentrations of the weak acid and conjugate base in the buffer solution

  18. Choosing Concentrations for a High Buffer Capacity [CH3CH2COONa] = 0.13 M; [CH3CH2COOH] = 0.10 M NOT ACCEPTABLE [CH3CH2COONa] = 0.065 M; [CH3CH2COOH] = 0.050 M NOT ACCEPTABLE [CH3CH2COONa] = 0.39 M; [CH3CH2COOH] = 0.30 M REASONABLE [CH3CH2COONa] = 0.65 M; [CH3CH2COOH] = 0.50 M GOOD CHOICE [CH3CH2COONa] = 1.3 M; [CH3CH2COOH] = 1.0 M GOOD CHOICE

  19. Adding Acid to Buffer Solutions What happens when we add strong acid solutions to the buffer? [H+] increased and the basic part of the buffer goes to work H+ (aq) + CH3COO- (aq)  CH3COOH (aq) This is the reverse of the usual acid dissociation equilibrium, hence, the reaction essentially goes to completion

  20. Adding Base to Buffer Solutions What happens when we add the base to the buffer? The [OH-] increases and the acid part of the buffer goes to work OH- (aq) + CH3COOH (aq)  CH3COO- (aq) + H2O (l) This is the reverse of the usual base dissociation equilibrium, hence, the reaction essentially goes to completion

  21. Acid-base Titration Curves • Chapter 4 – acid-base titrations • A known (standard) basic solution is slowly added to an unknown acid solution • What if we monitored the pH of the solution as a function of added titrant  acid base titration curve is generated • Three cases to consider

  22. Strong Acid/strong Base • When n(H+) = n(OH-), we are at the equivalence point of the titration • Product of reaction is a strong acid/strong base salt. The pH at the equivalence point is 7.00. HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l) Net ionic H+ (aq) + OH- (aq)  H2O (l)

  23. The Titration Curve

  24. Weak Base /Strong Acid • Equivalence point, n(NH3) = n(H+). • pH of the solution < 7.00. Determined by the ionization of the conjugate acid NH4+ (aq) ⇌ NH3 (aq) + H+ (aq) NH3 (aq) + HCl (aq)  NH4Cl (aq) Net ionic NH3 (aq) + H+ (aq)  NH4+ (aq)

  25. The Titration Curve 11.00 pH 9.00 7.00 Eq. point 5.00 3.00 V (strong acid) / mL

  26. Weak Acid/Strong base. NaOH (aq) + CH3COOH (aq)  CH3COONa (aq) + H2O (l) Net ionic OH- (aq) + CH3COOH (aq)  CH3COO- (aq) + H2O (l)

  27. The Titration Curve

  28. Equivalence point when n(CH3COOH) = n(OH-), • At the equivalence point • pH of the solution is determined by the ionization of the conjugate base of the weak acid CH3COO- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH- (aq) • Therefore, the pH at the equivalence point is > 7.00!

  29. Comparison Between Strong/Weak Acid Titrations

  30. Indicators Also usually coloured, but the colour is different than for the acid form of the indicator. Usually coloured Indicators are used to detect the endpoint of the acid-base titration. Indicators are weak acids. Their ionization can be represented by the following reaction. HIn (aq) ⇌ H+ (aq) + In- (aq)

  31. Choose the indicator whose transition range (i.e., the pH range where it changes colour) matches the steep part of the titration curve Note: we can use the following ratios as a guide. [HIn] / [In-] > 10  acid colour dominates. [In-] / [HIn] > 10  base colour dominates.

  32. Strong Acid/Strong Base  steep part of titration curve pH 4-10. A number of indicators change colour in this range Weak Acid/Strong Base  steep part of titration curve pH >7.0. The indicator colour change must occur in this range Strong Acid/Weak Base  steep part of titration curve pH <7.0. The indicator colour change must occur in this range

  33. Indicators in Titrations

  34. Solubility Equilibria Examine the following systems AgCl (s) ⇄ Ag+ (aq) + Cl- (aq) BaF2 (s) ⇄ Ba2+ (aq) + 2 F- (aq) Using the principles of chemical equilibrium, we write the equilibrium constant expressions as follows

  35. The Definition of the Ksp Don’t confuse the solubility of the solid with the Ksp. These quantities are related, but they are not the same. Ksp the solubility product constant. The product of the molar concentrations of the dissolved ions in equilibrium with the undissolved solid at a particular temperature.

  36. Examples of Ksp Calculations • Calculate the solubility of a solid in the presence of a common ion. • Calculate the solubility of a solid as a function of the pH of the solution. Calculate the solubility of a sparingly soluble solid in water.

  37. Solubility of Sparingly Soluble Solids in Water AgCl (s) ⇄ Ag+ (aq) + Cl- (aq) • We approach this using the principles of chemical equilibrium. • equilibrium data table, establish and solve for our unknown quantity!

  38. The Common Ion Effect Equilibrium is displaced to the left by LeChatelier’s principle (an example of the common ion effect). What about the solubility of AgCl in solution containing NaCl (aq)? AgCl (s) ⇄ Ag+ (aq) + Cl- (aq) NaCl (aq) Na+ (aq) + Cl- (aq) AgCl (s) ⇄ Ag+ (aq) + Cl- (aq)

  39. Solubility and pH What happens when we try to dissolve a solid like Mn(OH)2 in solutions of varying pH? We first calculate the pH of the saturated solution of the Mn(OH)2.

  40. Equilibrium is displaced to the right by LeChatelier’s principle. Suppose that we try to dissolve Mn(OH)2 in an acidic solution Mn(OH)2 (s) ⇄ Mn2+ (aq) + 2 OH- (aq) H+ (aq) + OH- (aq)  H2O (l) Mn(OH)2 (s) ⇄ Mn2+ (aq) + 2 OH- (aq)

  41. Solubility of CaF2 vs. pH

  42. Equilibrium is displaced to the left by LeChatelier’s principle. • Increase the [OH-] in the solution • looking at an example of the common ion effect Mn(OH)2 (s) ⇄ Mn2+ (aq) + 2 OH- (aq) NaOH(aq)  Na+ (aq) + OH- (aq) Mn(OH)2 (s) ⇄ Mn2+ (aq) + 2 OH- (aq)

  43. Any solid that produces a moderately basic ion on dissociation (e.g., CaF2, MgCO3). • solubility will decrease as the pH is increased (i.e., the [OH-] in the solution is increased). • solubility will increase as the pH is decreased (i.e., the [H+] in the solution is increased).

  44. What about solids whose anions do not exhibit basic tendencies? • PbCl2, the Cl- has no tendency to react with added acid. • solubility of PbCl2, does not depend on the solution pH!

  45. Predicting Precipitation: the Qsp Value Let’s examine the following equilibrium system. AgCl (s) ⇄Ag+ (aq) + Cl- (aq) Let’s say that we two solutions so that they would have the following concentrations. [NaCl]o = 1.0x 10-5 M  [Cl-]o = 1.0 x 10-5 M [AgNO3] = 1.0 x 10-6 M  [Ag+]o = 1.0 x 10-6 M Would we be able to predict whether or not a precipitate will occur?

  46. The QspValue We now examine the magnitude of the solubility product quotient (Qsp) with respect to the Ksp. Qsp < Ksp no precipitate will form Qsp > Ksp a precipitate will form Qsp = Ksp saturated solution. • Define the solubility product quotient  Qsp.

  47. The Formation of Metal Complexes We see that a number of metal anions can act as Lewis acids; therefore, these ions can react strongly with Lewis bases and form complex ions. AgCl (s) ⇄ Ag+ (aq) + Cl- (aq) Ag+ (aq) + 2 NH3 (aq) ⇄ Ag(NH3)2+ (aq) For the NH3 (aq) to increase the solubility of the metal salt, the NH3 (aq) must be a stronger Lewis base than the water molecules that it displaces.

  48. Solubility of many metal containing compounds increases markedly in the presence of suitable complexing species • NH3 (aq) • OH- (aq) • CN-(aq)

  49. is known as the formation constant for the complex ion. The species Ag(NH3)2+ (aq) is known as a complex ion. The equilibrium constant for the second reaction, Kf

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