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Applications of Aqueous Equilibria

Applications of Aqueous Equilibria

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Applications of Aqueous Equilibria

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  1. Applications of Aqueous Equilibria Chapter 17 Pg. 719

  2. Common Ion Effect • When a solution of a weak electrolyte is altered by adding one of its ions from another source, the ionization of the weak electrolyte is suppressed. This is referred to as the common ion effect. • A shift in equilibrium occurs because of the addition of an ion already involved in the equilibrium reaction.

  3. Effect on pH • Solutions that contain a weak acid plus a salt of a weak acid are always less acidic than solutions that contain the same concentration of the weak acid alone. (pH increases). • Solutions that contain a weak base plus the salt of a weak base are always less basic than solutions that contain the same concentration of the weak base alone. (pH decreases)

  4. Buffered Solutions • A buffered solution is one that resists a change in the pH when either hydroxide or hydronium ions are added. • A buffered solution may contain a weak acid and its salt (ex. HF and NaF) or a weak base and its salt (ex. NH3 and NH4Cl)

  5. pH of a Buffered Solution • Calculate the pH of a buffered solution containing 0.75 M lactic acid, HC3H5O3 (Ka=1.4 x 10-4) and 0.25 M sodium lactate, NaC3H5O3.

  6. Henderson-Hasselbalch Equation • pH = pKa + log [conjugate base/acid] • Using the information from the previous problem, calculate the pH using this equation.

  7. Selecting Buffers • When choosing the buffering components for a specific application, the ratio of [A-]/[HA] should be as close to 1 as possible. • The pKa of the acid to be used in the buffer should be as close as possible to the desired pH. • As long as the ratio of [A-]/[HA] remains the same, the pH will not be affected. • Increasing the concentrations of [A-] and [HA] determines the buffering capacity (the amount of H+ or OH- that can be absorbed without a significant change in pH.

  8. A chemist needs a solution buffered at pH 4.30 and can choose from the following acids (and their sodium salts): • Chloroacetic acid (Ka = 1.35 x 10-3) • Propanoic acid (Ka = 1.3 x 10-5) • Benzoic acid (Ka = 6.4 x 10-5) • Hypochlorous acid (Ka = 3.5 x 10-8) Calculate the ratio [HA]/[A-] required for each system to yield a pH of 4.30. Which system will work best?

  9. Solubility Equilibria • Calcium fluoride dissolves in water as follows: • CaF2 Ca2+ + 2F- • When the salt first dissolves, no ions are present. However, as the dissolution proceeds, the concentrations of the ions increase making it more likely that these ions will collide and reform the solid phase. • Ultimately, dynamic equilibrium is achieved. • CaF2 < -- > Ca2+ + 2F- At this point no more solid will dissolve and the solution is said to be saturated.

  10. Solubility Product Constant • An equilibrium expression can be written for the process: • Ksp = [Ca2+][F-]2 • The Ksp is called the solubility product constant or simply the solubility product. • CaF2 is a pure solid and is not included in the expression.

  11. Relative Solubilities • The greater the Ksp for a salt, the greater the solubility (if the salts being compared have the same number of ions). • The solubility of a salt is lowered if the solution already contains ions common to the solid (common ion effect).

  12. pH and Solubility • The pH can affect a salt’s solubility. • Mg(OH)2 < -- > Mg2+ + 2OH- • The addition of OH- (an increase in pH) will decrease the solubility (equilibrium shifts to the left). • The addition of H+ (a decrease in pH will increase the solubility (equilibrium shifts to the right)

  13. To precipitate or not precipitate • Consider the following expression: • Ksp = [Ca2+][F-] • If we add a solution containing Ca2+ to a solution containing F-, a precipitate may or may not form depending on the concentrations of the solutions. • To predict whether a precipitate will form, consider the relationship between Q and Ksp • If Q>Ksp , precipitation occurs and will continue until the concentrations are reduced to the point that they satisfy Ksp • If Q<Ksp , precipitation will not occur.

  14. Titrations and pH Curves • A pH curve is the plot of the pH of the solution being analyzed as a function of the amount of titrant added. • The titrant is the solution of known concentration. • The equivalence point occurs when the moles of acid are equal to the moles of base and is often signaled by a color change of an indicator.

  15. Titration Curves • A titration curve is a plot of pH vs. amount (volume) of acid or base added. • Used to find the molarity of an unknown solution. • Endpoint-the point at which the indicator changes color • Equivalence Point-point at which chemically equivalent amounts of acid and base have reacted. • Ideally, the endpoint and equivalence point should coincide.

  16. Acid-Base Indicators • Indicators are organic dyes whose color depends on the [H3O+] or pH. • Common examples include litmus and phenolphthalein. (see p. 715 for others) • Indicators typically change color over a range of 1.5-2.0 pH units.

  17. Interpreting Color Changes in Indicators • Bromthymol Blue HIn + H2O ↔ H3O+ + In- Ka = 1.0 x 10-7 yellow blue Color #1 Color #2 pH<6 pH>8 • General rule: [In-]/[HIn] > 10 color #2 [In-]/[HIn] < .10 color #1

  18. Universal Indicators • Universal indicators are a mixture of several indicators. • These indicators display a continuous range of colors over a wide range of pH values.

  19. Strong Acid-Strong Base Titrations • The net ionic equation for a strong acid-strong base titration is: H+ + OH- H2O • To determine the concentration of H+ at any given point, the amount of H+ that remains must be divided by the total volume of the solution. • Since titrations usually involve small quantities, we will use the millimole as our unit. • Number of millimoles = volume in mL x molarity

  20. Case Study-Strong Acid-Strong Base Titration • Situation: 50.0 mL of 0.200 M HNO3 is being titrated with 0.100 M NaOH. • We will calculate the pH of the solution at various points during the course of the titration. • We will then draw a pH curve of the data.

  21. A. No NaOH has been added. • HNO3 H+ + NO3- • 0.200 M 0.200 M + 0.200M • HNO3 is a strong acid so it completely ionizes. • pH = -log(0.200) = 0.699

  22. Additional Steps • B: 10.0 mL of 0.100 M NaOH has been added. • C: 20.0 mL of 0.100 M NaOH has been added. • D: 50.0 mL of 0.100 M NaOH has been added. • E: 100.0 mL of 0.100 M NaOH has been added. • F: 150.0 mL of 0.100 M NaOH has been added. • G: 200.0 mL of 0.100 M NaOH has been added.

  23. Characteristics of the pH curve for the titration of a strong acid with a strong base. • Before the equivalence point, the [H+] can be calculated by dividing the number of millimoles of H+ remaining by the total volume of the solution. • At the equivalence point, the pH = 7.00 • After the equivalence point, the [OH-] can be calculated by dividing the number of millimoles of excess OH- by the total volume of the solution. The H+ (and pH) can be determined using Kw. • The titration of a strong base with a strong acid requires similar reasoning except it will be reversed.

  24. Titration Curve for the reaction of a strong acid and strong base.

  25. Affect of procedural errors during titrations How would the following lab errors affect the calculated values of the molarity of the NaOH? • If the buret was rinsed with distilled water immediately prior to the titration with the NaOH. • If bromophenol blue was used as the indicator instead of phenolphthalein.

  26. Titrations of Weak Acids and Strong Bases • Calculation of the pH curve for a titration of a weak acid and strong base really amounts to a series of buffer problems. • Reminder: Even though the acid is weak, it reacts essentially to completion with the strong base.

  27. Case Study: Weak Acid and Strong Base Titration • Consider the titration of 50.0 mL of 0.10 M HC2H3O2 (Ka= - 1.8 x 10-5) with 0.10 M NaOH. Calculate the pH at the following points in the titration: A: No NaOH has been added. B: 10.0 mL of 0.10 M NaOH has been added. C: 25.0 mL of 0.10 M NaOH has been added. D: 40.0 mL of 0.10 M NaOH has been added. E: 50.0 mL of 0.10 M NaOH has been added. F: 60.0 mL of 0.10 M NaOH has been added. G: 75.0 mL of 0.10 M NaOH has been added.

  28. Conclusions • It is the amount of acid, not its strength, that determines the equivalence point. • The pH value at the equivalence point is affected by the strength. (The stronger the acid, the lower the pH at the equivalence point). (see p. 707) • For the titration of a weak acid and strong base, the equivalence point will be greater than 7. • At the halfway point in the titration, the pH = pKa.

  29. pH curve for strong base and weak acid.

  30. Titrations of Weak Bases and Strong Acids • See the example on page709-710. • For the titration of a weak base and strong acid, the equivalence point will be lower than 7. • At the halfway point in the titration, the pOH = pKb

  31. pH curve for titration of strong acid and weak base.