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Applications of Aqueous Equilibria

Applications of Aqueous Equilibria. Solutions of Acids or Bases Containing a Common Ion. Solution containing weak acid HA and its salt NaA. Salt dissolves in water and breaks up completely into its ions-it is a strong electrolyte NaA(s)  Na+(aq) + A-(aq). Common Ion Effect.

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Applications of Aqueous Equilibria

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  1. Applications of Aqueous Equilibria

  2. Solutions of Acids or Bases Containing a Common Ion

  3. Solution containing weak acid HA and its salt NaA • Salt dissolves in water and breaks up completely into its ions-it is a strong electrolyte • NaA(s)  Na+(aq) + A-(aq)

  4. Common Ion Effect • When AgNO3 is added to a saturated solution of AgCl, it is often described as a source of a common ion, the Ag+ ion. • By definition, a common ion is an ion that enters the solution from two different sources. • Solutions to which both NaCl and AgCl have been added also contain a common ion; in this case, the Cl- ion. • There is an effect of common ions on solubility product equilibria.

  5. The common-ion effect can be understood by considering the following question: What happens to the solubility of AgCl when we dissolve this salt in a solution that is already 0.10 M NaCl? As a rule, we can assume that salts dissociate into their ions when they dissolve. A 0.10 M NaCl solution therefore contains 0.10 moles of the Cl- ion per liter of solution. Because the Cl- ion is one of the products of the solubility equilibrium, LeChatelier's principle leads us to expect that AgCl will be even less soluble in an 0.10 M Cl- solution than it is in pure water.

  6. Calculate the solubility of AgCl in 0.10 M NaCl. In pure water:Cs = 1.3 x 10-5M In 0.10 M NaCl: Cs = 1.8 x 10-9M These calculations show how the common-ion effect can be used to make an "insoluble" salt even less soluble in water.

  7. The common ion effect can be applied to other equilibria, as well. Consider what happens when we add a generic acid (HA) to water. We now have two sources of a common ion ? the H3O+ ion. HA(aq) + H2O(l) !H3O+(aq) + A-(aq) 2 H2O(l)! H3O+(aq) + OH-(aq) Thus, it isn't surprising that adding an acid to water decreases the concentration of the OH- ion in much the same way that adding another source of the Ag+ ion to a saturated solution of AgCl decreases the concentration of the Cl- ion.

  8. The solubility of a solid is lowered if the solution already contains ions common to the solid Dissolving silver chloride in a solution containing silver ions Dissolving silver chloride in a solution containing chloride ions The common-ion effect can also be used to prevent a salt from precipitating from solution. Instead of adding a source of a common ion, we add a reagent that removes the common ion from solution.

  9. Ksp (Solubility Product Constant, Solubility Product) Ksp = [Ca2+][F-]2 Experimentally determined solubility of an ionic solid can by used to calculate its Ksp value The solubility of an ionic solid can be calculated if its Ksp value is known Relative Solubilities IF the salts being compared produce the same number of ions in solution, Ksp can be used to directly compare solubility NaCl(s), KF(s) Ksp = [cation][anion] = x2 IF the salts being compared produce different numbers of ions, Ksp cannot be directly compared Ag2S(s) Ksp = [2x]2[x] Bi2S3(s) = [2x]2[3x]3

  10. Common Ion-ion produced by both acid and its salt • Ion provided in solution by an aqueous acid (or base) as well as a salt • HF(aq) and NaF (F-in common) • HF(aq) H+(aq) + F-(aq) • NaF(s) (in water) Na+(aq) + F-(aq) • Excess F-added by NaF • Equilibrium shifts away from added component. • Fewer H+ ions present, making solution less acidic • pH is higher than expected. • NH4OH and NH4Cl (NH4+in common) • NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) • NH4Cl(s) (in water) NH4+(aq) + Cl-(aq) • Equilibrium shifts to the left. • pH of solution decreases due to decrease in OH-conc. • Common ion effect-shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction

  11. Equilibrium Calculations • Consider initial concentration of ion from salt when calculating values for H+ and OH- • Calculate the pH of a solution that contains 0.10 M HC2H3O2 and 0.050 M NaC2H3O2. (The pH of 0.10 M HC2H3O2 is 2.9. The addition of the common ion would shift the equilibrium to the left. As a result the [H3O+] would decrease and the pH would rise.)

  12. Ka = [H3O+][C2H3O2-]/[HC2H3O2] 1.8 x 10-5 = x(0.050 + x)/0.10 – x ≈0.050x/0.10 x = [H3O+] = 3.6 x 10-5 M pH = -log[H3O+] = -log(3.6 x 10-5) = 4.4 as we predicted, the pH rose from 2.9 to 4.4.

  13. Buffered Solutions

  14. Until now, we have considered solutions containing only pure weak acids, weak bases, or their salts dissolved in distilled water • A buffer is a solution with a very stable pH • You can add acid or base to a buffer solution without greatly affecting the pH of the solution • The pH of a buffer will also remain unchanged if the solution is diluted with water or if water is lost through evaporation • A mixture that contains a conjugate acid-base pair is known as buffer solution because the pH changes by a relatively small amount if a strong acid or base is added to it • Buffer systems are used to control pH in many biological and chemical reactions

  15. A buffer is created by placing a large amount of a weak acid or base into a solution along with its conjugate • A weak acid and its conjugate base (B-) will remain in solution together without neutralizing each other • A weak base and its conjugate acid (A+) will do the same

  16. When both the acid and the conjugate base are together in the solution, any hydrogen ions that are added will be neutralized by the base while any hydroxide ions that are added will be neutralized by the acid without having much of an effect on the solution’s pH • A buffer system is a chemical “sponge” for H3O+ or OH- ions that may be produced within a solution • When additional H3O+ ions are added, they react with the conjugate base component of the buffer system to produce H2O and a weak acid that is only slightly ionized, so the pH drops only minimally • H3O+(aq) + conjugate base-(aq)  weak acid(aq) + H2O(l) • When additional OH- ions are added, the basic OH- ions react with the acid component of the buffer system to produce H2O and a weak base, so the pH rises only minimally • OH-(aq) + conjugate acid (aq)  weak base-(aq) + H2O(l)

  17. These solutions are governed by the same equilibrium law as are weak acids and weak bases • Buffered solutions-Equilibrium systems that resist changes in acidity and maintain constant pH when acids or bases are added to them • Most effective pH range for any buffer is at or near the pH where the acid and salt concentrations are equal (pKa) • [H3O+] of weak acid/conjugate base buffer equals the Ka of the weak acid • [OH-] of weak base/conjugate acid buffer equals the Kb of the weak base • In general the most effect pH range of a buffer system is (optimum pH  1.0 pH unit

  18. Since Ka for a weak acid is small, equilibrium concentrations of the weak acid and its conjugate base are very nearly their initial concentrations All weak acid-conjugate base buffer systems have general expression [H3O+] = Ka x [weak acid]0/[conjugate base]0 Convert the buffer relationship into its logarithmic form pH for buffer-pH = pKa + log[A-]/[HA] = pKa + log [conjugatebase]0/[weak acid]0-obtained from equation for weak acid equilibrium (Ka = [H+][A-]/[HA]) this relationship is the Henderson-Hasselbalch equation [OH-] = Kb x [weak base]0/[conjugate acid]0 pOH = pKb + log x [conjugate acid]0/[weak base]0 Buffered solutions contain either: A weak acid and its salt A weak base and its salt

  19. Preparing a buffer • A buffer can be prepared by dissolving 0.200 mole of HF and 0.100 mole of NaF in water to make 1.00 liter of solution • Dissociation reaction of HF is • HF(aq) ⇋ F-(aq) + H+(aq) • Equilibrium law is • Ka = [F-][H+]/[HF] = 6.8 x 10-4 • Set up an equilibrium table using the given information

  20. Substituting the information from the equilibrium line into the equilibrium law yields Ka = (0.100 + x)(x)/0.200 – x = 6.8 x 10-4 Assuming that x is small compared to both 0.100 and 0.200, we simplify the equation to Ka = (0.100)(x)/0.200 = 6.8 x 10-4 x = 1.36 x 10-3 This value of x satisfies the assumptions made, and the last line of the equilibrium table can be completed The last column gives us [H+] = 1.36 x 10-3 M The pH of the buffer solution is calculated as 2.87 For almost all buffer solutions the assumptions hold true, and this type of problem is solved by simply entering the given concentrations of the conjugate acid and conjugate base directly into the equilibrium law

  21. Calculate the pH of the following buffer solutions • 0.250 M acetic acid and 0.150 M sodium acetate • Ka = [C2H3O2-][H+]/[HC2H3O2] • 1.8 x 10-3 = (0.150)[H+]/0.250 • [H+] = 3.0 x 10-3, and pH = 2.52 • A solution of 10.0 g each of formic acid and potassium formate dissolved in 1.00 L of H2O • Ka = [CHO2-][H+]/[HCHO2] • 1.8 x 10-4 = (0.417)[H+]/0.217 • [H+] = 9.4 x 10-5, and pH = 4.03

  22. 0.0345 M ethylamine and 0.0965 M ethyl ammonium chloride Kb = [C2H5NH3+][OH-]/[C2H5NH2] 4.3 x 10-4 = (0.0965)[OH-]/0.0345 [OH-] = 1.5 x 10-4, pOH = 3.82 and pH = 10.18 0.125 M hydrazine and 0.321 M hydrazine hydrochloride Kb = [N2H5+][OH-]/[N2H4] 9.6 x 10-7 = (0.321)[OH-]/0.125 [OH-] = 3.7 x 10-7, pOH = 6.43 and pH = 7.57

  23. Shortcuts in buffer calculations • Equilibrium law used for buffers involves a ratio of the concentrations of the conjugate acid and conjugate base • We may use the moles of conjugate acids and moles of conjugate base instead of the acid and base molarities • If the concentration of the conjugate acid and conjugate base are equal, their ratio is exactly 1.00 • With equal molarities or equal numbers of moles of conjugate acid and conjugate base • Ka = [H+] and pKa = pH for a buffer made from a weak acid and its conjugate base • Kb = [OH-] and pKb = pOH for a buffer made from a weak base and its conjugate acid

  24. pH changes in buffers • Addition of a strong acid to a buffer will decrease the pH slightly • Addition of a strong base to a buffer will increase the pH slightly • To determine the amount that the pH changes when a strong acid or base is added to a buffer, we must calculate the change in concentration of the conjugate acid or base in the buffer

  25. Steps Determine either the molarity or the number of moles of the conjugate acid and conjugate base in the original buffer Determine the amount of strong acid or base added in the same units as the conjugate acid and base in step 1 If a strong acid is added to the buffer, add the value from step 2 to the conjugate acid and subtract it from the conjugate base If a strong base is added to the buffer, add the value from step 2 to the conjugate base and subtract it fro the conjugate acid Substitute the new values for the conjugate acid and conjugate base into the equilibrium law and calculate the pH as shown before

  26. In an acetate buffer, acetic acid, HC2H3O2, is the conjugate acid, and the acetate ion, C2H3O2-, is the conjugate base Addition of a strong acid increases the concentration of acetic acid and decreases the concentration of acetate ions. Addition of a strong base to this buffer increases the acetate ion concentration and decreases the acetic acid concentration HC2H3O2 + OH- C2H3O2- An acetate buffer is prepared with 0.250 M acetic acid and 0.100 M NaC2H3O2. If 0.002 mol of solid NaOH is added to 100 mL of this buffer, calculate the change in pH of the buffer due to the addition of the NaOH. To calculate the change in pH, the pH of the original buffer is needed, along with the final pH after the base is added Dissociation reaction for acetic acid HC2H3O2⇋ H+ + C2H5O2- Equilibrium law Ka = [H+][ C2H5O2-]/[ HC2H3O2] Value for Ka and the acetate and acetic acid concentrations are substituted into the equilibrium law 1.8 x 10-5 = [H+])(0.100)/0.250

  27. Hydrogen ion concentration is calculated as [H+] = 4.5 x 10-5 M, and the pH is 4.35 To calculate the pH of the buffer after the NaOH is added, we must first convert either the molarities of the conjugate acids and bases to moles or the moles of NaOH to molarity so that all the units are the same Molarity NaOH = 0.00200 mol NaOH/0.100 L = 0.0200 M Adding this value to the acetate concentration gives us 0.120 M C2H3O2- Subtracting it from the acetic acid concentration yields 0.230 M HC2H3O2 Substituting these new values into the equilibrium law 1.8 x 10-5 = [H+](0.120)/0.230 [H+] = 3.45 x 10-5 M pH = 4.46 There is an increase of 0.11 pH units when the NaOH is added The answer is reasonable since the pH is expected to rise slightly when a base is added

  28. Calculations Involving Buffered Solutions Containing Weak Acids • "Buffered solutions are simply solutions of weak acids or bases containing a common ion. The pH calculations on buffered solutions require exactly the same procedures introduced in Chapter 14. This is not a new type of problem." • “When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium calculations. This procedure can be presented as follows:

  29. OH- + HA A- + H2O weak acid conjugate base OH- ions are not allowed to accumulate but are replaced by A- ions

  30. Ka = [H+][A-] / [HA] [H+] = Ka x [HA] / [A-] If the amounts of HA and A- originally present are very large compared with the amount of OH- added, the change in [HA]/[A-] will be small. Therefore the pH change will be small

  31. Buffering also works for addition of protons instead of hydroxide ions H++ A-HA Conjugate base Weak acid Buffering with a Weak Base and Its Conjugate Acid Weak base B reacts with any H+ added B + H+  BH+ Base Conjugate acid Conjugate acid BH+ reacts with any added OH BH+ + OH- B + H2O Conjugate acid Base

  32. Summary • Buffered solutions contain relatively large concentrations of a weak acid and the corresponding weak base. They can involve a weak acid HA and the conjugate base A- or a weak base and the conjugate acid BH+ • When H+ is added to a buffered solution, it reacts essentially to completion with the weak base present • When OH- is added to a buffered solution, it reacts essentially to completion with the weak acid present • The pH in the buffered solution is determined by the ratio of the concentrations of the weak acid and the weak base. The pH remains relatively unchanged as long as the concentrations of buffering materials are large compared with the amounts of H+ or OH- added

  33. Buffer Capacity

  34. Buffering Capacity • The amount of protons or hydroxide ions the buffer can absorb without a significant change in pH. • The pH of a buffered solution is determined by the ratio [A-]/[HA] • The capacity of a buffered solution is determined by the magnitudes of [HA] and [A-]

  35. Preparing a Buffer • Preparation of a buffer starts with the selection of the desired pH • Then a table of Ka and Kb values for weak acids and bases is consulted to find an appropriate conjugate acid-base pair to use • These 2 steps determine the ratio of the salt to the weak acid or base • Next, the moles of acid or base that need to be buffered are estimated • The total number of moles of the conjugate acid-base pair should be at least 20 times the amount of the acid or base that needs to be buffered • The volume of buffer solution needed is determined next

  36. Method for preparing the buffer is decided on Conjugate acid and conjugate base are measured and dissolved to the desired volume Measuring the desired amount of conjugate acid and then adding the appropriate amount of a strong base to convert some of the conjugate acid into its conjugate base Conjugate base is measured and the necessary amount of strong acid is added to convert some of the conjugate base into its conjugate acid Optimal buffering occurs when [HA] is equal to [A-] ( [A-]/[HA] = 1 ) The pKa of the weak acid to be used should be as close as possible to the desired pH

  37. Titration Titration technique used for chemical analysis utilizing reactions of 2 solutions

  38. One reactant solution is placed in a beaker and the other in a buret (long, graduated tube with stopcock) The stopcock (valve that allows chemist to add controlled amounts of solution from buret to beaker) An indicator is added to solution in beaker (substance which undergoes a color change in the pH interval of the equivalence point)

  39. Chemist reads volume of solution in buret at start of experiment and again at point where indicator changes color Difference in these volumes represents the volume of reactant delivered from the buret Controlled addition of a solution of known concentration (titrant) in order to determine concentration of solution of unknown concentration The crucial point about the titration experiment is that the indicator is designed to change color when the amount of reactant delivered from the buret is exactly the amount needed to react with the solution in the beaker (Equivalence Point (Stoichiometric Point)-point in a titration at which the reaction between titrant and unknown has just been completed.) Classic chemical reaction-Fe2+ and permanganate ion MnO4- 5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2O The purple permanganate ion is the indicator of the point where the correct amount has been added to completely react all of the Fe2+ ions in the sample

  40. Titration Curve (pH Curve)-plotting of the pH of the solution as a function of the volume of titrant added • Monitors the pH of the solution as an acid-base titration proceeds from beginning to well beyond the equivalence point • pH is plotted on the y-axis and volume of base (or acid) delivered is plotted on the x-axis • Shape of titration curve makes it possible to identify the equivalence point and is an aid in selecting an appropriate indicator • It can also be used to calculate the Ka or Kb of a weak acid or base

  41. There are four major points of interest during a titration experiment Start of the titration, where the solution contains only one acid or base pH is calculated as previously described Region where titrant is added up to the end point, and the solution now contains a mixture of unreacted sample and products Mixture of a conjugate acid and its conjugate base If weak acid or base is involved, this is a buffer solution If only strong acids and bases are used, this region is unbuffered End point, where all the reactant has been converted into product Solution is the salt of the acid or base pH is calculated as previously described titration curve used to determined pH during titration Region after end point, where solution contains product and excess titrant pH depends on excess titrant used

  42. Points on titration curve Midpoint of most vertical part of graph corresponds to exact endpoint. This will also correspond to the equivalence point, or the point at which the equivalents of acid equals the equivalents of base. In addition, the midpoint will also determine the pH of the salt that was formed during the titration.

  43. Half-equivalence point is at the center of the buffer region pH increases more quickly at first, then levels out into buffer region At the half-equivalence point, enough base has been added to convert exactly half of the acid into conjugate base The concentration of the acid is equal to the concentration of the conjugate base The curve remains fairly flat until just before the equivalence point, when the pH increases sharply

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