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Aqueous Equilibria

Aqueous Equilibria. Chapter 16. pH. The pH of aqueous solutions not only blood plasma, but seawater, detergents, sap and reaction mixtures is controlled, by the transfer of protons between ions and water molecules. Ions as acids and bases.

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Aqueous Equilibria

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  1. Aqueous Equilibria Chapter 16

  2. pH • The pH of aqueous solutions not only blood plasma, but seawater, detergents, sap and reaction mixtures is controlled, by the transfer of protons between ions and water molecules.

  3. Ions as acids and bases • Any cation that is the conjugate acid of a weak base functions as an acid and lowers the pH of the solution. Ammonium ion the conjugate acid of weak base ammonia, is an acid. • NH4 +(aq) +H2O(l) H3O+(aq) +NH3(aq) • Small highly charged metal cations that can act as Lewis acids in water, such as Al3+ and Ti3+ also produce acidic solutions, even though the cations don’t have any protons to donate. The protons come from the water molecules that hydrate the ions in solution.

  4. Salts that contain the conjugate acids of weak bases produce acidic aqueous solutions; so do salts that contain small, highly charged metal cations. Salts that contain the conjugate bases of weak acids produce basic aqueous solutions.

  5. pH of a salt solution • Aquoeus solutions of salts with acidic cations have a pH lower than 7; salts with basic anions produce a pH higher than 7 in aqueous solutions.

  6. Class Practice • Estimate the pH of 0.15M Ca(CH3COO)2(aq).

  7. pH of mixed solutions • The pH of a solution of a weak acid increases when a salt containing its conjugate base is added. The pH of a solution of a weak base decreases when a salt containing its conjugate acid is added.

  8. Class Practice • Calculate the pH of a solution that is 0.500 M HNO2(aq) and 0.100M KNO2(aq) • Ka=4.3x10-4 for HNO2

  9. Titrations • Strong acid strong base titrations

  10. Strong base and strong acid

  11. In the titration of a strong acid with a strong base (or a strong base with a strong acid), the pH increases (or decreases) slowly initially, increases (or decreases) rapidly through pH=7 at the stoichiometric point, and then increases (or decreases) slowly again.

  12. Class Practice • Suppose we carry out a titration in which 0.34M HCl (aq) is the titrant and the analyte initially consists of 25 ml of 0.25M NaOH (aq). What is the pH of the analyte solution after the addition of 5 ml of titrant?

  13. Calculating the pH at the stochiometric point of the titration of weak acid with strong base Calculate the pH of the solution resulting when 5 ml of 0.150 M NaOH (aq) is added to 25 ml of 0.100M HCOOH (aq) . Use Ka=1.8x10-4 for HCOOH (aq).

  14. Buffer solutions Solutions that resist the change in pH when small amounts of strong acids or base are added are called as buffers.

  15. Acid Buffer An acid buffer solution consists of a weak acid and its conjugate base; it has pH<7. CH₃COOH (aq) + H₂O(l)↔H₃O+(aq) + CH₃CO₂- (aq) Acid Buffer action: The weak acid transfers protons to the OH- ions from added strong base. The conjugate base of the weak acid accepts protons from the H₃O+ ions supplied by a strong acid.

  16. Base Buffer A base buffer solution consists of a weak base and its conjugate acid ; it has pH >7. NH₃(aq) + H₂O(l)↔NH₄+ (aq) + OH- (aq) Base Buffer action: The weak base accepts protons from the H₃O+ ions supplied by a strong acid. The conjugate acid of the weak base transfers protons to the OH- ions from added strong base.

  17. Calculate the pH of a buffer solution For a solution containing a weak acid HA and a salt that provides the conjugate base anion, A- , the proton transfer equilibrium is; HA(aq) +H2O(l)↔H3O+ (aq) +A- (aq) Ka=[H3O+ ][A-]/[HA] Rearranging the equation 1/[H3O+ ] =1/Ka x[A-]/[HA]

  18. Taking logarithm of both sides: log1/[H₃O+ ] = log1/ Ka + log[A-]/[HA] -log[H₃O+ ] = -log[Ka] + log[A-]/[HA] That is pH = pka + log[A-]/[HA] pH =pKa + log ([base]initial/[acid]initial) This relation is called the Henderson Hassel balch equation.

  19. Class Practice Calculate the pH of a buffer solution that is 0.40 M NaCH₃CO₂ (aq) and 0.80M CH₃COOH(aq) at 25⁰C. Calculate the ratio of the molarities of CO₃2- and HCO₃- ions required to achieve buffering at pH =9.50. The pka of H₂CO₃ is 10.25.

  20. Solubility product The solubility product is the equilibrium constant for the equilibrium between an un dissolved salt and its ions in a saturated solution.

  21. Class Practice The molar solubility of silver chromate Ag2CrO4 is 6.5x10-5 mol/L. determine the value of Ksp. The Ksp for lead(II)iodide in water is 1.4x10-8 . Estimate its molar solubility.

  22. The common ion effect The common-ion effect is a term used to describe the effect on a solution of two dissolved solutes that contain the same ion.Thepresence of a common ion suppresses the ionization of a weak acid or a weak base.

  23. If both sodium acetate and acetic acid are dissolved in the same solution they both dissociate and ionize to produce acetate ions. Sodium acetate is a strong electrolyte so it dissociates completely in solution. Acetic acid is a weak acid so it only ionizes slightly. NaC2H3O2(s) → Na+(aq) + C2H3O2-(aq) HC2H3O2(l) ↔ H+(aq) + C2H3O2-(aq)

  24. According to Le Chatelier's principle, the addition of acetate ions from sodium acetate will suppress the ionization of acetic acid and shift its equilibrium to the left. This will decrease the hydrogen ion concentration and thus the common-ion solution will be less acidic than a solution containing only acetic acid.

  25. Class Practice What is the approximate molar solubility of silver chloride in 0.10M NaCl (aq)?

  26. Predicting precipitation In the Ksp expression the right hand side of the expression is known as the "ion product". At saturation when the ions in solution are in equilibrium with the solid slightly soluble salt, the ion product is equal to a fixed value called the solubility product constant: Ksp = ion product

  27. Two further cases exist: ion product < Ksp ion product > Ksp If the ion product < Ksp then no precipitation will occur even though the salt may be insoluble according to the solubility rules.If, on the other hand, the ion product > Ksp the ion concentration will be large enough for precipitation to occur

  28. The concentration of Calcium ion in blood plasma is 0.0025 M. If the concentration of Oxalate ion is 1 X 10-8 M. Will Calcium Oxalate, CaC2O4precipitate? Ksp = 2.3 X 10-9.

  29. Write the slightly soluble salt equilibria. CaC2O4 = Ca+2 + C2O4-2 Write the ion product ion product = [Ca+2] [C2O4-2] Identify the given ion concentrations [Ca+2] = .0025 = 2.5 X 10-3 M [C2O4-2] = 1 X 10-8 M Plug in the given ion concentrations into the ion product equation

  30. ion product = [Ca+2] [C2O4-2] = (2.5 X 10-3) ( 1 X 10-8) ion product = 2.5 X 10-11 Compare the ion product value with Ksp value and make a conclusion. Since the ion product value (2.5 X 10-11) is less than the Ksp (2.3 X 10-9) we have to conclude that no precipitation will take place.

  31. Class Practice Lead II Chromate, PbCrO4, is used in yellow paint pigment("chrome yellow"). When the solution is 5.0 X 10-4 M in Pb+2 ion and 5.0 X 10-5 M in Chromate ion. Would you expect some of the Lead Chromate to precipitate. Ksp PbCrO4 = 1.8 X 10-14.

  32. Homework Page 744 16.22 all 16.30 all 16.55 all 16.85

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