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Applications of Aqueous Equilibria: The Common Ion Effect

Discover how the addition of salts with weak acid anions or weak base cations can affect the dissociation of acids and the equilibrium position of reactions. Learn how to calculate pH, percent dissociation, and buffer capacity using the Henderson-Hasselbalch equation. Explore pH changes in buffered solutions and understand titration and pH curves.

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Applications of Aqueous Equilibria: The Common Ion Effect

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  1. Chapter 15 Applications of Aqueous Equilibria

  2. The Common Ion Effect • When the salt with the anion of a weak acid is added to an acid, • It reverses the dissociation of the acid. • Lowers the percent dissociation of the acid. • The same principle applies to salts with the cation of a weak base. • Common Ion Effect: Shift in equilibrium position that occurs when ion is added that is already part of equilibrium reaction. • Example: NaF is added to HF solution. • The calculations are the same as last chapter. Thinkwell: The Common Ion Effect

  3. Common Ion Effect 0.25 M Acetic Acid pH 2.67 0.10 M Sodium Acetate pH 9.0 Mixture pH 4.35

  4. Common Ion Effect HF(aq)  H+(aq) + F-(aq) Expected Shift in Weak Acid HF(aq)  H+(aq) + F-(aq) NaF(s)  Na+(aq) + F-(aq) Equilibrium Shift from added Common Ion

  5. Acidic Solutions Containing Common Ions Calculate [H+] and percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF • Write the major species • Consider chemical properties of each • Write equilibrium reaction for [H+] • Write equilibrium expression for [H+] • Create ICE Table • Determine Equilibrium concentration in terms of x • Notice initial [F-]o comes from NaF • Since x is small, it has little change on [F-] & [HF] (5% rule) • Solve for x in terms of [H+] • Calc. percent dissociation = [H+]/[HFo] x 100 CT #23 & 25 p. 740

  6. Buffered Solutions • A buffered solution resists a change in pH when [H+] or [OH-] are added • Buffers are: • weak acid + salt Ex. HF & NaF • weak base + salt Ex. NH3 & NH4Cl • Salt has “Common Ion Effect” • Make a buffer by varying the concentrations • Use same steps to solve as before. Thinkwell Acidic Buffers Carnegie Time #33 p. 741

  7. pH Changes in Buffered Solutions Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of buffered solution of 0.50 M acetic acid (Ka = 1.8 x 10-5) and 0.50 M sodium acetate. • ID major species, write equation, write expression, • Do the stoichiometry first to find initialconcentrations • A strong base will grab protons from weak acid reducing [HA]0 • A strong acid will add protons to anion of the salt reducing [A-]0 • Build ICE table • Then do the equilibrium problem.

  8. How Does A Buffer Work • Buffered solution contains a large quantity of weak acid (HA) and conjugate base (A-) • When OH- are added the HA is only source of protons you get the following reaction OH- + HA  A- + H2O • Similar reasoning applies when you add protons to a buffered solution • When H+ are added the A- you get more weak acid H+ + A-  HA

  9. When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base is consumed in the reaction.

  10. Derivation of Henderson-Hasselbalch Equation The stability of pH can be understood by examining the equilibrium expression of HA p.687 Rearranged The [H+] depends on the ratio [HA]/[A-] Another useful form of this equation is arrived at by taking the negative log of both sides -log [H+] = -log Ka –log [HA]/[A-] Substituting pH & pKa and inverting the final log term to get Henderson-Hasselbalch equation

  11. Henderson-Hasselbach equation Use it to calculate pH • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate Ka=1.4x10-4 • 0.25 M NH3 and 0.40 M NH4Cl Kb = 1.8 x 10-5 p.689-691 Note there are 2 ways to solve.

  12. Summary of Characteristics of Buffered Solutions Buffered Solutions contain large conc. of weak acid and corresponding weak base. Resists change in pH when [H+] or [OH-] are added pH in solution is determined by Henderson-Hasselbalch p stands for potential H stands for Hydrogen But the jury is still out on the origin of the term

  13. Buffer capacity • Buffering Capacity is a measure of the amount of [H+] or [OH-] a buffer can absorb without altering the pH • A buffer with a large capacity contains large concentrations of component. • pH of buffered solution is determined by [A-]/[HA] ratio. • Capacity of buffer is determined by magnitude(how big) [HA] and [A-] are.

  14. Buffer Capacity • Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0 L of each of the following (Ka= 1.8x10-5 for acetic acid) • 5.00 M HC2H3O2 and 5.00 M NaC2H3O2 • 0.050 M HC2H3O2 and 0.050 M NaC2H3O2

  15. Buffer Capacity Table 15.1 Study the diagrams in the left column on p.694 • The best buffers have a ratio [A-]/[HA] = 1 • This is most resistant to change • Optimal buffering occurs when [A-] = [HA]

  16. Titrations and pH Curves • Titration is adding a solution of known concentration (titrant) from buret into a solution of unknown concentration (analyte) until the substance being analyzed is consumed. • Ex.: Put an acid of knownconcentration (titrant) into a base of unknownconcentration (analyte) until all the base is neutralized by the acid. • To see that point we use indicators which turn colors at the Stoichiometric / EquivalencePoint. • Endpoint is point where indicator actually changes color – NOT always same as Equivalence Point

  17. Titrations • Millimole (mmol) = 1/1000 mol • Molarity = mmol/mL = mol/L • Number of mmol = volume of solution (mL) x Molarity • Volume = original volume + volume of added titrant • Makes calculations easier because we will rarely add Liters of solution. • Graph of pH (y axis) vs. mL titrant (x axis) is called a Titration Curve or pH Curve.

  18. Titrating a Strong Acid with Strong Base • There is no equilibrium because they both dissociate completely • The net ionic reaction for a strong acid-base titration is: H+(aq) + OH-(aq)  H2O(l) • The titration of 50.0 mL of 0.200 M HNO3 (analyte) with 0.100 M NaOH (titrant) Case Study, beginning on p.697, walks through the steps and calculates the pH at seven points during the titration.

  19. Titration of 50.0 mL of 0.2 M HNO3 w/ 0.1 M NaOH

  20. Strong Acid Fig. 15.1 Strong Base Fig. 15.2 The following graphs are the result of SA – SB Titrations You should recognize these graphs on test.

  21. Weak acid with Strong base • Similar to series of buffer problems • Involves a 2 step procedure: • Do stoichiometry to determine concentrations • Do equilibrium to determine the weak acid equilibrium and pH Titrate 50.0 mL of 0.10 M H(C2H3O2) (Ka=1.5x10-5) with 0.10 M NaOH(Case Study p.700)

  22. Weak Acid Weak BaseYou should recognize these graphs on test.

  23. Titration Of Weak Acid • Hydrogen Cyanide gas (HCN) is a very weak acid (Ka=6.2 x 10-10) when dissolved in water. If a 50.0 mL sample of 0.100 M HCN is titrated with 0.10 M NaOH, calculate the pH of the solution: • After 8.00 mL of 0.100 M NaOH has been added • At the halfway point of the titration • At the equivalence point of the titration

  24. Indicators • Most common Indicators are weak acids that change color when they become bases. • Weak Acid Indicators are written HIn • Weak base In- HIn H+ + In- Phenolphthalein: clear pink • Equilibrium is controlled by pH • End point-when indicator changes color and stays • Equivalence Point = Stoichiometric Point is not necessarily the End point

  25. Indicators • Since it is at equilibrium the color change is gradual. • It is noticeable when the ratio is • Choose the indicator with a pKa 1 less than the pH at equivalence point if you are titrating with base. • Choose the indicator with a pKa 1 greater than the pH at equivalence point if you are titrating with acid.

  26. Solubility Equilibria • Any dissociating problem is an equilibrium problem • If there is not much solid it will all dissociate • As more solid is added the solution will become saturated • Solid  dissolved • The solid will precipitate as fast as it dissolves • Resulting in Equilibrium which can be used to set up an equilibrium constant expression

  27. Solubility Product Constant Ksp • M+ stands for the cation (metal). • Nm-stands for the anion (nonmetal). MaNmb(s)  aM+(aq) + bNm-(aq) • Because concentration of a solid doesn’t change: • Called the Solubility Productfor each compound.

  28. Solubility Equilibria • Solubility product = product of the concentration of ions in equation, raised to power of coefficients • Solubilityis not the same as solubility product • Solubility product is an equilibriumconstant • It doesn’t change except with temperature • Solubilityis an equilibrium position for how much can dissolve at a certain temperature. • A common ion can change Solubility • See p. 718 & appendix p. A24 for Table of Ksp values

  29. Calculating Ksp • Copper (I) Bromide has a measured Solubilityof 2.0 X 10-4 at 25oC. Calculate it’s Ksp value. • Write the balanced ionic reation • Equilibrium Concentrations can be obtained from measured solubility of CuBr

  30. Calculating Ksp • Build ICE Table • Plug the concentrations into Equilibrium Expression Study Ex 15.13 & 15.14 p.719-721

  31. Relative Solubility • Relative Solubility will only allow us to compare the Solubilityof solids that fall apart into the same total number of ions. See description p.721 • 3 FACTORS AFFECT SolubilityOF IONIC SUBSTANCES: • COMMON ION • pH • COMPLEX IONS Common Ion Effect • If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve, equilibrium will shift to the left. Study Ex. 15.15 p723

  32. pH and solubility • pH greatly affects salt’s solubility • Adding OH-or removing [H+]increases pHdecreases solubility, forces equilibrium to left • Removing OH- or adding [H+]decreases pHincreasessolubility, forces equilibrium to right • For other anions if they come from a weak acid they are more soluble in a acidic solution than in water

  33. Precipitation • Will a solid (precipitate) form from solution? • Ion Product, uses initial concentrations • If Q>Kspprecipitation forms until Q = Ksp • If Q<Kspsolid dissolves until Q = Ksp • If Q = Ksp equilibrium. Study Ex. 15.16 p725

  34. Selective Precipitations • Used to separate mixtures of metal ions in solutions. • Add anions that will only precipitate certain metals at a time. • Used to purify mixtures. Study Ex 15.18 p.728

  35. Complex Ion Equilibria • Complex Ion is a charged species consisting of metal ion surrounded by ligand • Ligand is a Lewis base using their lone pair to stabilize the charged metal ions • Common ligands are NH3, H2O, Cl-, CN- • Coordination number is the number of attached ligands, most common numbers are 6, 4 & 2 • Ex. Co(H2O)6+2 has a coordination # of 6 • Ex. Cu(NH3)4+2 has a coordination # of 4 • Ex. Ag(NH3)2+has a coordination # of 2

  36. Complex Ion Equilibria • Usually the ligand is in large excess. • Metal ions usually add Ligands 1 step at a time: Example Ag+ + NH3 Ag(NH3)+ K1 = 2.1 x 103 Ag(NH3)++ NH3 Ag(NH3)2+ K2 = 8.2 x 103 • The formation constant or stability constant is Kx where x = step number • Individual K’s will be large so treat them as if reactions goes to equilibrium. • The complex ion will be the biggest ion in solution.

  37. Summary • Strong acid and base just stoichiometry. • Weak acid before equivalence point • Stoichiometry • Henderson-Hasselbach • Weak acid at equivalence point Kb • Weak base after equivalence - leftover strong base.

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