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Chapter 16: Applications of Aqueous Equilibria

Chapter 16: Applications of Aqueous Equilibria. Renee Y. Becker Valencia Community College. Acid-base Neutralization Reaction. Acid-base neutralization reaction 1.      Products are water and a salt 2.      Four types a)      Strong Acid-Strong Base b)     Weak Acid-Strong Base

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Chapter 16: Applications of Aqueous Equilibria

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  1. Chapter 16: Applications of Aqueous Equilibria Renee Y. Becker Valencia Community College

  2. Acid-base Neutralization Reaction Acid-base neutralization reaction 1.      Products are water and a salt 2.      Four types a)      Strong Acid-Strong Base b)     Weak Acid-Strong Base c)      Strong Acid-Weak Base d)     Weak Acid-Weak Base

  3. Strong Acid-Strong Base HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) 1. Because HCl is a strong acid and NaOH is a strong base they are both strong electrolytes, they dissociate nearly 100% 2.      Complete ionic equation H+ + Cl- + Na+ + OH- H2O(l) + Na+ + Cl- 3.      Net ionic equation H3O+ + OH- 2 H2O(l)

  4. Strong Acid-Strong Base 4.      Kn = 1 / [H3O+] [OH-] = 1 / Kw = 1 / 1 x 10-14 = 1 x 1014 a)      Kn is the equilibrium constant with respect to neutralization 1 x 1014 is a large # and means that for a strong acid-strong base reaction proceeds essentially 100% to completion 5.      pH = 7

  5. Weak Acid-Strong Base HF(aq) + NaOH(aq) H2O(l) + NaF 1. Because HF is a weak acid-weak electrolyte it does not dissociate well and will not be ionized on the reactant side 2. Complete ionic equation HF + Na+ + OH- H2O+ Na+ + F- 3. Net Ionic equation HF + OH- H2O + F-

  6. Weak Acid-Strong Base 4. To obtain the equilibrium constant, Kn we need to multiply known equilibrium constants for reactions that add to give the net ionic equation for the neutralization HF(aq) + H2O(l) H3O+(aq) + F-(aq) Ka = 3.5 x 10-4 H3O+(aq) + OH- 2 H2O(l) 1/Kw = 1 x 1014 Net: HF + OH- H2O(l) + F- Kn = Ka (1/Kw) = 3.5 x 10-4 (1 x 1014) = 3.5 x 1010

  7. Weak Acid-Strong Base For any weak acid-strong base reaction 1. Kn = Ka (1/Kw) 2.  100% completion because of OH- strong affinity for protons 3. The pH will be > 7, due to basicity of conjugate base

  8. Strong Acid-Weak Base NH3(aq) + HCl(aq) NH4+(aq) + Cl- 1. A strong acid is completely dissociated into H3O+ and A- ions 2. Neutralization reaction is a proton transfer 3. Net ionic equation H3O+ + NH3 H2O + NH4+

  9. Strong Acid-Weak Base • Just as before we can obtain the equilibrium constant for the neutralization reaction by multiplying known equilibrium constants for reactions that add to give the net ionic equation NH3 + H2O  NH4+ + OH- Kb = 1.8 x 10-5 H3O+ + OH- 2 H2O 1/Kw = 1 x 1014 Net: H3O+ + NH3 H2O + NH4+ Kn =Kb(1/Kw) = 1.8 x 109

  10. Strong Acid-Weak Base For any strong acid-weak base reaction 1.  Kn = Kb(1/Kw) 2.  100% completion because H3O+ is a powerful proton donor 3. The pH will be < 7, due to the conjugate acid

  11. Weak Acid-Weak Base CH3CO2H + NH3 NH4+ + CH3CO2- 1. Both acid and base are largely undissociated 2.  Neutralization reaction is a proton transfer from the weak acid to the weak base 3. The equilibrium constant can be obtained by adding equations for the acid dissociation, the base protonation and the reverse of the dissociation of water

  12. Weak Acid-Weak Base CH3CO2H + H2O  H3O+ + CH3CO2- Ka = 1.8 x 10-5 NH3 + H2O  NH4+ + OH- Kb = 1.8 x 10-5 H3O+ + OH- 2 H2O 1/Kw = 1 x 1014 Net: CH3CO2H + NH3 NH4+ + CH3CO2- Kn = Ka(Kb)(1/Kw) = 3.2 x 104 For any weak acid-weak base reaction 1.      Kn = Ka(Kb)(1/Kw) Kn is smaller so the reaction does not go to completion

  13. Example 1: The Common-Ion Effect Calculate the pH of a solution prepared by dissolving .10 mol acetic acid and .10 mol sodium acetate in water, and then diluting the solution to a volume of 1.00 L

  14. Buffer Solutions 1.  A weak acid and it’s conjugate base 2.  Resist drastic changes in pH 3.   If a small amount of OH- is added, the pH increases but not by much because the acid component of the buffer neutralizes the OH- 4. If a small amount of H3O+ is added the pH decreases but not by much because the conjugate base in the buffer neutralizes the added H3O+

  15. Buffer Solutions 5.  Examples a) CH3CO2H + CH3CO2- b) HF + F- c) NH4+ + NH3 d) H2PO4- + HPO42- 6. Very important in biological systems (blood is a buffer) (H2CO3 + HCO3-)

  16. Example 2: Addition of OH- to a buffer, we add 0.01 mol solid NaOH to 1.00 L of a 0.10 M acetic acid-0.10 M sodium acetate solution. What is the pH? Because solutions involving a strong acid or base go to nearly 100% completion we must account for the neutralization before we can calculate H3O+

  17. Example 3: Addition of H3O+ to a buffer, we add 0.01 mol HCl to 1.00 L of a 0.10 M acetic acid-0.10 M sodium acetate solution. What is the pH?

  18. Buffer capacity 1. Measure of the amount of acid or base that a solution can absorb without a significant change in pH 2.    Measure of how little the pH changes with the addition of a given amount of acid or base 3.   Depends on how many moles of weak acid and conjugate base are present 4.  The more concentrated the solution (acid & conj. base), the greater the buffer capacity 5. The greater the volume (acid and conj. Base), the greater the buffer capacity

  19. Henderson-Hasselbalch Equation Henderson-Hasselbalch Equation pH = pKa + log [base]/[acid] • Tells us how the pH affects the % dissociation of a weak acid 2. Also tells us how to prepare a buffer solution with a given pH. a) Pick a weak acid that has a pKa close to the desired pH b) Adjust the [base]/[acid] ratio to the value specified by the HH equation The pKa of the weak acid should be within ± 1 pH unit of desired pH

  20. Example 4: I want to prepare a buffer solution with a pH of 7.00 and one of a pH of 9.00 which of the following pairs of weak acid-conj. bases should I use? CH3CO2H + CH3CO2- Ka = 1.8 x 10-5 pKa = 4.74 HF + F- Ka = 3.5 x 10-4 pKa = 3.46 NH4+ + NH3 Ka = 5.56 x 10-10 pKa = 9.25 H2PO4- + HPO42- Ka = 6.2 x 10-8 pKa = 7.21

  21. Example 5: Use the HH equation to calculate the pH of a buffer solution prepared by mixing equal volumes of 0.20 M NaHCO3 and 0.10 M Na2CO3 (Ka = 5.6 x 10-11 for HCO3-)

  22. Example 6: Give a recipe for preparing a NaHCO3-Na2CO3 buffer solution that has a pH = 10.40

  23. pH Titration Curves 1.  A plot of the pH of the solution as a function of the volume of the added titrant 2.  A solution of a known concentration of base or acid is added slowly from a buret to a second solution with an unknown concentration of acid or base 3.  Progress is monitored with a pH meter or by color of indicator 4. Equivalence point is the point at which stoichiometrically equivalent quantities of acid and base have been mixed together

  24. pH Titration Curves 5. Endpoint is when the color of the acid-base indicator changes 6. There are four important types of titration curves a)      Strong Acid-Strong Base b)      Weak Acid-Strong Base c)      Weak Base-Strong Acid d)      Polyprotic Acid-Strong Base We will only be calculating for Strong acid-strong base titrations but you are responsible to be able to recognize and label the titration curves for all

  25. Strong Acid-Strong Base

  26. Weak Acid-Strong Base

  27. Strong Acid-Weak Base

  28. Polyprotic Acid-Strong Base

  29. Strong Acid-Strong Base Titrations Titration of a strong acid (50 mL of a 0.02 M HCl) by a strong base (0.030 M NaOH) There are four main calculations for this type of titration 1.  Before any Base has been added 2.  Before the equivalence point 3.   At the equivalence point 4. After the equivalence point

  30. 1. Before any base has been added 1.  Since HCl is a strong acid the initial concentration of H3O+ = initial molarity = 0.02 M pH = -log[0.02] = 1.70

  31. 2. Before the equivalence point Let’s say we have added 10 mL of 0.03 M NaOH The added OH- ions will neutralize some of the H3O+ ions Moles of H3O+ ions = M*V = 0.02 * .05 L = .001 mol H3O+ Moles of OH- ions = M*V = 0.03 * .01L = .0003 moles Molarity of H3O+ after addition of NaOH =[Moles H3O+ - moles OH-] /total volume,L M = (.001 - .0003) / (.05 L + .01 L) = 1.2 x 10-2 M = [H3O+] pH = - log[1.2 x 10-2] = 1.92

  32. 3. At the equivalence point At the equivalence point the pH = 7 To find the volume of NaOH would give you the equivalence point use equation: M*V = M*V .02(50 mL) = .03(x mL) 33.3 mL NaOH

  33. 4. After the equivalence point After the equivalence point you have neutralized all of the acid and now you have excess base, 45 mL NaOH Find the moles of acid & base Moles of acid = M*V = .001 Moles H3O+ Moles of base = M*V = .03 * .045 L = .00135 moles OH-

  34. 4. After the equivalence point [OH-] = (moles of base – moles of acid)/ total volume, L = (.00135 - .001) / (.05 L + .045 L) = 3.7 x 10-3 M Kw = [H3O+] [OH-] [H3O+] = 1 x 10-14 / 3.7 x 10-3 = 2.7 x 10-12 pH = -log [ 2.7 x 10-12] = 11.57

  35. Solubility Equilibria Solubility Product Constant, Ksp Same as Kc, Kp, Kw, Ka, & Kb Prod / reactant Coefficients are exponents, omit solids and pure liquids CaF2(s) Ca2+(aq) + 2 F-(aq) Ksp = [Ca2+][F-]2

  36. Example 7: Measuring Ksp and Calculating Solubility from Ksp A saturated solution of Ca3(PO4)2 has [Ca2+] = 2.01 x 10-8 M and [PO43-] = 1.6 x 10-5 M. Calculate Ksp for Ca3(PO4)2

  37. Factors that Affect Solubility 1. The Common ion effect MgF2(s) Mg2+(aq) + 2 F-(aq) If we try dissolve this in a aqueous solution of NaF the equilibrium will shift to the left. This will make MgF2 less soluble 2. Formation of Complex ions Complex ion: An ion that contains a metal cation bonded to one or more small molecules or ions, NH3, CN- or OH- AgCl(s) Ag+ + Cl- Ag+ + 2 NH3 Ag(NH3)2+ Ammonia shifts the equilibrium to the right by tying up Ag+ ion in the form of a complex ion

  38. Factors that Affect Solubility 3. The pH of the solution a) An ionic compound that contains a basic anion becomes more soluble as the acidity of the solution increases CaCO3(s) Ca2+ + CO32- H3O+ + CO32- HCO3- + H2O Net: CaCO3(s) + H3O+ Ca2+ + HCO3- + H2O Solubility of calcium carbonate increases as the pH decreases because the CO32- ions combine with protons to give HCO3- ions. As CO32- ions are removed from the solution the equilibrium shifts to the right to replenish the carbonate PH has no effect on the solubility of salts that contain anions of strong acids because these anions are not protonated by H3O+

  39. Precipitation of Ionic Compounds Ion Product (IP) Same as Ksp but at some time, t, snapshot like Qc, reaction quotient CaF2(s) Ca2+ + 2 F- IP = [Ca2+][F-]2 If IP > Ksp solution is supersaturated and precipitation will occur If IP = Ksp the solution is saturated and equilibrium exists If IP< Ksp the solution is unsaturated and ppt will not occur

  40. Example 8: Will a precipitate form on mixing equal volumes of the following solutions? • 3.0 x 10-3 M BaCl2 and 2.0 x 10-3 M Na2CO3 (Ksp = 2.6 x 10-9 for BaCO3) • 1.0 x 10-5 M Ba(NO3)2 and 4.0 x 10-5 M Na2CO3

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