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Chapter 3: Acid – Base Equilibria

Chapter 3: Acid – Base Equilibria. HCl + KOH  KCl + H 2 O acid + base  salt + water. What is an acid? The Arrhenius concept proposed that acids are substances that produce hydrogen ions (H + ) in aqueous solutions. If this: HA (aq) + H 2 O (l)  H 3 O (aq) + A - (aq)

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Chapter 3: Acid – Base Equilibria

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  1. Chapter 3: Acid – Base Equilibria HCl + KOH  KCl + H2O acid + base  salt + water

  2. What is an acid? The Arrhenius concept proposed that acids are substances that produce hydrogen ions (H+) in aqueous solutions. If this: HA(aq) + H2O(l)  H3O(aq) + A-(aq) is the general form of the acid reaction, then we can calculate an equilibrium constant (Ka) for the reaction. Ka is called the acid dissociation constant. Ka = [H3O+][A-] / [HA] which equals [H+][A-] / [HA]. A strong acid is one that undergoes significant dissociation and has a very large Ka. A weak acid only partially dissociates and has a relatively small Ka. pKa = -logKa

  3. As you noticed from the previous slide (table 3-1), acids can contain more than one acidic proton. i.e. H2SO4 H+ + HSO4- HSO4-  H+ + SO42- Polyprotic acids will have more than one pKa Remember: pKa = -logKa Acids that undergo significant dissociation have a negative pKa, and acids that only partially dissociate have a positive pKa.

  4. What is a base? The Arrhenius concept proposed that a base is a substance that produces OH- ions in aqueous solution. If this: B(aq) + H2O(l)  BH+(aq) + OH-(aq) is the general form of the base reaction, then we can calculate an equilibrium constant (Kb) for the reaction. Kb is called the base dissociation constant. Kb = [BH+][OH-] / [B]. A strong base is one that undergoes essentially complete dissociation and has a large Kb. A weak base only partially dissociates and has a relatively small Kb. pKb = -logKb

  5. *As it was with acids, base may also require more than one step to complete dissociation. You may also notice the absence of NaOH. For our purposes, you may assume this base completely dissociates.

  6. The dissociation of water and pH H2O  H+ + OH- and Kw = [H+][OH-] / [H2O] = [H+][OH-] Where Kw is the equilibrium constant for water. Kw varies as a function of temperature. (See table 3-3.) Kw(25°C)= 10-14 =[H+][OH-] *Remember that Kw is a constant, therefore if you know the Kw and the concentration of either H+ or OH-, you can find the remaining unknown concentration. http://www.fishdoc.co.uk/water/cocktail04.html http://www.whoi.edu/oceanus/viewArticle.do?archives=true&id=65266

  7. For example: If [H+] for a solution is 10-3, what is [OH-] in that solution? [OH-] = 10-11 The pH scale was designed to simplify the description of the acidity of a solution. The pH is often called the hydrogen ion exponent. pH = -log[H+] Therefore, if, for example [H+] = 10-3, then the pH = 3 Pure water at 25°C has a pH of 7. This means that the number of H+ ions and the number of OH- ions are equal at 10-7 a piece.

  8. Why is knowing the specific [H+] important?

  9. Let’s do some examples… Determine the pH of a solution with a [OH-] of 3 X 10-5. Recall: Kw(25°C)= 10-14 =[H+][OH-] 10-14 / 3 X 10-5 = [H+] [H+] = 3.33 X 10-9.  pH = -log(3.33 X 10-9) = 8.48 Is this solution acidic or basic?

  10. pH of Natural Waters

  11. pH of river systems? http://www.ecy.wa.gov/programs/wq/tmdl/littlespokane/impairments.html

  12. Not so natural water….acid mine drainage.

  13. It is very important to be familiar with the carbonate-carbonic acid system. It is one of the most important systems in the environment. http://bholistic.typepad.com/b_holistic/2010/04/b-refreshed-with-sparkling-mineral-water.html

  14. You might want to write these equations down. You will, without a doubt, see them again…. CO2 + H2O → H2CO3 H2CO3 → H+ + HCO3- HCO3- → H+ + CO32- These describe the formation and dissociation of carbonic acid. Theoretically, what would happen to the concentration of CO2 if more H+ were added to the system? We will address this later.

  15. CO2(g) + H2O  H2CO3(aq) where [H2CO3(aq)] = KCO2PCO2 and Ka1 = [H+][HCO3-] / KCO2PCO2 The dissociation of H2CO3 is a 2-step process! H2CO3(aq)  H+ + HCO3-  HCO3-  H+ + CO32- Ka1 = [H+][HCO3-] / [H2CO3(aq)] and Ka2 = [H+][CO32-] / [HCO3-] Further at 25°C: [H2CO3(aq) ] / [HCO3-] = [H+] / Ka1 = [H+] / 10-6.35 And [HCO3-] / [CO32-] = [H+] / Ka2 = [H+] / 10-10.33

  16. http://www.phschool.com/science/biology_place/biocoach/photosynth/overview.htmlhttp://www.phschool.com/science/biology_place/biocoach/photosynth/overview.html CO2 + H2O → CH2O + O2 You should all be familiar with the basic equation for photosynthesis. How is it similar to the carbonic acid equation? More accurate equation for photosynthesis: 6CO2 + 6H2O → C6H12O6 + 6O2 What does this imply?

  17. You can consider respiration to be the opposite of photosynthesis using the simplest of terms… CH2O + O2→ CO2 + H2O http://sbi3u1tdoust.edublogs.org/2010/06/01/respiration/

  18. Given: [H2CO3(aq) ] / [HCO3-] = [H+] / Ka1 = [H+] / 10-6.35 [HCO3-] / [CO32-] = [H+] / Ka2 = [H+] / 10-10.33 How would you expect H2CO3 to affect the pH of natural waters? It depends upon the level of dissociation.

  19. How does the weathering of calcium, silica and carbonate species change the atmospheric concentration of carbon dioxide? Images from midfind.com

  20. CaSiO3 + 3H2O + 2CO2 Ca2+ + 2(HCO3- ) + H4SiO4 wollastonite + water + carbon dioxide  calcium + bicarbonate + silicic acid From pore spaces &/or atmosphere and: Ca2+ + 2(HCO3- )  CaCO3 + H2O + CO2 and: CaCO3 + SiO2 CaSiO3 + CO2 calcite + quartz  wollastonite + carbon dioxide ________

  21. What other factors can change the pH of natural waters?

  22. Lets consider the effects of water in contact with atmospheric carbon dioxide. CO2(aq) + H2O(l) H2CO3(aq) Given this relationship between CO2 in solution and water, would you expect the surface waters of the ocean to be more or less acidic than the ocean waters out of contact with the atmosphere? In general, surface waters are more acidic at the surface and in coastal waters (for another reason…) With the CO2—H2O system there are two end-member cases: An open system: in equilibrium with atmospheric CO2. A closed system: isolated from atmospheric CO2.

  23. Let’s recall problem 25 from the problem set 2…. What did we discover about the ionic charges in a solution? The total positive (cation) charge and total negative (anion) charge in a solution mustbeequal. For example: for the CO2—H2O system, the charge balance equation is written: mH+ = mHCO3- + 2mCO32- + mOH- Where m is the molar concentration of each species in solution.

  24. The Carbonic acid – Carbonate System is regulated by not only the presence of H2CO3, but also the presence of the minerals calcite or aragonite (CaCO3). CaCO3calcite Ca2+ +CO32- Ksp = [Ca2+][CO32-] Final charge balance equation becomes: mH+ + 2mCa2+ = mHCO3- + 2mCO32- + mOH-

  25. Now let’s do some problems from the book……

  26. Example 3-3 (pg. 66) Calculate the pH of rainwater in equilibrium with atmospheric CO2. First of all, this book was first published in 2004, thus the data may be as much as10 years old. What are the atmospheric CO2 levels today? For the present day atmosphere = (396ppm) ⟹ PCO2 = 10-3.40atm. [H2CO3] = KCO2PCO2 = (10-1.47)(10-3.40) = 10-4.87. Ka=[H+][HCO3-] / [H2CO3] ∴ [H+][HCO3-] = Ka[H2CO3] = (10-6.35)(10-4.87) = [10-11.22] If you assume that [H+] = [HCO3-] then [H+] = (10-11.22)0.5 = 10-5.61 this means that the pH of “acid rain” is less than 5.61. It is important to note that the book suggest acid rain has a pH of 5.66. What does this imply if the atmospheric CO2 concentrations approach 500ppm? Remember this is a logarithmic scale….. The pH would be 5.56.

  27. Example 3-4 A groundwater sample has a measured pH of 6.84 and HCO3- of 460mg L-1. We will assume that activity equals concentration. At 25⁰C, calculate the PCO2 for this groundwater sample. We must first convert the measured concentration of HCO3- to moles per liter. The atomic weight of HCO3- = 61.0g [HCO3- ] = 460 x 10-3 g L-1 / 61g/mol = 7.54 x 10-3 mol L-1 The using equation 3-15: logPCO2 = -pH + log([HCO3-]/Ka1KCO2) logPCO2 = -6.84 + log([7.54 x 10-3] / 10-6.3510-1.47) = -1.14 PCO2 = 10 -1.14 How does this compare with atmospheric concentrations? What does this imply about the groundwater? Is it close to a pollution source, anaerobic or aerobic decay or in supersaturated in CaCO3?

  28. Example 3-5 Calculate the concentration of each carbonate species in a solution at 25⁰C when CT = 1 x 10-3 mol L-1 and pH = 5.7. Which species do you expect to be dominate? aH = 1 + Ka1/[H+] + Ka1Ka2/[H+]2 = 1 + (10-6.35)/(10-5.7) + (10-6.35)(10-10.33)/(10-5.7)2 = 1.224 [H2CO3] = CT/aH = 1 x 10-3 mol L-1/1.224 = 8.17 x 10-4 mol L -1 [HCO3-] = CTKa1/[H+]aH = (1 x 10-3)(10-6.35)/(10-5.7)(1.224) = 1.83 x 10-4 mol L-1 [CO32-]= CTKa1Ka2/[H+]2aH = (1 x 10-3)(10-6.35)(10-10.33)/(10-5.7)2(1.224) = 4.29 x 10-9 mol L-1 Do these concentrations make sense given a pH of 5.7

  29. pH = 5.7: [H2CO3] = 8.17 x 10-4, [HCO3-] = 1.83 x 10-4, [CO32-] = 4.29 x 10-9

  30. Acidity and Alkalinity Acidity is the capacity of water to donate protons. Also described as the ability of a solution to neutralize bases. Alkalinity is the capacity of water to accept protons. Also described as the ability of a solution to neutralize acids. Nonconservative species: species whose abundances vary as a function of pH or some other intensive variable (i.e.P & T). Conservative species: species whose abundances do not vary as a function of pH or some other intensive variable (i.e.P & T).

  31. What is the easiest way to determine the concentration of either H+ or OH-? Titration. Ct X Vt = Cs X Vs Where Ct is the concentration of the titrant, Vt is the volume of the titrant, Cs is the concentration (acidity or alkalinity) of the unknown solution and Vs is the volume of the unknown solution.

  32. Example 3-11 100mL of an acidic solution is titrated with a 100 meq L-1 NaOH solution. Neutrality (pH = 7) is achieved after 50mL of titrant have been added to the acid solution. Calculate the acidity of the solution. Cs = (Ct x Vt) / Vs = ((100meq L-1)(50 x 10-3L)) /100 x 10-3L = 50 meq L-1 = total acidity of the unknown.

  33. Salts • strong acid with a strong base. –makes a neutral solution. • weak acid with a strong base. –makes a basic solution. • weak base with a strong acid. –makes an acidic solution. • weak acid with a weak base. –makes either a basic or an acidic solution depending on the relative strength of the ions. **Most, but not all, minerals can be considered to be salts of weak acids and strong bases.

  34. In every family there’s always an oddball…. Amphoteric Hydroxides: hydroxides that can behave as either and acid or a base. This behavior varies as a function of pH. For example: Al(OH)3 behaves as a base in an acidic solution and forms aluminum salts. In a basic solution, it behaves as an acid H3AlO3 and forms salts with AlO33-. KA is the equilibrium constant for an amphoteric reaction. A table of these constants is listed on page 75 of the text.

  35. Buffers A buffered solution is a solution that resists changes in pH when either hydrogen or hydroxyl ions are added to the solution. A buffer is a weak acid and its salt or a weak base and its salt. Let’s go back to our buddy LeChâtelier’s principle and look at the following reactions. NaC2H3O2 Na+ + C2H3O2- HC2H3O2 H+ + C2H3O2- What would happen if a strong acid were added to a solution containing both NaC2H3O2 andHC2H3O2?

  36. NaC2H3O2 Na+ + C2H3O2- HC2H3O2 H+ + C2H3O2- HCl  H+ + Cl- When you add additional H+ to this buffered solution, what happens to the concentration of: HC2H3O2? C2H3O2- ? NaC2H3O2? Increases Decreases Decreases What ultimately happens to the H+ added to the solution? It combines with C2H3O2- to make HC2H3O2 until all the is NaC2H3O2 used up.

  37. Example 3-13 What would happen to the pH of a 1 liter solution, containing carbonic acid, if 10-4 mol of H+ ions are added to the solution? At pH = 7, T = 25°C, [HCO3-] = 10-3 mol/L, pKa1 = 6.35, therefore, [H2CO3] = 10-3.65 mol/L H+ + HCO3- H2CO3(aq) According to LeChâtelier’s principle the concentration of will HCO3-decrease by 10-4 mol/L, and the concentration of H2CO3(aq) will increase by 10-4 mol/L. pH = pKa1 + log([HCO3-] / [H2CO3]) = 6.35 + log([10-3 – 10-4] / [10-3.65 + 10-4]) = 6.79 If the pH of the solution would have been 4 without the H2CO3; and with the H2CO3 the pH is 6.79, then it can be said that the solution was successfully buffered.

  38. The previous exercise demonstrated the Henderson-Hasselbalch equation. Consider the generic: H+ + A-  HA. pH = -logKa + log ([A-] / [HA]) Remember that buffers are important in the natural environment because they control the impact of acid or base additions on natural waters. This is an acidic lake in Norway. Norway’s geology consists primarily of crystallines and metamorphites and little limestone. The acidity was caused by acid rain. http://www.lifeinfreshwater.org.uk/Web%20pages/ponds/Pollution.htm

  39. Buffering Capacities of Natural Waters Water is an effective buffer only at very high or very low pH. Buffering capacity: a measure of the amount of H+ or OH- ions a solution can absorb without significant change in pH. http://www.coztarica.com/2009/12/23/poas-volcano-costa-rica/

  40. Buffering Capacity of H2CO3 Buffering capacity of CaCO3 – H2CO3 system http://www.ourtravelpics.com/?place=shilin&photo=3

  41. Figure 3-9 summarizes the buffering capacities of waters in contact with minerals Does the fate of the Norwegian acid lake seem clearer?

  42. Exercise 3-15 takes a look at the buffering index for the calcite-carbonic acid system. Calculate the buggering index for the system calcite-carbonic acid at pH = 6, CT = 1 x 10-2, and T = 25⁰C. Using equation 3-54…. After several steps and plugging and chugging… B = 2.34 x 10-6 + 4.91 x 10-3 + 0.16994 = 0.17485 eq L-1pH-1 = 174.9 meq L-1pH-1

  43. Before we leave the discussion on acids and bases, we should also mention the importance of silicic acid (H4SiO4). This important acid is a produced during the weathering of silicate minerals – the most abundant mineral group in the crust of the earth. We need only consider the first two dissociation stops with this tetraprotic acid. At pH = 9.83, [H4SiO4] = [H3SiO4-]. This implies that for pH < 9.83, H4SiO4 is the dominant species. Only at pH > 13.17—rarely if ever found in nature—is H2SiO42- the dominant species. http://accessscience.com/studycenter.aspx?main=7&questionID=4438&expandType=1 2KAlSi3O8 + 9H2O + 2H+ Al2Si2O5(OH)4 + 2K+ + 4H4SiO4 http://www.mii.org/Minerals/

  44. Problem Set 3 Chapter 3 #s: 3, 8, 14, 19, 21, 24, 27, 35, 36 These are due Friday, October 5 at the start of class. http://blog.richmond.edu/psyc200/2009/12/01/turn-the-frown-upside-down/

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