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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. The Br önstead – Lowry model describes an acid simply as a proton donor

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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. The Brönstead – Lowry model describes an acid simply as a proton donor Consider the following reaction: HCl(aq) + H2O(l) H3O+(aq) + Cl- Here the proton is transferred from the HCl molecule to the water molecule to for the hydronium ion (H3O+). A general form of this equation is: HA(aq) + H2O(l)  H3O(aq) + A-(aq) Acid + Base Conjugate + Conjugate Acid Base

  3. 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

  4. As you noticed from the previous slide (table), acids can contain more than one acidic proton. A diprotic acid contains 2 acidic protons and undergoes a two-step dissociation. i.e. H2SO4 H+ + HSO4- HSO4-  H+ + SO42- A triprotic acid (such as phosphoric acid H3PO4) has three acidic protons, and must undergo a three-step dissociation. Recall: pKa = -logKa Acids that undergo significant dissociation have a negative pKa, and acids that only partially dissociate have a positive pKa.

  5. What is a base? The Arrhenius concept proposed that a base is a substance that produces OH- ions in aqueous solution. The Brönstead – Lowry model describes a base simply as a proton acceptor. Consider the following reaction: NH3(aq) + H2O(l) NH4+(aq) + OH- Here the OH- ion is derived from the H2O molecule and the NH3 molecule acts as a proton acceptor. A general form of this equation is: B(aq) + H2O(l)  BH+(aq) + OH-(aq) Base + Acid Conjugate + Conjugate Acid Base

  6. 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

  7. As it was with acids, base may also require more than one step to complete dissociation.

  8. The dissociation of water and pH H2O  H+ + OH- and Kw = [H+][OH-] / [H2O] which equals [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.

  9. 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.

  10. Here are a few common solutions and their corresponding pH. Note the concentration of H+ compared to pure H2O.

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

  12. 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?

  13. 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.

  14. For example: 30 mL of 0.10M NaOH neutralised 25.0mL of hydrochloric acid. Determine the concentration of the acid. NaOH(aq) + HCl(aq) -----> NaCl(aq) + H2O(l) NaOH: V = 30mL , M = 0.10M; HCl: V = 25.0mL, M = ? NaOH: V = 30 x 10-3L , M = 0.10M; HCl: V = 25.0 x 10-3L, M = ? # of moles(NaOH) = M x V = 0.10 x 30 x 10-3 = 3 x 10-3 moles From the balanced chemical equation find the mole ratio NaOH:HCl equals: 1:1 Moles(NaOH): Moles(HCl) is 1:1 Therefore, at the equivalence point, there are 3 X 10-3 moles(HCl) Calculate concentration of HCl: M = n / V n = 3 x 10-3 mol, V = 25.0 x 10-3L M(HCl) = 3 x 10-3 / 25.0 x 10-3 = 0.12M or 0.12 mol L-1

  15. Ok, now what is the pH of this .12M(HCl) solution? –before the titration. (We know the pH of the tested solution is 7 as the added NaOH neutrallized the solition.) 0.12mol(HCL) / L What is the pKa for HCl at 25°C? -3 Ka = [H+][Cl-] / [HCl] we know that [H+] = [Cl-] and we know that the amount a hydrochloric acid left over after dissociation is 0.12 – [H+]. Also if pKa = -3 thenKa = 1000. If we substitute x for [H+] we could set up a quadratic formula and solve for x. Huh?

  16. Time for a quick cleansing breath……

  17. Ok, now, where were we… If Ka = [x][x] / 0.12 – x, then we can solve for x using the quadratic formula: Ka = 1000 = x2 / (0.12 – x)  x2 + 1000x – 120 = 0 = -1000 [(10002 – 4(1)(-120)]1/2 / 2(1) = either –1000 or 0.12 Since –1000 makes no sense, we can assume x = 0.12

  18. Oh, did I forget to tell you that with HCl, the reaction goes into, for all our purposes, full dissociation… So, what is the pH of [H+] = 0.12 moles/L solution? 0.92 Now that you have had a review of the math, then exercise 3-1 in the text shouldn’t seem so insane!

  19. pH of Natural Waters

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

  21. 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 ________ CO2(g) + H2O  H2CO3(aq) where [H2CO3(aq)] = KCO2PCO2 and Ka1 = [H+][HCO3-] / KCO2PCO2

  22. H2CO3(aq)  H+ + HCO3-  HCO3-  H+ + CO32- This implies that the complete dissociation of carbonic acid is a two-step process. Such that: 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

  23. 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.

  24. 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.

  25. 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.

  26. 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-

  27. 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.

  28. 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-. Think Le Châtelier’s principle. KA is the equilibrium constant for an amphoteric reaction. A table of these constants is listed on page 75 of the text.

  29. 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).

  30. 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? • What would happen if a strong base were added to a solution containing both NaC2H3O2 andHC2H3O2?

  31. Question 1: 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.

  32. Question 2: NaC2H3O2 Na+ + C2H3O2- HC2H3O2 H+ + C2H3O2- NaOH  Na+ + OH- When you add additional OH- to this buffered solution, what happens to the concentration of: HC2H3O2? C2H3O2- ? NaC2H3O2? Decreases Increases Increases What ultimately happens to the OH- added to the solution? It combines with H+ to make H2O. More H+ is created until all of the HC2H3O2 is used up.

  33. 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 decrease by 10-4 mol/L, and the concentration of 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.

  34. 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.

  35. 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.

  36. Buffering Capacity of H2CO3 Buffering capacity of CaCO3 – H2CO3 system

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

  38. I urge you all to read the case studies in this chapter. It puts things in perspective. Next week we will start with Chapter 4: Oxidation – Reduction Reactions.

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